Use Theorem 2.1.1 to verify the logical equivalence. Give a reason for each step. -(pv –q) v(-p^q) = ~p

Answer:The statement [tex]\lnot(p\lor\lnot q)\lor(\lnot p \land \lnot q)[/tex] is equivalent to [tex]\lnot p[/tex], [tex]\lnot(p\lor\lnot q)\lor(\lnot p \land \lnot q) \equiv \lnot p [/tex]Step-by-step explanation:We need to prove that the following statement [tex]\lnot(p\lor\lnot q)\lor(\lnot p \land \lnot q)[/tex] is equivalent to [tex]\lnot p[/tex] with the use of Theorem 2.1.1.So[tex]\lnot(p\lor\lnot q)\lor(\lnot p \land \lnot q) \equiv[/tex][tex]\equiv (\lnot p \land \lnot(\lnot q))\lor(\lnot p \land \lnot q)[/tex] by De Morgan's law.[tex]\equiv (\lnot p \land q)\lor(\lnot p \land \lnot q)[/tex] by the Double negative law[tex]\equiv \lnot p \land (q \lor \lnot q)[/tex] by the Distributive law[tex]\equiv \lnot p \land t[/tex] by the Negation law[tex]\equiv \lnot p [/tex] by Universal bound lawTherefore [tex]\lnot(p\lor\lnot q)\lor(\lnot p \land \lnot q) \equiv \lnot p [/tex]