How can this be proved to be a tautology using laws of logical equivalence?((x ∨ y) ∧ (x → z) ∧ (¬z)) → y

Accepted Solution

Answer:Step-by-step explanation:If we assume that [tex][(x \vee y) \wedge (x \rightarrow z) \wedge (\neg z)][/tex] is true, then:[tex](x \vee y)[/tex] is true[tex](x \rightarrow z)[/tex] is true[tex](\neg z)[/tex] is trueIf [tex](\neg z)[/tex] is true, then [tex]z[/tex] is false.[tex](x \rightarrow z) \equiv (\neg x \vee z)[/tex], since [tex](x \rightarrow z)[/tex] is true, then [tex](\neg x \vee z)[/tex] is trueIf [tex]z[/tex] is false and [tex](\neg x \vee z)[/tex] is true, then [tex]\neg x[/tex] is true.If [tex]\neg x[/tex] is true, then [tex]x[/tex] is false, as [tex](x \vee y)[/tex] is true and [tex]x[/tex] is false, then [tex]y[/tex] is true.Conclusion [tex]y[/tex] it's true.