MAT102 Tutorial W3B

• 102_F2025_Tutorial2.pdf
• 1
  • Let's see.
  • 1
    • $\mathrm{\neg }(\mathrm{\forall }x\in \mathbb{Z},(\mathrm{\neg }P(x)⟹Q(x)))$
    • $\mathrm{\neg }(\mathrm{\forall }x\in \mathbb{Z},(P(x)\vee Q(x)))$
    • $(\mathrm{\exists }x\in \mathbb{Z},(\mathrm{\neg }P(x)\wedge \mathrm{\neg }Q(x))))$
  • 2
    • $(\mathrm{\exists }x\in \mathbb{Z},(\mathrm{\neg }P(x)\wedge \mathrm{\neg }Q(x)))$
  • See that the negation of $\ast$ is $\ast \ast$. So there is no way that these are true at the same time.
• 2
  • a
    • $\ast$
      • $\mathrm{\forall }x\in \mathbb{Z},\mathrm{\neg }(x\in A)⟹(x\in B)$
      • $\mathrm{\forall }x\in \mathbb{Z},x\notin A⟹x\in B$
    • $\ast \ast$
      • $\mathrm{\exists }x\in \mathbb{Z},\mathrm{\neg }(x\in A\cup x\in B)$
      • $\mathrm{\exists }x\in \mathbb{Z},x\notin A\wedge x\notin B$
      • $\mathrm{\exists }x\in \mathbb{Z},x\in A^c\wedge x\in B^c$
      • $\mathrm{\exists }x\in \mathbb{Z},x\in A^c\cap B^c$
    • Notice $A=\{x\in \mathbb{Z}:\mathrm{\exists }y\in \mathbb{Z},xy>0\}$
    • $0\notin A$
    • Notice $B=\{x\in \mathbb{Z}:\mathrm{\forall }y\in \mathbb{Z},xy=0\}$
    • $B=\{0\}$
    • So $A\cup B=\mathbb{Z}$
  • b
    • $\ast$
      • It's true
      • If $x\notin A$ then $x\in B$.
      • True
      • $A^c=B$
      • And $B^c=A$
    • $\ast \ast$
      • This one says there is an $x$ such that it's not in $A$ and not in $B$.
      • False
      • There is no intersection between $A\cap B$ so there is also no intersection between $A^c\cap B^c$ too.
• 3
  • $\ast$
    • The things that make this true
    • $(P\wedge R)⟺S$
    • So $P\wedge R$ must be true
    • All True or all False
  • $\ast \ast$
    • $\mathrm{\neg }(P⟹(Q\vee W))$
    • $\mathrm{\neg }(\mathrm{\neg }P\vee (Q\vee W))$
    • $P\wedge \mathrm{\neg }(Q\vee W)$
    • $P\wedge \mathrm{\neg }Q\wedge \mathrm{\neg }W$
    • $P=T,Q=F,W=F$
  • $\ast \ast \ast$
    • $(S⟹W)\wedge (\mathrm{\neg }R⟹\mathrm{\neg }W)$
    • $(\mathrm{\neg }S\vee W)\wedge (R\vee \mathrm{\neg }W)$
    • $\mathrm{\neg }S\vee W=T$
    • $R\vee \mathrm{\neg }W=T$
    • Cases:
      • $W=F$ from $\ast \ast$
      • $S=F$
      • $R=F$
  • $\begin{array}{c}P=T\\ Q=F\\ R=F\\ S=F\\ W=F\end{array}$