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@ -127,9 +127,9 @@ middle_, these comprise the axioms of a system of propositional logic.
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+ $or$ connects the smallest propositions surrounding it.
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]
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= Notes on Logic and Proofs, 1.2
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= Lecture #datetime(day: 8, month: 1, year: 2025).display()
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_Prototypical example for this section:_ If $sin pi = 1$, then $6$ is prime.
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== More propositional forms
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#definition[
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For a *antedecent* $P$ and *consequent* $Q$, the *conditional sentence* $P =>
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@ -143,9 +143,6 @@ Q$ is the proposition "If $P$, then $Q$."
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A conditional may be true even when the antedecent and consequent are unrelated.
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= Lecture #datetime(day: 8, month: 1, year: 2025).display()
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== More propositional forms
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#definition[
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Let $P$ and $Q$ be propositions. The *biconditional sentence*
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@ -996,7 +993,7 @@ iota$ which is an inclusion function $iota' : A -> C$.
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#definition[
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Let $R$ be an equivalence relation on $A$. Recall $A\/R$ is the set of all
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equivalence classes. The *canonical map* is
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equivalence classes under $R$. The *canonical map* is
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$
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f : A -> A\/R, f(x) = overline(x), forall x in A
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@ -10,7 +10,7 @@
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#show heading.where(level: 2): it => [#it.body.]
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#show heading.where(level: 3): it => [#it.body.]
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#set par(first-line-indent: 0pt, spacing: 1em)
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#set par(spacing: 1em)
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Problems:
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@ -19,8 +19,6 @@ Problems:
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2.5: \#1abc, 3, 10
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#outline()
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= 2.4
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== 4
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@ -228,43 +226,69 @@ If a set $A$ has $n$ elements, then $cal(P) (A)$ has $2^n$ elements.
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Ok now let's do it the annoying way.
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#mitext(`
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Base Case ($n=0$). \\
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If $A$ has $0$ elements, then $A = \varnothing$. Its power set is
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\[
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\mathcal{P}(A) = \{\varnothing\},
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\]
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which has exactly one element. Hence
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\[
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|\mathcal{P}(A)| = 1 = 2^0.
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\]
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So the statement holds for $n=0$.
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Consider the set with $0$ elements, $X = {}$. Then $cal(P)(X) = {emptyset}$
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so indeed it has cardinality $2^0 = 1$. Otherwise suppose that any set $X$
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with $n$ elements indeed has $|cal(P)(X)| = 2^n$.
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Now proceed by induction. Assume the statement is true for some integer $k \ge
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0$. That is, suppose that for any set $A$ with $k$ elements, we have
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\[
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|\mathcal{P}(A)| = 2^k.
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\]
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Then, consider a set $Y$ of $n+1$ elements. For each subset of $cal(P)(Y)$,
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we choose an arbitrary element $k in Y$, and define $Z := Y backslash {k}$.
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Note that $Z$ has $n$ elements so our inductive hypothesis says it has $2^n$
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elements. We seek to show that this implies $cal(P)(Y)$ has $2^(n+1)$
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elements.
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Let $B$ be a set with $k+1$ elements. Choose one element $x \in B$, and let $A = B \setminus \{x\}$. Then $A$ has $k$ elements. By the inductive hypothesis,
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\[
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|\mathcal{P}(A)| = 2^k.
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\]
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Now observe that any subset of $B$ is either:
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\begin{itemize}
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\item A subset of $A$ (does not contain $x$), or
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\item Of the form $S \cup \{x\}$ where $S \subseteq A$ (does contain $x$).
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\end{itemize}
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Thus every subset of $A$ gives rise to exactly one subset of $B$ that excludes $x$, and exactly one subset of $B$ that includes $x$. Therefore,
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\[
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|\mathcal{P}(B)| = |\mathcal{P}(A)| + |\mathcal{P}(A)|
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= 2^k + 2^k
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= 2 \cdot 2^k
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= 2^{k+1}.
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\]
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Every subset of $Y$ in $cal(P)(Y)$ satisfies the following: either the subset
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does not contain $k$ and so it's also a subset of $Z$, or its only difference
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from a subset of $Z$ is the addition of $k$.
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Since $B$ was any set with $k+1$ elements, the statement holds for $k+1$. By the principle of mathematical induction, the proof is complete.
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`)
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Therefore, for each subset of $Z$, we can introduce $k$, and it's still a
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subset of $Y$, and in fact completes $cal(P)(Y)$ from $cal(P)(Z)$, due to our
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previous assertion. So for each element in $cal(P)(Z)$, there is one
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additional element in $cal(P)(Y)$. In other words,
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$
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|cal(P)(Z)| dot 2 = |cal(P)(Y)|
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$
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Recall our inductive hypothesis states $|cal(P)(Z)| = 2^n$ because $Z$ has
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$n$ elements, so $|cal(P)(X)| = 2 dot |cal(P)(Z)| = 2^(n+1)$, completing our
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induction. Therefore for any arbitrary set $X$ with $n$ elements, its power
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set $cal(P)(X)$ has $2^n$ elements.
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// #mitext(`
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// Base Case ($n=0$). \\
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// If $A$ has $0$ elements, then $A = \varnothing$. Its power set is
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// \[
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// \mathcal{P}(A) = \{\varnothing\},
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// \]
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// which has exactly one element. Hence
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// \[
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// |\mathcal{P}(A)| = 1 = 2^0.
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// \]
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// So the statement holds for $n=0$.
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//
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// Now proceed by induction. Assume the statement is true for some integer $k \ge
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// 0$. That is, suppose that for any set $A$ with $k$ elements, we have
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// \[
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// |\mathcal{P}(A)| = 2^k.
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// \]
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//
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// Let $B$ be a set with $k+1$ elements. Choose one element $x \in B$, and let $A = B \setminus \{x\}$. Then $A$ has $k$ elements. By the inductive hypothesis,
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// \[
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// |\mathcal{P}(A)| = 2^k.
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// \]
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// Now observe that any subset of $B$ is either:
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// \begin{itemize}
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// \item A subset of $A$ (does not contain $x$), or
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// \item Of the form $S \cup \{x\}$ where $S \subseteq A$ (does contain $x$).
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// \end{itemize}
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// Thus every subset of $A$ gives rise to exactly one subset of $B$ that excludes $x$, and exactly one subset of $B$ that includes $x$. Therefore,
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// \[
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// |\mathcal{P}(B)| = |\mathcal{P}(A)| + |\mathcal{P}(A)|
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// = 2^k + 2^k
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// = 2 \cdot 2^k
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// = 2^{k+1}.
