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Formulas $φ$ and $ψ$ are said to be logically equivalent, which we write as $φ ≡ ψ$, if for all truth assignments $v$, $v(φ) = v(ψ)$.

Which seems a lot like a biconditional, since $ φ↔ψ$ if $v(φ) = v(ψ)$.

In fact, my textbook states the following theorem:

$φ ≡ ψ$ if and only if $(φ ↔ ψ)$ is a tautology, for all formulas $φ, ψ$.

So I am sure they are related. I successfully proved the theorem, but why is this true only when $(φ ↔ ψ)$ is a tautology? From my guts feeling, is it because logical equivalence depends entirely on the meaning of the formula, i.e. the way the truth tables of connectives are set? In other words, this is a necessary relation once the meaning/truth table of both formulae are laid out? (Like the identity relation between a bachelor and an unmarried man?)

Whereas for a biconditional, it is not a necessary relation: for a given biconditional that is true, if it is contingent - there are truth assignments that can make it false (unless it is a tautology of course, hence the theorem)- thus not qualified for a necessary relation. Am I right? But what would be an example?

Also, in my textbook, while proving $(φ ∨ (ψ ∨ θ)) ≡ ((φ ∨ ψ) ∨ θ)$, it simply assumed $(φ ∨ (ψ ∨ θ))$=F and proved $((φ ∨ ψ) ∨ θ)$=F and vice versa. Why is it not necessary to assume $v(φ ∨ (ψ ∨ θ))$ or $v((φ ∨ ψ) ∨ θ)$ to be true and prove it the same way?

Constantly confused
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1 Answers1

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Note that $\varphi$ ans $\psi$ are logically equivalent iff $\varphi \vDash \psi$ and $\psi \vDash \varphi$.

But we have that $\varphi \vDash \psi$ iff $\vDash \varphi \to \psi$, i.e. iff $(\varphi \to \psi)$ is a tautology.

Thus, $\varphi$ ans $\psi$ are logically equivalent iff $(\varphi \leftrightarrow \psi)$ is a tautology.

"a given biconditional that is true, if it is contingent [correct terminology: satisfiable but not a tautology] ... thus not qualified for a necessary relation."

If a formula $(\varphi \leftrightarrow \psi)$ is not a tautology, the two sub-formulas are not logically equivalent, exactly because there is a valuation $v$ satisfying one of them and falsifying the second one [such that e.g. $v(\varphi)=\text{True}$ and $v(\psi)=\text{False}$].

This is the case with $(p \lor q) ↔ p$ is not a tautology: for $v(q)=\text{True}$ and $v(p)=\text{False}$, the formula is $\text{False}$.

And thus, $p ∨ q$ is not logically equiv to $p$. This explain the iff between logical equivalence and the tautologueness of the corresponding bi-conditional.

As an aside, we can use the above facts to show that $\varphi \vDash \bot$ iff $\varphi$ is unsatisfiable (i.e. always false).

We have that $\varphi \vDash \bot$ iff $(\varphi \to \bot)$ is a tuatology.

But $v(\bot)=\text{False}$, for every $v$, and thus, by the truth table for the conditional, we have that $v(\varphi)=\text{False}$, for every $v$.

In concluson: if we have $\varphi \vDash \bot$, necessarily $\varphi$ must be unsatisfiable (like e.g $p ∧ ¬p$).

Mauro ALLEGRANZA
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