Part (a)

Problem

Prove or disprove: If either \(\Sigma\vDash\alpha\) or \(\Sigma\vDash\beta\), then \(\Sigma\vDash(\alpha\lor\beta)\)

Solution

I will prove the claim is true. Suppose \(v\) is a truth assignment which satisfies \(\Sigma\), then since \(\Sigma\vDash\alpha\) or \(\Sigma\vDash\beta\), we will have that \(\overline{v}(\alpha) = T\) or \(\overline{v}(\beta) = T\). This implies that \(\overline{v}(\alpha\lor\beta) = T\). Therefore, \(\Sigma\vDash (\alpha\lor\beta)\).

Part (b)

Problem

Prove or disprove: If \(\Sigma\vDash(\alpha\lor\beta)\), then either \(\Sigma\vDash\alpha\) or \(\Sigma\vDash\beta\).

Solution

I will disprove the claim. To do this, I provide a counterexample.

Let \(\alpha = (A\leftrightarrow B)\), let \(\beta = (B\leftrightarrow C)\), and let \(\Sigma = ((A\land B)\land(\lnot C))\lor((\lnot A)\land(B\land C))\)

I show that \(\Sigma\vDash(\alpha\lor\beta)\) with a truth table \begin{array}{|c|c|} A & B & C&(A\land B) & (B\land C)&(\lnot A)& (\lnot C) & \Sigma&\alpha = (A\leftrightarrow B)& \beta = (B\leftrightarrow C)& (\alpha\lor\beta)
\hline T & T & T & T & T & F & F & F & T & T & T
T & F & F & F & F & F & T & F & F & T & T\ F & T & F & F & F & T & T & F & F & F & F\ F & F & T & F & F & T & F & F & T & F & T\ T & T & F & T & F & F & T & T & T & F & T
T & F & T & F & F & F & F & F & F & F & F\ F & T & T & F & T & T & F & T & F & T & T\ F & F & F & F & F & T & T & F & T & T & T\ \end{array}

The truth table reveals that \(\Sigma\vDash(\alpha\lor\beta)\), however it also shows that \(\Sigma\not\vDash\alpha\) and \(\Sigma\not\vDash\beta\)

Thus, the counterexample disproves the claim.