auto-update(nvim): 2025-01-13 02:08:30
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1 changed files with 186 additions and 27 deletions
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@ -13,6 +13,8 @@
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),
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),
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)
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)
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#set par(first-line-indent: 0pt, spacing: 1em)
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Problems:
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Problems:
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1.1: \#1ceij, 2c, 3cdeghjL, 4cdefh, 6, 7cg, 10ce, 11bei, 12a, 13
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1.1: \#1ceij, 2c, 3cdeghjL, 4cdefh, 6, 7cg, 10ce, 11bei, 12a, 13
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@ -27,16 +29,19 @@ Problems:
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True.
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True.
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] \
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] \
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e. #[
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e. #[
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Either $pi$ is rational and $17$ is a prime, or $7 < 13$ and $81$ is a perfect square.
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Either $pi$ is rational and $17$ is a prime, or $7 < 13$ and $81$ is a perfect square.
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True.
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True.
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] \
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] \
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i. #[
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i. #[
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It is not the case that $39$ is prime, or that 64 is a power of 2.
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It is not the case that $39$ is prime, or that 64 is a power of 2.
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False.
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False.
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] \
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] \
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j. #[
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j. #[
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There are more than three false statements in this book, and this statement is one of them.
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There are more than three false statements in this book, and this statement is one of them.
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@ -160,35 +165,9 @@ Problems:
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6. \
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6. \
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a. $not P and not Q$, $not (P and not Q)$ are not equivalent because we can choose $P$ and $Q$ to both be true which gives false for proposition 1 and true for proposition 2. \
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a. $not P and not Q$, $not (P and not Q)$ are not equivalent because we can choose $P$ and $Q$ to both be true which gives false for proposition 1 and true for proposition 2. \
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b. $(not P) or (not Q), not (P or not Q)$ are not equivalent because we can choose $P$ to be true and $Q$ to be false, and proposition 1 is true while proposition 2 is false. \
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b. $(not P) or (not Q), not (P or not Q)$ are not equivalent because we can choose $P$ to be true and $Q$ to be false, and proposition 1 is true while proposition 2 is false. \
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c. $(P and Q) or R$, $P and (Q or R)$ are not equivalent, by a truth table.
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c. $(P and Q) or R$, $P and (Q or R)$ are not equivalent
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#proof[
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#proof[
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Direct computation, by a truth table. Hopefully this suffices to show the absurdity of computing nontrivial boolean results by truth table.
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#table(
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align: center,
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columns: (1fr, 1fr, 1fr, 1fr, 1fr, 2fr, 2fr),
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[$P$],
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[$Q$],
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[$R$],
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[$P and Q$],
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[$Q or R$],
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[$(P and Q) or R$],
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[$P and (Q or R)$],
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[T], [T], [T], [T], [T], [T], [T],
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[T], [F], [T], [F], [T], [T], [T],
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[T], [T], [F], [F], [T], [T], [T],
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[T], [F], [F], [F], [F], [F], [F],
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[F], [T], [T], [F], [T], [T], [F],
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[F], [F], [T], [], [], [], [],
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[F], [T], [F], [], [], [], [],
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[F], [F], [F], [], [], [], [],
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)
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We need not complete the table as we have already identified a contradiction for when $P$ and $Q$ are false, $R$ true.
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]
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#remark(numbering: none)[
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We can also show this result directly by
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We can also show this result directly by
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$
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$
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@ -358,3 +337,183 @@ $ not (not P) or (not Q and not S) $
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]
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]
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= Exercises
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= Exercises
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1. \
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b. If #math.underbrace("the moon is made of cheese", "antedecent"), then #math.underbrace("triangles have four sides", "consequent").
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d. #math.underbrace([The differentiability of $f$], "antedecent") is sufficient for #math.underbrace[$f$ to be continuous][consequent].
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2. \
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b.
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Converse: If triangles have four sides, then squares have three sides.
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Contrapositive: If triangles don't have four sides, then squares don't have three sides.
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#linebreak()
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d.
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Converse: The continuity of $f$ is sufficient for $f$ to be differentiable.
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Contrapositive: If $f$ is not continuous, then $f$ is not differentiable.
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3. \
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a. $Q$ must be true.
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#linebreak()
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b. $Q$ must be true.
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#linebreak()
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c. $Q$ is false.
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#linebreak()
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d. $Q$ is false.
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#linebreak()
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e. $Q$ is true.
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5. \
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c. True.
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#linebreak()
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f. True.
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#linebreak()
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g. True.
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6. \
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b. False.
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#linebreak()
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c. False
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#linebreak()
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g. False.
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#linebreak()
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7. \
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b.
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#table(
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columns: 6,
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align: center,
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[$P$],
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[$Q$],
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[$not P$],
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[$not P => Q$],
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[$Q <=> P$],
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[$(not P => Q) or (Q <=> P)$],
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[T], [T], [F], [T], [T], [T],
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[F], [T], [T], [T], [F], [T],
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[T], [F], [F], [T], [F], [T],
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[F], [F], [T], [F], [T], [T],
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)
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#linebreak()
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e.
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#table(
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columns: 5,
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align: center,
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[$P$], [$Q$], [$not Q$], [$Q <=> P$], [$not Q => (Q <=> P)$],
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[$T$], [$T$], [F], [T], [T],
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[$F$], [$T$], [F], [F], [T],
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[$T$], [$F$], [T], [F], [F],
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[$F$], [$F$], [T], [T], [T],
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)
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10. \
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b. If $n$ is prime, then $n = 2$ or $n$ is odd.
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$
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n "is prime" => (n=2) or (n mod 2 = 1)
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$
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f.
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$
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2 < n - 6 => 2n < 4 or n > 4
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$
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g.
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$
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6 >= n - 3 <=> n > 4 or n > 10
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$
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12. \
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b.
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#table(
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columns: 9,
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align: center,
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[$P$],
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[$Q$],
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[$R$],
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[$not R$],
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[$not Q$],
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[$P and Q$],
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[$P and not R$],
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[$(P and Q) => R$],
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[$(P and not R) => not Q$],
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[T], [T], [T], [F], [F], [T], [F], [T], [T],
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[T], [F], [T], [F], [T], [F], [F], [T], [T],
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[T], [T], [F], [T], [F], [T], [T], [F], [F],
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[T], [F], [F], [T], [T], [F], [T], [T], [T],
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[F], [T], [T], [F], [F], [F], [F], [T], [T],
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[F], [F], [T], [F], [T], [F], [F], [T], [T],
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[F], [T], [F], [T], [F], [F], [F], [T], [T],
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[F], [F], [F], [T], [T], [F], [F], [T], [T],
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)
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So they are equivalent.
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#linebreak()
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c.
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#proof[
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$
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P => (Q and R) &<=> not (Q and R) => not P &&("contrapositive") \
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not (Q and R) => not P &<=> (not Q or not R) => not P space &&("DeMorgan's") \
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$
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]
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13. \
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a. Continuity implies differentiability.
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#linebreak()
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b. Differentiability implies continuity.
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#linebreak()
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c. It is not possible.
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#linebreak()
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d. $P => Q$, where $P$ is true and $Q$ is true.
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16. \
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c. Contradiction.
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#linebreak()
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d. Neither. Statement is equivalent to $P => Q$.
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#linebreak()
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e. Tautology. $P and (Q or not Q)$ is independent of $Q$ so this reduces to
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$P <=> P$.
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