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// \]
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//
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// Since $B$ was any set with $k+1$ elements, the statement holds for $k+1$. By the principle of mathematical induction, the proof is complete.
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// `)
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]
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@ -272,224 +296,261 @@ Since $B$ was any set with $k+1$ elements, the statement holds for $k+1$. By the
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=== c
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#mitext(`
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Let \( n = 5 \). Then:
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\[
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(5+1)! = 6! = 720 \quad\text{and}\quad 2^{5+3} = 2^8 = 256.
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\]
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Since \( 720 > 256 \), the inequality holds for \( n = 5 \).
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$(n+1)! > 2^(n+3)$ for all $n >= 5$.
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Assume that for some natural number \( n \ge 5 \) the inequality holds for all integers \( m \) with \( 5 \le m \le n \). In particular, assume
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\[
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(n+1)! > 2^{n+3}.
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\]
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We must show that
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\[
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(n+2)! > 2^{(n+1)+3} = 2^{n+4}.
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\]
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Consider $n = 5$. Then $6! > 2^(8)$. Now proceed by induction. Suppose that
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$(n+1)! > 2^(n + 3)$.
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Starting with the left-hand side, we have:
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\[
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(n+2)! = (n+2)(n+1)!.
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\]
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Using the inductive hypothesis,
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\[
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(n+2)! > (n+2) \cdot 2^{n+3}.
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\]
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Since \( n \ge 5 \), it follows that \( n+2 \ge 7 \). Therefore,
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\[
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(n+2) \cdot 2^{n+3} \ge 7 \cdot 2^{n+3}.
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\]
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But clearly,
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\[
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7 \cdot 2^{n+3} > 2 \cdot 2^{n+3} = 2^{n+4}.
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\]
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Thus,
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\[
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(n+2)! > 2^{n+4},
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\]
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which completes the inductive step.
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Then $(n + 1 + 1)! > 2^(n+3 + 1)$ so $(n + 2)(n+1)! > 2 dot 2^(n + 3)$. We note
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that $n+2$ is always at least $2$, so this is true by our inductive hypothesis.
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By the generalized principle of mathematical induction, the inequality
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\[
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(n+1)! > 2^{n+3}
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\]
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holds for all natural numbers \( n \ge 5 \).
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If we consider $n = 1$, then $2! < 2^(4)$ so the PMI does not hold for all $NN$.
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The inequality is stated to hold for all \( n \ge 5 \). However, for \( n < 5 \) the inequality fails. For instance, when \( n = 4 \):
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\[
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(4+1)! = 5! = 120 \quad\text{and}\quad 2^{4+3} = 2^7 = 128.
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\]
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Since \( 120 \) is not greater than \( 128 \), the inequality is false for \( n = 4 \).
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`)
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// #mitext(`
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// Let \( n = 5 \). Then:
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// \[
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// (5+1)! = 6! = 720 \quad\text{and}\quad 2^{5+3} = 2^8 = 256.
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// \]
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// Since \( 720 > 256 \), the inequality holds for \( n = 5 \).
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//
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// Assume that for some natural number \( n \ge 5 \) the inequality holds for all integers \( m \) with \( 5 \le m \le n \). In particular, assume
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// \[
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// (n+1)! > 2^{n+3}.
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// \]
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// We must show that
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// \[
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// (n+2)! > 2^{(n+1)+3} = 2^{n+4}.
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// \]
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//
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// Starting with the left-hand side, we have:
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// \[
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// (n+2)! = (n+2)(n+1)!.
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// \]
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// Using the inductive hypothesis,
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// \[
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// (n+2)! > (n+2) \cdot 2^{n+3}.
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// \]
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// Since \( n \ge 5 \), it follows that \( n+2 \ge 7 \). Therefore,
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// \[
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// (n+2) \cdot 2^{n+3} \ge 7 \cdot 2^{n+3}.
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// \]
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// But clearly,
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// \[
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// 7 \cdot 2^{n+3} > 2 \cdot 2^{n+3} = 2^{n+4}.
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// \]
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// Thus,
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// \[
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// (n+2)! > 2^{n+4},
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// \]
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// which completes the inductive step.
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//
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// By the generalized principle of mathematical induction, the inequality
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// \[
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// (n+1)! > 2^{n+3}
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// \]
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// holds for all natural numbers \( n \ge 5 \).
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//
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// The inequality is stated to hold for all \( n \ge 5 \). However, for \( n < 5 \) the inequality fails. For instance, when \( n = 4 \):
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// \[
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// (4+1)! = 5! = 120 \quad\text{and}\quad 2^{4+3} = 2^7 = 128.
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// \]
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// Since \( 120 \) is not greater than \( 128 \), the inequality is false for \( n = 4 \).
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// `)
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=== e
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#mitext(`
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Let \( n = 4 \). Then:
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\[
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4! = 24 \quad\text{and}\quad 3 \times 4 = 12.
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\]
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Since \( 24 > 12 \), the inequality holds for \( n = 4 \).
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$(n! > 3n)$ for all $n >= 4$.
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Assume that for some natural number \( n \ge 4 \) the inequality holds for all integers \( m \) with \( 4 \le m \le n \). In particular, assume that
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\[
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n! > 3n.
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\]
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We must show that
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\[
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(n+1)! > 3(n+1).
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\]
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Consider $n = 4$. Then $4! > 3(4)$. Suppose $(n! > 3n)$. Then
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$
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(n+1)! > 3(n+1) \
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(n+1)n! > 3n + 3
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$
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Assume $n >= 4$. Then the above statement holds true for any $n$. So we
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conclude the statement is always true for $n >= 4$.
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Starting with the left-hand side:
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\[
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(n+1)! = (n+1) \cdot n!.
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\]
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By the inductive hypothesis, we have:
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\[
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(n+1)! > (n+1) \cdot 3n.
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\]
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Since \( n \ge 4 \) (so \( n \ge 2 \)), it follows that:
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\[
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(n+1) \cdot 3n \ge 3(n+1).
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\]
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To see this, note that for \( n \ge 2 \) we have:
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\[
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3n(n+1) = 3(n+1)n > 3(n+1),
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\]
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because \( n > 1 \). Therefore,
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\[
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(n+1)! > 3(n+1),
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\]
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which completes the inductive step.
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By the generalized principle of mathematical induction, the inequality
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\[
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n! > 3n
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\]
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holds for all natural numbers \( n \ge 4 \).
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The claim is that \( n! > 3n \) for all \( n \ge 4 \). However, for some smaller natural numbers the inequality is false. For instance, when \( n = 3 \):
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\[
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3! = 6 \quad\text{and}\quad 3 \times 3 = 9.
