Convert the following expression to CNF (Conjunctive Normal Form) format using the propositional laws:(p → q) → rIn your answer make sure that you (1) Show your steps (2) Do not skip any steps and (3) Justify each step with the name of the law being used.Part(b) [50 points].(i)(25 points) Prove that the Method of Affirming (Modus Ponens) is sound using truth tables.(ii)(25 points) Prove that the Method of Denying (ModusTollens) is sound using truth tables.

Answer:See steps belowStep-by-step explanation:a)[tex](p\rightarrow q)\rightarrow r\Leftrightarrow \neg(\neg p\vee q)\vee r[/tex] equivalence of (r implies s) with (not r or s)[tex] \neg(\neg p\vee q)\vee r\Leftrightarrow (\neg \neg p\wedge \neg q)\vee r[/tex] De Morgan's Law[tex] (\neg \neg p\wedge \neg q)\vee r\Leftrightarrow (p\wedge \neg q)\vee r[/tex] Double negation[tex](p\wedge \neg q)\vee r\Leftrightarrow (p\vee r)\wedge (\neg q\vee r)[/tex] Distributive LawThe last expression is in CNF.b)i)Modus Ponens states the following,If (p implies q) is true and p is true, then q is true.By watching the truth table of implication[tex]\left[\begin{array}{ccc}p&q&p\rightarrow q\\T&T&T\\T&F&F\\F&T&T\\F&F&T\end{array}\right][/tex]We can notice that the only row that satisfies  (p implies q) is true and p is trueis the first row, so q must be true.ii)Modus Tollens states that if (p implies q) is true and (not q) is true, then (not p) is true.By watching the following truth table[tex]\left[\begin{array}{ccccc}p&q&\neg p&\neg q&p\rightarrow q\\T&T&F&F&T\\T&F&F&T&F\\F&T&T&F&T\\F&F&T&T&T\end{array}\right][/tex]We can notice that the only row that satisfies (p implies q) is true and (not q) is true, is the fourth row, so (not p) must be true.