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As per the title, im asked to Prove that $p \implies (q \lor (\neg r \implies p)) \equiv q \lor r$.

However, im stuck on this logical equivalence question, my working so far is as follows;

$$p \implies (q \lor (\neg r \implies p)) \equiv \neg p \lor (q \lor r\lor p)$$

$$\equiv (\neg p \lor p) \lor (q \lor r)$$

Would this not be just a tautology?, whereas $q \lor r$ is not?

log x log log y
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2 Answers2

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$p \implies (q \lor (\neg r \implies p)) \equiv q \lor r$

im stuck on this logical equivalence question

Neither $$P → \bigg((Q ∨ (¬R → P)) ↔ Q ∨ R\bigg)\tag1$$ nor $$\bigg(P → (Q ∨ (¬R→P))\bigg) ↔ Q∨R.\tag2$$ is actually logically valid.

P.S. I'm suggesting sentence $(1)$ in case the author is treating $→/\Rightarrow$ and $↔/≡$ as being collectively right-associative.

ryang
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\begin{align} p \implies (q \lor (\neg r \implies p)) &\equiv \neg p \lor (q \lor r\lor p) \hspace{0.4cm} \text{Using }(p\implies q)\iff (\neg p\lor q)\\ &\equiv (\neg p \lor p) \lor (q\lor r) \hspace{0.4cm} \text{Associativity and commutativity of }\lor\\ \end{align}

$(\neg p\lor p)$ is a tautology, if $p$ is true, its negation is wrong, and so the ($\neg p\lor p$) is always true. Meaning that the formula is a tautology.

Thanks to @insipidintegrator for the remark.

Hamdiken
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