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\]
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`)
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// #mitext(`
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// Let \( n = 4 \). Then:
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// \[
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// 4! = 24 \quad\text{and}\quad 3 \times 4 = 12.
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// \]
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// Since \( 24 > 12 \), the inequality holds for \( n = 4 \).
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//
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// Assume that for some natural number \( n \ge 4 \) the inequality holds for all integers \( m \) with \( 4 \le m \le n \). In particular, assume that
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// \[
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// n! > 3n.
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// \]
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// We must show that
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// \[
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// (n+1)! > 3(n+1).
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// \]
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//
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// Starting with the left-hand side:
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// \[
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// (n+1)! = (n+1) \cdot n!.
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// \]
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// By the inductive hypothesis, we have:
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// \[
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// (n+1)! > (n+1) \cdot 3n.
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// \]
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// Since \( n \ge 4 \) (so \( n \ge 2 \)), it follows that:
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// \[
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// (n+1) \cdot 3n \ge 3(n+1).
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// \]
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// To see this, note that for \( n \ge 2 \) we have:
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// \[
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// 3n(n+1) = 3(n+1)n > 3(n+1),
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// \]
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// because \( n > 1 \). Therefore,
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// \[
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// (n+1)! > 3(n+1),
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// \]
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// which completes the inductive step.
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//
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// By the generalized principle of mathematical induction, the inequality
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// \[
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// n! > 3n
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// \]
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// holds for all natural numbers \( n \ge 4 \).
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//
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// The claim is that \( n! > 3n \) for all \( n \ge 4 \). However, for some smaller natural numbers the inequality is false. For instance, when \( n = 3 \):
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// \[
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// 3! = 6 \quad\text{and}\quad 3 \times 3 = 9.
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// \]
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// `)
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== 7
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=== a
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#mitext(`
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Let \(\{A_i : i \in \mathbb{N}\}\) be an indexed family of sets. Then for every natural number \( n\ge1 \),
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\[
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\left(\bigcup_{i=1}^{n} A_i\right)^c = \bigcap_{i=1}^{n} A_i^c
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\]
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and
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\[
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\left(\bigcap_{i=1}^{n} A_i\right)^c = \bigcup_{i=1}^{n} A_i^c.
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\]
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$
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(sect.big^n_(i=1) A_i)^c = union.big^n_(i=1) A_i^c
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$
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For \( n=1 \), we have
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\[
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\left(\bigcup_{i=1}^{1} A_i\right)^c = A_1^c \quad \text{and} \quad \bigcap_{i=1}^{1} A_i^c = A_1^c.
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\]
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Thus, the identity holds for \( n=1 \).
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First consider $sect.big^n_(i=1) A_i$. Let $X := sect.big^n_(i=1) A_i$.
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Then $x in X$ if and only if $x in A_i$ for every $A_i$.
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Assume that for some \( n \ge 1 \) the statement holds; that is, assume
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\[
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\left(\bigcup_{i=1}^{n} A_i\right)^c = \bigcap_{i=1}^{n} A_i^c.
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\]
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We need to show that
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\[
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\left(\bigcup_{i=1}^{n+1} A_i\right)^c = \bigcap_{i=1}^{n+1} A_i^c.
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\]
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Now consider $Y := (sect.big^n_(i=1) A_i)^c$. Then $y in Y$ if and only if $y
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in.not X$, that is, if and only there exists some $A_i$ such that $y in.not
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A_i$. Now let $Z := union.big^n_(i=1) A_i^c$. Every element $z in Z$ is in some
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$A^c _i$, that is, $z in Z$ if and only if there exists some $A_i$ such that $z
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in.not A_i$. Now note this is actually precisely the definition of $Y$. Hence
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$Z = Y$.
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Notice that
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\[
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\bigcup_{i=1}^{n+1} A_i = \left(\bigcup_{i=1}^{n} A_i\right) \cup A_{n+1}.
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\]
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Taking the complement of both sides, and using De Morgan's Law for two sets, we obtain:
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\[
|
||||
\left(\bigcup_{i=1}^{n+1} A_i\right)^c = \left(\left(\bigcup_{i=1}^{n} A_i\right) \cup A_{n+1}\right)^c = \left(\bigcup_{i=1}^{n} A_i\right)^c \cap A_{n+1}^c.
|
||||
\]
|
||||
By the induction hypothesis,
|
||||
\[
|
||||
\left(\bigcup_{i=1}^{n} A_i\right)^c = \bigcap_{i=1}^{n} A_i^c.
|
||||
\]
|
||||
Thus, we have:
|
||||
\[
|
||||
\left(\bigcup_{i=1}^{n+1} A_i\right)^c = \left(\bigcap_{i=1}^{n} A_i^c\right) \cap A_{n+1}^c = \bigcap_{i=1}^{n+1} A_i^c.
|
||||
\]
|
||||
|
||||
This completes the inductive step.
|
||||
|
||||
By the PMI, we conclude that for every natural number \( n\ge1 \),
|
||||
\[
|
||||
\left(\bigcup_{i=1}^{n} A_i\right)^c = \bigcap_{i=1}^{n} A_i^c.
|
||||
\]
|
||||
|
||||
Thus, De Morgan's Laws hold for any indexed family \(\{A_i : i \in \mathbb{N}\}\).
|
||||
`)
|
||||
// #mitext(`
|
||||
// Let \(\{A_i : i \in \mathbb{N}\}\) be an indexed family of sets. Then for every natural number \( n\ge1 \),
|
||||
// \[
|
||||
// \left(\bigcup_{i=1}^{n} A_i\right)^c = \bigcap_{i=1}^{n} A_i^c
|
||||
// \]
|
||||
// and
|
||||
// \[
|
||||
// \left(\bigcap_{i=1}^{n} A_i\right)^c = \bigcup_{i=1}^{n} A_i^c.
|
||||
// \]
|
||||
//
|
||||
// For \( n=1 \), we have
|
||||
// \[
|
||||
// \left(\bigcup_{i=1}^{1} A_i\right)^c = A_1^c \quad \text{and} \quad \bigcap_{i=1}^{1} A_i^c = A_1^c.
|
||||
// \]
|
||||
// Thus, the identity holds for \( n=1 \).
|
||||
//
|
||||
// Assume that for some \( n \ge 1 \) the statement holds; that is, assume
|
||||
// \[
|
||||
// \left(\bigcup_{i=1}^{n} A_i\right)^c = \bigcap_{i=1}^{n} A_i^c.
