The statement (p wedge (p rightarrow q) wedge | vedclass

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(A) Let the statement be $S = (p \wedge (p$ $\rightarrow q) \wedge (q$ $\rightarrow r))$ $\rightarrow r$. Using the implication law $a \rightarrow b \

Vedclass · Accepted Answer

a tautology

Vedclass · Answer

(A) Let the statement be $S = (p \wedge (p$ $\rightarrow q) \wedge (q$ $\rightarrow r))$ $\rightarrow r$. Using the implication law $a \rightarrow b \equiv \sim a \vee b$,we have: $S \equiv \sim (p \wedge (p$ $\rightarrow q) \wedge (q$ $\rightarrow r)) \vee r$ Using De Morgan's law $\sim (A \wedge B \wedge C) \equiv \sim A \vee \sim B \vee \sim C$: $S \equiv \sim p \vee \sim (p$ $\rightarrow q) \vee \sim (q$ $\rightarrow r) \vee r$ Since $\sim (a \rightarrow b) \equiv a \wedge \sim b$: $S \equiv \sim p \vee (p \wedge \sim q) \vee (q \wedge \sim r) \vee r$ Using distributive law $\sim p \vee (p \wedge \sim q) \equiv (\sim p \vee p) \wedge (\sim p \vee \sim q) \equiv T \wedge (\sim p \vee \sim q) \equiv \sim p \vee \sim q$: $S \equiv (\sim p \vee \sim q) \vee (q \wedge \sim r) \vee r$ Using associative and distributive laws: $S \equiv \sim p \vee \sim q \vee (q \vee r) \wedge (\sim r \vee r)$ $S \equiv \sim p \vee \sim q \vee (q \vee r) \wedge T$ $S \equiv \sim p \vee (\sim q \vee q) \vee r$ $S \equiv \sim p \vee T \vee r \equiv T$ Since the result is always true,the statement is a tautology.

The statement $(p \wedge (p$ $\rightarrow q) \wedge (q$ $\rightarrow r))$ $\rightarrow r$ is :

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