Question #306327
  1. –p → q
  2.  (p → –q) ∨ –q
  3.  p → –(p ∨ q)
  4. (p ∧ q) ∨ (–p ∨ q)
  5. [ (p ∧ q) ∨ –p ] ∧ –q
1
Expert's answer
2022-03-08T11:59:04-0500

1)


pqppq0010011110011101\begin{matrix} \hline p & q & -p & -p\rightarrow q \\ \hline 0 & 0 & 1 & 0 \\ \hline 0 & 1 & 1 & 1 \\ \hline 1 & 0 & 0 & 1 \\ \hline 1 & 1 & 0 & 1 \\ \hline \end{matrix}

2)


pqqpq(pq)q00111010111011111000\begin{matrix} \hline p & q & -q & p\rightarrow -q & (p\rightarrow-q)\vee-q \\ \hline 0 & 0 & 1 & 1 & 1 \\ \hline 0 & 1 & 0 & 1 & 1 \\ \hline 1 & 0 & 1 & 1 & 1 \\ \hline 1 & 1 & 0 & 0 & 0 \\ \hline \end{matrix}

3)

pqpq(pq)p(pq)00011011011010011100\begin{matrix}\hline p & q & p\vee q & -(p\vee q) & p\rightarrow -(p \vee q) \\ \hline 0 & 0 & 0 & 1 & 1 \\ \hline 0 & 1 & 1 & 0 & 1 \\ \hline 1 & 0 & 1 & 0 & 0 \\ \hline 1 & 1 & 1 & 0 & 0 \\ \hline \end{matrix}

4)

pqpqppq(pq)(pq))000111010111100000111011\begin{matrix}\hline p & q & p \land q & -p & -p \vee q & (p \land q)\vee(-p \vee q)) \\ \hline 0 & 0 & 0 & 1 & 1 & 1 \\ \hline 0 & 1 & 0 & 1 & 1 & 1 \\ \hline 1 & 0 & 0 & 0 & 0 & 0 \\ \hline 1 & 1 & 1 & 0 & 1 & 1 \\ \hline \end{matrix}

5)


pqpqp(pq)pq[(pq)p]q0001111010110010000101110100\begin{matrix}\hline p & q & p \land q & -p & (p \land q)\vee-p & -q & {[}(p \land q) \vee -p{]} \land -q \\ \hline 0 & 0 & 0 & 1 & 1 & 1 & 1 \\ \hline 0 & 1 & 0 & 1 & 1 & 0 & 0 \\ \hline 1 & 0 & 0 & 0 & 0 & 1 & 0 \\ \hline 1 & 1 & 1 & 0 & 1 & 0 & 0 \\ \hline \end{matrix}


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