|
||||
// \]
|
||||
// We need to show that
|
||||
// \[
|
||||
// \left(\bigcup_{i=1}^{n+1} A_i\right)^c = \bigcap_{i=1}^{n+1} A_i^c.
|
||||
// \]
|
||||
//
|
||||
// Notice that
|
||||
// \[
|
||||
// \bigcup_{i=1}^{n+1} A_i = \left(\bigcup_{i=1}^{n} A_i\right) \cup A_{n+1}.
|
||||
// \]
|
||||
// Taking the complement of both sides, and using De Morgan's Law for two sets, we obtain:
|
||||
// \[
|
||||
// \left(\bigcup_{i=1}^{n+1} A_i\right)^c = \left(\left(\bigcup_{i=1}^{n} A_i\right) \cup A_{n+1}\right)^c = \left(\bigcup_{i=1}^{n} A_i\right)^c \cap A_{n+1}^c.
|
||||
// \]
|
||||
// By the induction hypothesis,
|
||||
// \[
|
||||
// \left(\bigcup_{i=1}^{n} A_i\right)^c = \bigcap_{i=1}^{n} A_i^c.
|
||||
// \]
|
||||
// Thus, we have:
|
||||
// \[
|
||||
// \left(\bigcup_{i=1}^{n+1} A_i\right)^c = \left(\bigcap_{i=1}^{n} A_i^c\right) \cap A_{n+1}^c = \bigcap_{i=1}^{n+1} A_i^c.
|
||||
// \]
|
||||
//
|
||||
// This completes the inductive step.
|
||||
//
|
||||
// By the PMI, we conclude that for every natural number \( n\ge1 \),
|
||||
// \[
|
||||
// \left(\bigcup_{i=1}^{n} A_i\right)^c = \bigcap_{i=1}^{n} A_i^c.
|
||||
// \]
|
||||
//
|
||||
// Thus, De Morgan's Laws hold for any indexed family \(\{A_i : i \in \mathbb{N}\}\).
|
||||
// `)
|
||||
|
||||
== 9
|
||||
|
||||
#mitext(`
|
||||
Given \(n\) points \(P_1, P_2, \dots, P_n\) in a plane with no three collinear, the number of line segments joining every pair of points is
|
||||
\[
|
||||
\frac{n^2 - n}{2}.
|
||||
\]
|
||||
Suppose there was one point. Then there would be $(1^2 - 1)/2 = 0$ line
|
||||
segments which is correct. Now we proceed by induction. Suppose that for
|
||||
$n$ points, the number of line segments joining all pairs of points are $(n^2 -
|
||||
n)/2$. Now introduce an additional point such that there are $n+1$ points. For
|
||||
each of the previous $n$ points, we draw a line connecting it to the newly
|
||||
added point. So we introduce $n$ additional lines. Therefore we have
|
||||
$
|
||||
(n^2 - n) / 2 + n
|
||||
$
|
||||
lines, which we can rewrite
|
||||
$
|
||||
(n^2 + n) / 2 = ((n+1)^2 - (n+1)) / 2
|
||||
$
|
||||
So we conclude that our hypothesis holds for $n+1$ points and therefore holds
|
||||
for all $n in NN$ points.
|
||||
|
||||
For \(n=2\), there is exactly 1 line segment joining the two points. The formula gives
|
||||
\[
|
||||
\frac{2^2 - 2}{2} = \frac{4-2}{2} = 1.
|
||||
\]
|
||||
So the statement holds for \(n=2\).
|
||||
|
||||
Assume that for some \(n \ge 2\), the number of line segments joining \(n\) points is
|
||||
\[
|
||||
\frac{n^2 - n}{2}.
|
||||
\]
|
||||
Now consider \(n+1\) points. When we add a new point \(P_{n+1}\), this new point can be connected to each of the \(n\) existing points, thereby adding \(n\) new segments. Hence, the total number of segments becomes
|
||||
\[
|
||||
\frac{n^2 - n}{2} + n.
|
||||
\]
|
||||
Simplify the expression:
|
||||
\[
|
||||
\frac{n^2 - n}{2} + n = \frac{n^2 - n + 2n}{2} = \frac{n^2 + n}{2} = \frac{n(n+1)}{2}.
|
||||
\]
|
||||
Therefore, by the PMI, the number of line segments joining all pairs of \(n\) points is
|
||||
\[
|
||||
\frac{n^2 - n}{2}
|
||||
\]
|
||||
for all \(n \ge 2\).
|
||||
`)
|
||||
// #mitext(`
|
||||
// Given \(n\) points \(P_1, P_2, \dots, P_n\) in a plane with no three collinear, the number of line segments joining every pair of points is
|
||||
// \[
|
||||
// \frac{n^2 - n}{2}.
|
||||
// \]
|
||||
//
|
||||
// For \(n=2\), there is exactly 1 line segment joining the two points. The formula gives
|
||||
// \[
|
||||
// \frac{2^2 - 2}{2} = \frac{4-2}{2} = 1.
|
||||
// \]
|
||||
// So the statement holds for \(n=2\).
|
||||
//
|
||||
// Assume that for some \(n \ge 2\), the number of line segments joining \(n\) points is
|
||||
// \[
|
||||
// \frac{n^2 - n}{2}.
|
||||
// \]
|
||||
// Now consider \(n+1\) points. When we add a new point \(P_{n+1}\), this new point can be connected to each of the \(n\) existing points, thereby adding \(n\) new segments. Hence, the total number of segments becomes
|
||||
// \[
|
||||
// \frac{n^2 - n}{2} + n.
|
||||
// \]
|
||||
// Simplify the expression:
|
||||
// \[
|
||||
// \frac{n^2 - n}{2} + n = \frac{n^2 - n + 2n}{2} = \frac{n^2 + n}{2} = \frac{n(n+1)}{2}.
|
||||
// \]
|
||||
// Therefore, by the PMI, the number of line segments joining all pairs of \(n\) points is
|
||||
// \[
|
||||
// \frac{n^2 - n}{2}
|
||||
// \]
|
||||
// for all \(n \ge 2\).
|
||||
// `)
|
||||
|
||||
== 10
|
||||
|
||||
#mitext(`
|
||||
The Tower of Hanoi problem with \( n \) disks can be solved in exactly \( 2^n - 1 \) moves.
|
||||
|
||||
For \( n = 1 \):
|
||||
There is only one disk, and it can be moved directly from the source peg to the destination peg in one move.
|
||||
Since \( 2^1 - 1 = 1 \), the statement holds for \( n = 1 \).
