#import "./dvd.typ": *

#show: dvdtyp.with(
  title: "Pset 1",
  author: "Youwen Wu",
)

#set heading(
  numbering: (
    num => {
      return "1." + str(num)
    }
  ),
)

Problems:

1.1: \#1ceij, 2c, 3cdeghjL, 4cdefh, 6, 7cg, 10ce, 11bei, 12a, 13

1.2: \#1bd, 2bd, 3, 5cfgh, 6bcg, 7be, 10bfg, 12bc, 13, 16cde

= Exercises

1. \
  c. #[
  $x/2$ is a rational number.

  True.
  ] \
  e. #[
  Either $pi$ is rational and $17$ is a prime, or $7 < 13$ and $81$ is a perfect square.

  True.
  ] \
  i. #[
    It is not the case that $39$ is prime, or that 64 is a power of 2.

    False.
  ] \
  j. #[
  There are more than three false statements in this book, and this statement is one of them.

  False.
  ]

2c. #[
  $P$ is $5^2 + 12^2 = 13^2$ and $Q$ is $sqrt(2) + sqrt(3) = sqrt(2 + 3)$

  $P and Q: "false"$

  $P or Q: "true"$
]

3. #[
    \
    c. $P and not Q$
    #[
      #table(
        columns: 3,
        align: center,
        [$P$], [$Q$], [$P and not Q$],
        [T], [T], [F],
        [T], [F], [T],
        [F], [T], [F],
        [F], [F], [F],
      )
    ] \
    d. $P and (Q or not Q)$
    #[
      #table(
        columns: 5,
        align: center,
        [$P$], [$Q$], [$not Q$], [$Q or not Q$], [$P and (Q or not Q)$],
        [T], [T], [F], [T], [T],
        [T], [F], [T], [T], [T],
        [F], [T], [F], [T], [F],
        [F], [F], [T], [T], [F],
      )
    ] \
    e. $(P and Q) or not Q$ \
    #[
      #table(
        columns: 5,
        align: center,
        [$P$], [$Q$], [$P and Q$], [$not Q$], [$(P and Q) or not Q$],
        [T], [T], [T], [F], [T],
        [F], [T], [F], [F], [F],
        [T], [F], [F], [T], [T],
        [F], [F], [F], [T], [T],
      )
    ] \
    g. $(P or S) and (P or K)$
    #table(
      columns: (1fr, 1fr, 1fr, 1fr, 1fr, 2fr),
      align: center,
      [$P$], [$S$], [$K$], [$P or S$], [$P or K$], [$(P or S) and (P or K)$],
      [T], [T], [T], [T], [T], [T],
      [T], [T], [F], [T], [T], [T],
      [T], [F], [T], [T], [T], [T],
      [T], [F], [F], [T], [T], [T],
      [F], [T], [T], [T], [T], [T],
      [F], [T], [F], [T], [F], [F],
      [F], [F], [T], [F], [T], [T],
      [F], [F], [F], [F], [F], [F],
    )
  ] \
  h. $not P and not Q$
  #table(
    align: center,
    columns: 5,
    [$P$], [$Q$], [$not P$], [$not Q$], [$not P and not Q$],
    [T], [T], [F], [F], [F],
    [T], [F], [F], [T], [F],
    [F], [T], [T], [F], [F],
    [F], [F], [T], [T], [T],
  ) \
  j. $(P and Q) or (P and R)$
  #table(
    align: center,
    columns: 6,
    [$P$], [$Q$], [$R$], [$P and Q$], [$P and R$], [$(P and Q) or (P and R)$],
    [T], [T], [T], [T], [T], [T],
    [T], [F], [T], [F], [T], [T],
    [T], [T], [F], [T], [F], [T],
    [T], [F], [F], [F], [F], [F],
    [F], [T], [T], [F], [F], [F],
    [F], [F], [T], [F], [F], [F],
    [F], [T], [F], [F], [F], [F],
    [F], [F], [F], [F], [F], [F],
  ) \
  L. $(P and Q) or (P and not S)$
  #table(
    align: center,
    columns: 7,
    [$P$],
    [$Q$],
    [$S$],
    [$not S$],
    [$P and Q$],
    [$P and not S$],
    [$(P and Q) or (P and not S)$],

    [T], [T], [T], [F], [T], [F], [T],
    [T], [F], [T], [F], [F], [F], [F],
    [T], [T], [F], [T], [T], [T], [T],
    [T], [F], [F], [T], [F], [T], [T],
    [F], [T], [T], [F], [F], [F], [F],
    [F], [F], [T], [F], [F], [F], [F],
    [F], [T], [F], [T], [F], [F], [F],
    [F], [F], [F], [T], [F], [F], [F],
  )

4. \
  c. $(P or Q) and (R or S)$ is true. \
  d. $(not P or not Q) or (not R or not S)$ is true. \
  e. $not P or not Q$ is false. \
  f. $(not Q or S) and (Q or S)$ is false. \
  h. $K and not (S or Q)$ is false.

6. \
  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. \
  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. \
  c. $(P and Q) or R$, $P and (Q or R)$ are not equivalent, by a truth table.

