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Can any body explain the logical statement in common terms:

$$P \to (Q \lor R) ~~\iff ~~(P \land\lnot Q) \to R$$

  • $\lnot$ negation ("not")
  • $\land$ conjunction ("and")
  • $\lor$ disjunction ("or")
  • $\to$ condition ("implies")
  • $\iff$ logical equivalence ("if and only if")
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  • $\begingroup$ Does Q' mean Q is not real? $\endgroup$ – OrdinaryWitch Apr 1 '18 at 8:01
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Assume that $P\rightarrow Q\vee R$, we are to show that $P\wedge\neg Q\rightarrow R$. Now assume that $P\wedge\neg Q$ holds, in particular, $P$ holds, so by assumption, we have that $Q\vee R$ holds, we get that either $Q$ or $R$ holds. Since we have $\neg Q$ holds, which means that $Q$ does not hold, so from either $Q$ or $R$ holds, we get that $R$ must hold, this justifies that $P\wedge\neg Q\rightarrow R$.

Now assume that $P\wedge\neg Q\rightarrow R$, we are to show that $P\rightarrow Q\vee R$. So assume that $P$ holds but $Q\vee R$ does not hold, then neither $Q$ nor $R$ holds. In particular, $\neg Q$ holds, so $P\wedge\neg Q$ holds, so we get that $R$ holds, but we have both $R$ and $\neg R$ hold, this is a contradiction. The contradiction comes from that we have assumed that $Q\vee R$ does not hold, so $Q\vee R$ must hold, this justifies that $P\rightarrow Q\vee R$.

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Look at these two truth tables: $$\begin{array}{|c|c|c|c|c|} \hline % after \\: \hline or \cline{col1-col2} \cline{col3-col4} ... P & Q & R & Q \cup R =:I & P\Longrightarrow I\\ \hline 0 & 0 & 1 & 1 & 1 \\ 0 & 0 & 1 & 1 & 1 \\ 0 & 1 & 0 & 1 & 1 \\ 0 & 1 & 1 & 1 & 1 \\ 1 & 0 & 0 & 0 & 0 \\ 1 & 0 & 1 & 1 & 1 \\ 1 & 1 & 0 & 1 & 1 \\ 1 & 1 & 1 & 1 & 1 \\ \hline \end{array}$$

$$\begin{array}{|c|c|c|c|c|} \hline % after \\: \hline or \cline{col1-col2} \cline{col3-col4} ... P & Q' & R & P \cap Q' =:J & J\Longrightarrow R\\ \hline 0 & 1 & 1 & 0 & 1 \\ 0 & 1 & 1 & 0 & 1 \\ 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 & 1 \\ 1 & 1 & 0 & 1 & 0 \\ 1 & 1 & 1 & 1 & 1 \\ 1 & 0 & 0 & 0 & 1 \\ 1 & 0 & 1 & 0 & 1 \\ \hline \end{array}$$

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$$ P\Longrightarrow (Q\cup R) $$ is equivalent to $$ P'\cup (Q\cup R) $$ which is equivalent to $$ (P'\cup Q)\cup R $$ which is equivalent to $$ (P'\cup Q)'\Longrightarrow R $$ which is equivalent to $$ P\cap Q'\Longrightarrow R $$

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$P\to (Q\lor R)$ means that "At least one from $Q$ or $R$ is true, if $P$ is true."

$(P\land\lnot Q)\to R$ means that "$R$ is true, if $P$ is true but $Q$ is false."

Assume $P\to (Q\vee R)$ holds; that is at least one from $Q$ or $R$ is true, if $P$ is true.   Well, from that we can infer that $R$ will be true is $P$ is true but $Q$ is false.   So $P\to (Q\vee R)$ logically entails $(P\land\neg Q)\to R$.

Likewise if $(P\land\neg Q)\to R$ holds we can infer that $P\to(Q\vee R)$ also holds.

In summary $(P\land\neg Q)\to R$ is logically equivalent to $P\to (Q\vee R)$

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$P \to (Q \lor R)$

If it is the weekend, then it is Saturday or Sunday.

$(P \land \lnot Q) \to R$

If it is the weekend, and it is not Saturday, then it is Sunday.

Notice how both statements are equivalent to saying "the only possible weekend days are Saturday and Sunday", but neither necessarily implies that Saturday is a weekend or implies that Sunday is a weekend.

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You can get this from two simple equivalences:

  1. $P \vee Q \Leftrightarrow \neg P \rightarrow Q$

  2. $(P \wedge Q) \rightarrow R \Leftrightarrow P \rightarrow (Q \rightarrow R)$

Here $(P \wedge \neg Q) \rightarrow R$ is equivalent (by 2) to $P \rightarrow (\neg Q \rightarrow R)$ is equivalent (by 1) to $P \rightarrow (Q \vee R)$.

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