|
||||
|
||||
Assume that for some \( k \ge 1 \) the Tower of Hanoi with \( k \) disks can be solved in \( 2^k - 1 \) moves (this is the inductive hypothesis). We now show that a Tower of Hanoi with \( k+1 \) disks can be solved in \( 2^{k+1} - 1 \) moves.
|
||||
|
||||
1. Move the top \( k \) disks from the source peg to the auxiliary peg.
|
||||
By the inductive hypothesis, this takes \( 2^k - 1 \) moves.
|
||||
2. Move the largest disk (the \((k+1)^\text{th}\) disk) from the source peg to the destination peg.
|
||||
This requires 1 move.
|
||||
3. Move the \( k \) disks from the auxiliary peg to the destination peg.
|
||||
Again by the inductive hypothesis, this requires \( 2^k - 1 \) moves.
|
||||
|
||||
\[
|
||||
\text{Total Moves} = (2^k - 1) + 1 + (2^k - 1) = 2 \cdot 2^k - 1 = 2^{k+1} - 1.
|
||||
\]
|
||||
|
||||
This completes the inductive step.
|
||||
|
||||
By the PMI, the Tower of Hanoi with \( n \) disks can be solved in \( 2^n - 1 \) moves.
|
||||
`)
|
||||
First, note that with 1 disk, it takes $2^1 - 1 = 1$ moves to solve the puzzle.
|
||||
Now suppose that for $n$ disks, it takes $2^n - 1$ moves to solve. We proceed
|
||||
by induction on $n+1$ disks. First, ignore the disk at the very bottom of the
|
||||
stack, such that our situation is equivalent to when there are $n$ disks
|
||||
(because any disk can be stacked on top of the largest disk, our set of
|
||||
possible moves is unchanged from $n$ disks). Then we can move each of these
|
||||
disks to another peg in $2^n - 1$ moves. Now move the largest disk, which now
|
||||
no more disks above it, to the other free peg. Now we again can ignore the
|
||||
largest disk and move our $n$ disks on top of the largest disk in $2^n - 1$
|
||||
moves. So it took us a total of $2 dot (2^n - 1) + 1$ moves to move the entire
|
||||
stack to another peg, or $2^(n+1) - 1$ moves.
|
||||
|
||||
== 12
|
||||
|
||||
|
|
@ -522,48 +583,20 @@ is no reason why this should hold true for all $NN$.
|
|||
|
||||
=== a
|
||||
|
||||
#mitext(`
|
||||
Every natural number \( n \ge 11 \) can be written in the form
|
||||
\[
|
||||
n = 2s + 5t,
|
||||
\]
|
||||
for some nonnegative integers \( s \) and \( t \).
|
||||
Every natural number greater than or equal to 11 can be written in the form $2s
|
||||
+ 5t$ for some naturals $s$ and $t$.
|
||||
|
||||
We verify the statement for the initial numbers:
|
||||
- \( n = 11 \): \( 11 = 2\cdot 3 + 5\cdot 1 \)
|
||||
- \( n = 12 \): \( 12 = 2\cdot 1 + 5\cdot 2 \)
|
||||
- \( n = 13 \): \( 13 = 2\cdot 4 + 5\cdot 1 \)
|
||||
- \( n = 14 \): \( 14 = 2\cdot 2 + 5\cdot 2 \)
|
||||
- \( n = 15 \): \( 15 = 2\cdot 5 + 5\cdot 1 \)
|
||||
First, we see that $11 = 2(3) + 5(1)$, $12 = 2(2) + 5(2)$, $13 = 2(4) + 5(1)$,
|
||||
$14 = 2(2) + 5(2)$. Now suppose that we can write all naturals up to $11 <= k
|
||||
<= n$ as $2s + 5t$ for naturals $s,t$. Note that we have shown naturals from 11
|
||||
to 15 can be written in the $2s + 5t$ form. Then we note $n >= 15$ iff. $n - 4 >= 11$,
|
||||
and $n - 4$ is covered
|
||||
by our inductive hypothesis, so we can write
|
||||
$
|
||||
n + 1 = (n - 4) + 4 + 1 = 2s + 5t + 5 = 2s + 5(t + 1)
|
||||
$
|
||||
So indeed $n+1$ can be written as $2s + 5t$ for some naturals $s, t$.
|
||||
|
||||
Thus, the claim holds for all \( 11 \le n \le 15 \).
|
||||
|
||||
Assume as the induction hypothesis that for every integer \( m \) with \( 11 \le m < n \) (where \( n \ge 16 \)) there exist nonnegative integers \( s \) and \( t \) such that
|
||||
\[
|
||||
m = 2s + 5t.
|
||||
\]
|
||||
|
||||
Since \( n \ge 16 \), notice that:
|
||||
\[
|
||||
n - 2 \ge 14.
|
||||
\]
|
||||
Because \( n-2 \) is at least 11 (indeed, \( n-2 \ge 14 \)), the induction hypothesis applies. Therefore, there exist nonnegative integers \( s \) and \( t \) such that:
|
||||
\[
|
||||
n - 2 = 2s + 5t.
|
||||
\]
|
||||
|
||||
Then,
|
||||
\[
|
||||
n = (n - 2) + 2 = 2s + 5t + 2 = 2(s + 1) + 5t.
|
||||
\]
|
||||
If we set \( s' = s + 1 \) (which is clearly a nonnegative integer), we obtain:
|
||||
\[
|
||||
n = 2s' + 5t.
|
||||
\]
|
||||
Thus, \( n \) can be written in the desired form.
|
||||
|
||||
By the PCI, every natural number \( n \ge 11 \) can be written as \( 2s + 5t \) for some nonnegative integers \( s \) and \( t \).
|
||||
`)
|
||||
|
||||
=== b
|
||||
|
||||
|
|
@ -573,55 +606,32 @@ Every natural number \( n > 22 \) (i.e. every \( n \ge 23 \)) can be written in
|
|||
n = 3s + 4t,
|
||||
\]
|
||||
with integers \( s \ge 3 \) and \( t \ge 2 \).
|
||||
|
||||
We prove the statement by complete (strong) induction.
|
||||
|
||||
We explicitly verify the claim for a few numbers:
|
||||
- For \( n = 23 \):
|
||||
\( 23 = 3 \cdot 5 + 4 \cdot 2 \) (here, \( s = 5 \ge 3 \) and \( t = 2 \ge 2 \)).
|
||||
- For \( n = 24 \):
|
||||
\( 24 = 3 \cdot 4 + 4 \cdot 3 \) (here, \( s = 4 \ge 3 \) and \( t = 3 \ge 2 \)).
|
||||
- For \( n = 25 \):
|
||||
\( 25 = 3 \cdot 3 + 4 \cdot 4 \) (here, \( s = 3 \ge 3 \) and \( t = 4 \ge 2 \)).