  #proof[
    Direct computation, by a truth table. Hopefully this suffices to show the absurdity of computing nontrivial boolean results by truth table.
    #table(
      align: center,
      columns: (1fr, 1fr, 1fr, 1fr, 1fr, 2fr, 2fr),
      [$P$],
      [$Q$],
      [$R$],
      [$P and Q$],
      [$Q or R$],
      [$(P and Q) or R$],
      [$P and (Q or R)$],

      [T], [T], [T], [T], [T], [T], [T],
      [T], [F], [T], [F], [T], [T], [T],
      [T], [T], [F], [F], [T], [T], [T],
      [T], [F], [F], [F], [F], [F], [F],
      [F], [T], [T], [F], [T], [T], [F],
      [F], [F], [T], [], [], [], [],
      [F], [T], [F], [], [], [], [],
      [F], [F], [F], [], [], [], [],
    )

    We need not complete the table as we have already identified a contradiction for when $P$ and $Q$ are false, $R$ true.
  ]

  #remark(numbering: none)[
    We can also show this result directly by

    $
      (P and Q) or R &= (P or R) and (Q or R) \
      P and (Q or R) &= (P or Q) and (P or R)
    $

    Clearly the choices of $P$ and $Q$ are false, $R$ is true again yields a contradictory result where proposition 1 is true while proposition 2 is false.
  ]

  d. $not (P and Q), not P and not Q$ \

  The statements are not equivalent.

  #proof[
    By DeMorgan's,
    $ not (P and Q) = not P or not Q $
    Therefore, choosing $P,Q$ with $P != Q$ gives a contradictory result between the two propositions.
  ]

  e. $(P and Q) or R, P or (Q and R)$

  They are not equivalent.

  #proof[
    Symbolic manipulation gives
    $
      (P and Q) or R &= (P or R) and (Q or R) \
      P or (Q and R) &= (P or Q) and (P or R)
    $
    Then select the conjunctive statement in parentheses that occurs
    exclusively in one of the statements and set both of its propositions to
    false. In this case we take $Q$ and $R$ false while $P$ is true, which
    makes proposition 1 false while proposition 2 is true.
  ]

  f. $(P and Q) or P, P$

  They are equivalent, trivially. Clearly $Q$ is independent of the outcome's truth value which only depends on $P$.

7. \

  c. Julius Caesar was born in 1492 or 1493 and died in 1776. \

  This proposition is of the form $(P or Q) and R$, where

  $
    P &= #[Caesar was born in 1492] \
    Q &= #[Caesar was born in 1493] \
    R &= #[Caesar died in 1776] \
  $

  It is false, clearly, as Caesar certainly could not have died in 1776.

  g. It is not the case that $-5$ and $13$ are elements of $NN$, but $4$ is in the set of rational numbers.

  This proposition takes the form $not (P and Q) and R$, where

  $
    P &= -5 in NN \
    Q &= 13 in NN \
    R &= 4 in QQ
  $

  It is true, since $P$ is false, $Q$ is true, $R$ is true, which implies $not
  (P and Q)$ is true, thus making the proposition as a whole true.

10. \
  c. $(P and Q) or (not P or not Q)$ \

  This is a tautology that is true $forall P,Q$.

  #proof[
    #table(
      align: center,
      columns: 5,
      [$P$],
      [$Q$],
      [$P and Q$],
      [$not P or not Q$],
      [$(P and Q) or (not P or not Q)$],

      [T], [T], [T], [F], [T],
      [T], [F], [F], [T], [T],
      [F], [T], [F], [T], [T],
      [F], [F], [T], [T], [T],
    )
  ]

  e. $(Q and not P) and not (P and R)$ \

  This is neither a tautology nor a contradiction.

  #proof[
    #table(
      align: center,
      columns: 5,
      [$P$],
      [$Q$],
      [$Q and not P$],
      [$not (P and R)$],
      [$(Q and not P) and not (P and R)$],

      [T], [T], [F], [F], [F],
      [F], [T], [F], [T], [F],
      [T], [F], [T], [T], [T],
      [F], [F], [F], [T], [F],
    )
  ]

11. \

  b. Cleveland will win the first game or the second game.

  Cleveland loses both the first and second games.

  e. Roses are red and violets are blue.

  Roses are green.

  i. The function $g$ has a relative maximum at $x = 2$ or $x = 4$ and a relative minimum at $x = 3$.

  $ g(x) "is given by" integral^x_0 (t - 5)(t-6) dif t $

12a.

Restore parentheses to $not not P or not Q and not S$.

$ not (not P) or (not Q and not S) $

13. \

  a. Make a truth table for the exclusive or ($xor$)

  #table(
    align: center,
    columns: 3,
    [$P$], [$Q$], [$P xor Q$],
    [T], [T], [F],
    [F], [T], [T],
    [T], [F], [T],
    [F], [F], [F],
  )

  b. Show that $A xor B = (A or B) and not (A and B)$.

  Intuitively this fact makes sense. Either $A$ or $B$ are true, but $A$ and $B$ can't _both_ be true.

  #proof[
    By direct computation,

    #table(
      align: center,
      columns: 6,
      [$P$],
      [$Q$],
      [$P xor Q$],
      [$A or B$],
      [$not (A and B)$],
      [$(A or B) and not (A and B)$],

      [T], [T], [F], [T], [F], [F],
      [F], [T], [T], [T], [T], [T],
      [T], [F], [T], [T], [T], [T],
      [F], [F], [F], [F], [T], [F],
    )
  ]

= Exercises