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Can someone please verify whether my proof is okay? It looks very tedious and possibly stating unnecessary statements. I might also be missing some important facts.

Prove that $p$ and $q$ are logically equivalent if and only if $p\leftrightarrow q$ is a tautology.

Assume $p$ is logically equivalent to $q$. Then every truth value gives $p$ and $q$ the same truth values. Then $p\leftrightarrow q$ is a tautology.

Assume $p\leftrightarrow q$ is a tautology. Then $p$ and $q$ have the same truth values. Then it must be that $p$ is logically equivalent to $q$.

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    $\begingroup$ Can you explain what do you mean by logically equivalent? The statement you are trying to prove seems true to me by definition. $\endgroup$ – Hugo C Botós Jun 10 '18 at 1:59
  • $\begingroup$ @Hugocito Logically equivalent: $p \Leftrightarrow q$. Yes, the definition of logically equivalent proves it is a tautology already! So I am just wondering whether my proof is too tedious? $\endgroup$ – user482939 Jun 10 '18 at 4:03
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    $\begingroup$ Correct; see Logical equivalence. $\endgroup$ – Mauro ALLEGRANZA Jun 10 '18 at 9:28
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"Assume p is logically equivalent to q. Then every truth value gives p and q the same truth values."

That sounds fine. 'p' and 'q' have the same truth value by definition.

"Then p↔q is a tautology."

You might do better to refer to the values of ($\bot$↔$\bot$) and ($\top$↔$\top$) specifically. And you might do well to note that no other truth values than $\top$ and $\bot$ exist for how ↔ gets understood, because there exist three-valued logical systems where there exist connectives for a similar concept, but the same truth values for both 'p' and 'q' do not guarantee that (p↔q) is a tautology.

The same goes for the second part.

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