|
||||
|
||||
Thus, the statement holds for \( n = 23, 24, 25 \).
|
||||
|
||||
Assume that for every integer \( m \) with \( 23 \le m \le n \) (where \( n \ge 25 \)), there exist integers \( s \ge 3 \) and \( t \ge 2 \) such that
|
||||
\[
|
||||
m = 3s + 4t.
|
||||
\]
|
||||
|
||||
The number \( n+1 \) can also be written in the form
|
||||
\[
|
||||
n+1 = 3s' + 4t',
|
||||
\]
|
||||
with \( s' \ge 3 \) and \( t' \ge 2 \).
|
||||
|
||||
Since \( n \ge 25 \), observe that
|
||||
\[
|
||||
n+1 - 3 = n-2 \ge 23.
|
||||
\]
|
||||
By the inductive hypothesis, there exist integers \( s \ge 3 \) and \( t \ge 2 \) such that
|
||||
\[
|
||||
n-2 = 3s + 4t.
|
||||
\]
|
||||
Then
|
||||
\[
|
||||
n+1 = (n-2) + 3 = 3s + 4t + 3 = 3(s+1) + 4t.
|
||||
\]
|
||||
Define \( s' = s+1 \) (so that \( s' \ge 3+1 = 4 \ge 3 \)) and let \( t' = t \) (which satisfies \( t' \ge 2 \)). This shows that
|
||||
\[
|
||||
n+1 = 3s' + 4t',
|
||||
\]
|
||||
with the required conditions.
|
||||
|
||||
By the PCI, every natural number \( n > 22 \) (i.e. \( n \ge 23 \)) can be written in the form
|
||||
\[
|
||||
n = 3s + 4t,
|
||||
\]
|
||||
where \( s \ge 3 \) and \( t \ge 2 \) are integers.
|
||||
`)
|
||||
|
||||
#proof[
|
||||
First, let's consider a few base cases.
|
||||
$
|
||||
23 &= 3(5) + 4(2) \
|
||||
24 &= 3(4) + 4(3) \
|
||||
25 &= 3(3) + 4(4) \
|
||||
$
|
||||
So for 23, 24, and 25, we can write them in the form $3s + 4t$ such that $s
|
||||
>= 3$ and $t >= 2$. Now we proceed by strong
|
||||
induction. Suppose that any natural $23 <= k <= n$ can be written as $k = 3s
|
||||
+ 4t$ for $t >= 2$ and $s >= 3$. Note that when $n >= 26$, $n - 3 >= 23$
|
||||
(otherwise, it's already covered in our base case and we are done). Then we
|
||||
can assume $n-3$ is always covered by the inductive hypothesis and can be
|
||||
written in the desired form $3s + 4t$. Now we seek to show that this implies
|
||||
$n+1$ can also be written in this form.
|
||||
$
|
||||
(n+1) = (n-3) + 4 = 3s + 4t + 4 = 3s + 4(t+1)
|
||||
$
|
||||
And $s >= 3$, $t + 1 >= 2$. So indeed any natural number greater than 22 can
|
||||
be written as $3s + 4t$ for some naturals $t >= 2$ and $s >= 3$.
|
||||
]
|
||||
|
||||
|
||||
|
||||
=== c
|
||||
|
||||
#mitext(`
|
||||
|
|
@ -630,66 +640,32 @@ Every natural number \( n > 33 \) (i.e. every \( n \ge 34 \)) can be written in
|
|||
n = 4s + 5t,
|
||||
\]
|
||||
where \( s \) and \( t \) are integers with \( s \ge 3 \) and \( t \ge 2 \).
|
||||
|
||||
We prove the statement by complete induction.
|
||||
|
||||
We verify the claim for the first four numbers:
|
||||
|
||||
- \( n = 34 \):
|
||||
\( 34 = 4\cdot 6 + 5\cdot 2 \)
|
||||
(Here, \( s = 6 \ge 3 \) and \( t = 2 \ge 2 \).)
|
||||
|
||||
- \( n = 35 \):
|
||||
\( 35 = 4\cdot 5 + 5\cdot 3 \)
|
||||
(Here, \( s = 5 \ge 3 \) and \( t = 3 \ge 2 \).)
|
||||
|
||||
- \( n = 36 \):
|
||||
\( 36 = 4\cdot 4 + 5\cdot 4 \)
|
||||
(Here, \( s = 4 \ge 3 \) and \( t = 4 \ge 2 \).)
|
||||
|
||||
- \( n = 37 \):
|
||||
\( 37 = 4\cdot 3 + 5\cdot 5 \)
|
||||
(Here, \( s = 3 \ge 3 \) and \( t = 5 \ge 2 \).)
|
||||
|
||||
Thus, the statement holds for \( n = 34, 35, 36, \) and \( 37 \).
|
||||
|
||||
Inductive Hypothesis:
|
||||
Assume that for every integer \( m \) with \( 34 \le m \le n \) (where \( n \ge 37 \)) there exist integers \( s \ge 3 \) and \( t \ge 2 \) such that
|
||||
\[
|
||||
m = 4s + 5t.
|
||||
\]
|
||||
|
||||
The number \( n+1 \) can also be written in the form
|
||||
\[
|
||||
n+1 = 4s' + 5t',
|
||||
\]
|
||||
with \( s' \ge 3 \) and \( t' \ge 2 \).
|
||||
|
||||
Since \( n \ge 37 \), we have:
|
||||
\[
|
||||
n+1 - 4 = n - 3 \ge 37 - 3 = 34.
|
||||
\]
|
||||
Thus, \( n-3 \) is at least 34, and by the inductive hypothesis, there exist integers \( s \ge 3 \) and \( t \ge 2 \) such that
|
||||
\[
|
||||
n - 3 = 4s + 5t.
|
||||
\]
|
||||
Then,
|
||||
\[
|
||||
n+1 = (n-3) + 4 = 4s + 5t + 4 = 4(s+1) + 5t.
|
||||
\]
|
||||
Define \( s' = s+1 \) and \( t' = t \). Since \( s \ge 3 \), it follows that \( s' \ge 4 \ge 3 \), and \( t' = t \ge 2 \). Thus, we obtain the required representation:
|
||||
\[
|
||||
n+1 = 4s' + 5t',
|
||||
\]
|
||||
with \( s' \ge 3 \) and \( t' \ge 2 \).
|
||||
|
||||
By the PCI, every natural number \( n > 33 \) (i.e. \( n \ge 34 \)) can be written in the form
|
||||
\[
|
||||
n = 4s + 5t,
|
||||
\]
|
||||
where \( s \ge 3 \) and \( t \ge 2 \) are integers.
