Claim

\(\Sigma;\alpha \vDash \beta \text{ iff }\Sigma\vDash(\alpha\to\beta)\)

Proof

”\(\Rightarrow\)”

For the forward direction, assume that \(\Sigma;\alpha\vDash\beta\)

Let \(v\) be an arbitrary truth assignment which satisfies \(\Sigma\)

Case 1: \(v\) also satisfies \(\alpha\)

In this case, since we already know that \(\Sigma;\alpha\vDash\beta\), we know that the truth assignment \(v\) will also satisfy \(\beta\). Thus, \(\overline{v}(\alpha) = \overline{v}(\beta) =T\) so that \(\overline{v}(\alpha\to\beta) = T\)

**Case 2: \(v\) does not satisfy \(\alpha\)

In this case, \(\overline{v}(\alpha) = F\), so no matter the value of \(\overline{v}(\beta)\), \(\overline{v}(\alpha\to\beta) = T\)

Since \(v\) is arbitrary in both cases, we see that that \(\Sigma\vDash(\alpha\to\beta)\)

”\(\Leftarrow\)”

For the backwards direction, assume that \(\Sigma\vDash(\alpha\to\beta)\)

Let \(v\) be a truth assignment satisfying \(\Sigma;\alpha\). Then \(\overline{v}(\alpha)=T\) and \(\overline{v}(\alpha\to\beta)=T\), since \(\Sigma\vDash(\alpha\to\beta)\) Suppose \(\overline{v}(\beta)=F\), then since \(\overline{v}(\alpha)=T\) and by the axioms which \(\overline{v}\) must follow, it must be that \(\overline{v}(\alpha\to\beta) = F\), which is a contradiction. Thus, we must have that \(\overline{v}(\beta) = T\)

Since \(v\) was an arbitrary truth assignment satisfying \(\Sigma;\alpha\), we have that \(\Sigma;\alpha\vDash\beta\)

Part (b)

Claim: \(\alpha\) is tautologically equivalent to \(\beta\) if and only if \((\alpha\leftrightarrow\beta)\) is a tautology.

”\(\Rightarrow\)”

For the forward direction, assume \(\alpha\) and \(\beta\) are tautologically equivalent.

Now, let \(v\) be an arbitrary truth assignment.

Case 1: \(\overline{v}(\alpha) =F\) In this case, we must have that \(\overline{v}(\beta) = F\), since \(\beta\vDash\alpha\).

Case 2: \(\overline{v}(\alpha)=T\) In this case, we must have that \(\overline{v}(\beta) = T\), since \(\alpha\vDash\beta\)

Both cases show that for an arbitrary truth assignment \(v\), it will be that \(\overline{v}(\alpha) = \overline{v}(\beta)\). Thus, \((\alpha\leftrightarrow\beta)\) will be true for any truth assignment, meaning it is a tautology.

”\(\Leftarrow\)”

Now for the backwards direction, we assume that \((\alpha\leftrightarrow\beta)\) is a tautology.

Let \(v\) be a truth assignment such that \(\overline{v}(\alpha)=T\), then since \(\overline{v}(\alpha\leftrightarrow\beta) = T\), we must have that \(\overline{v}(\beta) = T\). Thus, \(\alpha\vDash\beta\)

Now, let \(v\) be a truth assignment such that \(\overline{v}(\beta) =T\). By the same logic as the preceding sentences, it must be that \(\overline{v}(\alpha) = T\), so that \(\beta \vDash\alpha\).

Therefore, \(\alpha\) and \(\beta\) are tautologically equivalent.