|
||||
`)
|
||||
|
||||
#proof[
|
||||
We seek to show the proposition. We first show a few base cases, for $n = 34,
|
||||
35, 36$.
|
||||
$
|
||||
34 &= 4(6) + 5(2) \
|
||||
35 &= 4(5) + 5(3) \
|
||||
36 &= 4(4) + 5(4)
|
||||
$
|
||||
Now we state our inductive hypothesis. Suppose that for some natural $n$, $34
|
||||
<= k <= n$ can be written in the form $k = 4s + 5t$, for some integral $s >=
|
||||
3$ and $t >= 2$. Then we proceed by strong induction. Note that if $n >= 37$,
|
||||
then $n - 3 >= 34$ so $n - 3$ is covered by our inductive hypothesis
|
||||
(otherwise $n$ would be covered by one of our base cases and we are done). So
|
||||
$n-3$ can be written in the desired $4s + 5t$ form. Now we seek to show that
|
||||
this implies $n+1$ can also be written in the desired form.
|
||||
|
||||
$
|
||||
n + 1 = (n - 3) + 4 = 4s + 5t + 4 = 4(s+1) + 5t
|
||||
$
|
||||
$s+1 >= 3$ because $s >= 3$ and $t >= 2$ is still true. So indeed $n+1$ can
|
||||
be written in the desired form. By strong induction, every natural $n >= 34$
|
||||
can be written as $n = 4s + 5t$ for some integral $s >= 3$ and $t >= 2$.
|
||||
]
|
||||
|
||||
== 3
|
||||
|
||||
#mitext(`
|
||||
|
|
@ -701,90 +677,40 @@ Then for all natural numbers \( n \),
|
|||
\[
|
||||
a_n = 2^n.
|
||||
\]
|
||||
|
||||
We will prove by complete (strong) induction that for every natural number \( n \), the equality
|
||||
\[
|
||||
a_n = 2^n
|
||||
\]
|
||||
holds.
|
||||
|
||||
- For \( n = 1 \):
|
||||
\[
|
||||
a_1 = 2 = 2^1.
|
||||
\]
|
||||
- For \( n = 2 \):
|
||||
\[
|
||||
a_2 = 4 = 2^2.
|
||||
\]
|
||||
|
||||
Thus, the statement is true for \( n = 1 \) and \( n = 2 \).
|
||||
|
||||
Inductive Hypothesis:
|
||||
Assume that for all natural numbers \( j \) with \( 1 \le j \le n \) (for some \( n \ge 2 \)), we have
|
||||
\[
|
||||
a_j = 2^j.
|
||||
\]
|
||||
|
||||
We need to show that \( a_{n+1} = 2^{n+1} \).
|
||||
|
||||
Notice that if \( n \ge 2 \), we can apply the recurrence relation with \( j = n-1 \) (since \( n-1 \ge 1 \)):
|
||||
\[
|
||||
a_{(n-1)+2} = a_{n+1} = 5a_n - 6a_{n-1}.
|
||||
\]
|
||||
By the inductive hypothesis, we know:
|
||||
\[
|
||||
a_n = 2^n \quad \text{and} \quad a_{n-1} = 2^{n-1}.
|
||||
\]
|
||||
Thus,
|
||||
\[
|
||||
a_{n+1} = 5(2^n) - 6(2^{n-1}).
|
||||
\]
|
||||
Factor \(2^{n-1}\) from the right-hand side:
|
||||
\[
|
||||
a_{n+1} = 2^{n-1}\Bigl(5\cdot 2 - 6\Bigr) = 2^{n-1}(10-6) = 2^{n-1} \cdot 4 = 2^{n+1}.
|
||||
\]
|
||||
|
||||
By the PCI, the equality
|
||||
\[
|
||||
a_n = 2^n
|
||||
\]
|
||||
holds for all natural numbers \( n \), completing the proof.
|
||||
`)
|
||||
|
||||
#proof[
|
||||
|
||||
We already have bases cases for $n = 1$ and $n = 2$, so we just need to show
|
||||
assume the inductive hypothesis for $n > 2$. Note that when $n > 2$,
|
||||
$a_(n+1)$
|
||||
is
|
||||
$
|
||||
a_(n+1) = 5a_(n) - 6a_(n-1)
|
||||
$
|
||||
|
||||
Suppose that all $a_k$ such that $3 <= k <= n$ are given by $a_k = 2^k$. We
|
||||
seek to show that $n+1$ is also given by this equation.
|
||||
|
||||
Because $a_n$ and $a_(n-1)$ are included in our inductive hypothesis,
|
||||
we have
|
||||
$
|
||||
a_(n+1) = 5(2^n) - 6(2^(n-1) = 5(2^n) - 3(2^n) = 2^(n+1)
|
||||
$
|
||||
Therefore indeed $a_(n+1)$ is given by $2^(n+1)$ which satisfies our
|
||||
proposition and so it is true for all natural numbers.
|
||||
]
|
||||
|
||||
== 10
|
||||
|
||||
#mitext(`
|
||||
Every nonempty subset \( S \) of \(\mathbb{Z}^-\) (the set of negative integers) has a largest element.
|
||||
|
||||
Let \( S \subseteq \mathbb{Z}^- \) be nonempty. Define the set
|
||||
\[
|
||||
T = \{ -s : s \in S \}.
|
||||
\]
|
||||
Since every element \( s \) in \( S \) is negative, each \( -s \) is a positive integer. Hence, \( T \) is a nonempty subset of the positive integers \(\mathbb{N}\).
|
||||
|
||||
By the well-ordering principle of \(\mathbb{N}\), the set \( T \) has a least element, say \( m \). Thus,
|
||||
\[
|
||||
m \in T \quad \text{and} \quad m \le t \quad \text{for all } t \in T.
|
||||
\]
|
||||
|
||||
Since \( m \in T \), there exists an element \( s_0 \in S \) such that
|
||||
\[
|
||||
m = -s_0.
|
||||
\]
|
||||
|
||||
We now claim that \( s_0 \) is the largest element of \( S \). To see this, let \( s \) be any element of \( S \). Then \(-s \in T\), and by the minimality of \( m \) we have
|
||||
\[
|
||||
m \le -s.
|
||||
\]
|
||||
Substituting \( m = -s_0 \), we obtain
|
||||
\[
|
||||
-s_0 \le -s.
|
||||
\]
|
||||
Multiplying both sides by \(-1\) (which reverses the inequality) gives
|
||||
\[
|
||||
s_0 \ge s.
|
||||
\]
|
||||
Since \( s \) was an arbitrary element of \( S \), it follows that \( s_0 \) is an upper bound of \( S \) and, being an element of \( S \), is the largest element of \( S \).
|
||||
|
||||
Thus, every nonempty subset of \(\mathbb{Z}^-\) has a largest element.
|
||||
`)
|
||||
|
||||
#proof[
|
||||
We construct a new set, $S' = {-x | x in S}$. Then we note that $S'$ is a set
|
||||
of positive integers, and in fact $S' subset.eq NN$. By the well ordering
|
||||
principle, $S'$ always has a least element. Call this least element $k in
|
||||
S'$. $k$ is the least element in $S'$ if and only if $-k$ is the greatest
|
||||
element in $S$. Thus, $S$ always has a largest element.
|
||||
]
|
||||
|
|
|
|||
|
|
@ -5,11 +5,11 @@
|
|||
"nixpkgs-lib": "nixpkgs-lib"
|
||||
},
|
||||
"locked": {
|
||||
"lastModified": 1736143030,
|
||||
"narHash": "sha256-+hu54pAoLDEZT9pjHlqL9DNzWz0NbUn8NEAHP7PQPzU=",
|
||||
"lastModified": 1738453229,
|
||||
"narHash": "sha256-7H9XgNiGLKN1G1CgRh0vUL4AheZSYzPm+zmZ7vxbJdo=",
|
||||
"owner": "hercules-ci",
|
||||
"repo": "flake-parts",
|
||||
"rev": "b905f6fc23a9051a6e1b741e1438dbfc0634c6de",
|
||||
"rev": "32ea77a06711b758da0ad9bd6a844c5740a87abd",
|
||||
"type": "github"
|
||||
},
|
||||
"original": {
|
||||
|
|
@ -20,11 +20,11 @@
|
|||
},
|
||||
"nixpkgs": {
|
||||
"locked": {
|
||||
"lastModified": 1736012469,
|
||||
"narHash": "sha256-/qlNWm/IEVVH7GfgAIyP6EsVZI6zjAx1cV5zNyrs+rI=",
|
||||
"lastModified": 1739866667,
|
||||
"narHash": "sha256-EO1ygNKZlsAC9avfcwHkKGMsmipUk1Uc0TbrEZpkn64=",
|
||||
"owner": "NixOS",
|
||||
"repo": "nixpkgs",
|
||||
"rev": "8f3e1f807051e32d8c95cd12b9b421623850a34d",
|
||||
"rev": "73cf49b8ad837ade2de76f87eb53fc85ed5d4680",
|
||||
"type": "github"
|
||||
},
|
||||
"original": {
|
||||
|
|
@ -36,14 +36,14 @@
|
|||
},
|
||||
"nixpkgs-lib": {
|
||||
"locked": {
|
||||
"lastModified": 1735774519,
|
||||
"narHash": "sha256-CewEm1o2eVAnoqb6Ml+Qi9Gg/EfNAxbRx1lANGVyoLI=",
|
||||
"lastModified": 1738452942,
|
||||
"narHash": "sha256-vJzFZGaCpnmo7I6i416HaBLpC+hvcURh/BQwROcGIp8=",
|
||||
"type": "tarball",
|
||||
"url": "https://github.com/NixOS/nixpkgs/archive/e9b51731911566bbf7e4895475a87fe06961de0b.tar.gz"
|
||||
"url": "https://github.com/NixOS/nixpkgs/archive/072a6db25e947df2f31aab9eccd0ab75d5b2da11.tar.gz"
|
||||
},
|
||||
"original": {
|
||||
"type": "tarball",
|
||||
"url": "https://github.com/NixOS/nixpkgs/archive/e9b51731911566bbf7e4895475a87fe06961de0b.tar.gz"
|
||||
"url": "https://github.com/NixOS/nixpkgs/archive/072a6db25e947df2f31aab9eccd0ab75d5b2da11.tar.gz"
|
||||
}
|
||||
},
|
||||
"root": {
|
||||
|
|
@ -62,11 +62,11 @@
|
|||
]
|
||||
},
|
||||
"locked": {
|
||||
"lastModified": 1735930054,
|
||||
"narHash": "sha256-30Q6QmUHT/RBoPrrGfSNE9mjub3qLHJ0IDPUxuyHDAQ=",
|
||||
"lastModified": 1738982361,
|
||||
"narHash": "sha256-QWDOo/+9pGu63knSlrhPiESSC+Ij/QYckC3yH8QPK4k=",
|
||||
"owner": "loqusion",
|
||||
"repo": "typix",
|
||||
"rev": "29144f9e4131628afb800cc54806da1a702f7c80",
|
||||
"rev": "bdb42d3e9a8722768e2168e31077129207870f92",
|
||||
"type": "github"
|
||||
},
|
||||
"original": {
|
||||
|
|
@ -78,11 +78,11 @@
|
|||
"typst-packages": {
|
||||
"flake": false,
|
||||
"locked": {
|
||||
"lastModified": 1736182943,
|
||||
"narHash": "sha256-w8acKWK2aKkCsaOJcR9GLtkrsPJxDrrPN/77h1OkhuM=",
|
||||
"lastModified": 1740119901,
|
||||
"narHash": "sha256-VzjqNki0yam7wrKbNkgMdyKyChEYBo0vjK0gwJZCFs4=",
|
||||
"owner": "typst",
|
||||
"repo": "packages",
|
||||
"rev": "820818337a0f2a27dfc880f5ed96634914ed592f",
|
||||
"rev": "92ae895b0076ac5d12ca81665b2af56f628b09f0",
|
||||
"type": "github"
|
||||
},
|
||||
"original": {
|
||||
|
|
@ -93,11 +93,11 @@
|
|||
},
|
||||
"zen-typ": {
|
||||
"locked": {
|
||||
"lastModified": 1737198395,
|
||||
"narHash": "sha256-V3runX3cITanGPVF/Sfjak2/FNs6DuAwayQ44Iedal8=",
|
||||
"lastModified": 1740095364,
|
||||
"narHash": "sha256-czzheiWIwJeze1aAzLmOgh+y9RoG8Qyzeicn1fd0ESY=",
|
||||
"owner": "youwen5",
|
||||
"repo": "zen.typ",
|
||||
"rev": "3564895682f394f3c9be291332cc19c6ab695a60",
|
||||
"rev": "d761990a004ad8dee8fc642a03fd4f11348a5fe2",
|
||||
"type": "github"
|
||||
},
|
||||
"original": {
|
||||
|
|
|
|||
Loading…
Reference in a new issue