Question

Which of the following propositions is tautology?

• A. (p ∨ q) → q
• B. p ∨ (q → p)
• C. p ∨ (p → q)
• D. Both (b) & (c)

Answer 1

Answer: (C)

(c)

Answer 2

To determine which proposition is a tautology, we analyze each one:

1. $(p \vee q) \rightarrow q$: This is not a tautology. For example, if $p$ is True and $q$ is False, the expression becomes $(T \vee F) \rightarrow F$, which simplifies to $T \rightarrow F$, which is False.

2. $p \vee (q \rightarrow p)$: This is not a tautology. If $p$ is False and $q$ is True, the expression becomes $F \vee (T \rightarrow F)$, which simplifies to $F \vee F$, which is False.

3. $p \vee (p \rightarrow q)$: This is a tautology. We can use logical equivalences: $p \vee (p \rightarrow q)$ $\equiv p \vee (\neg p \vee q)$ (Implication Law) $\equiv (p \vee \neg p) \vee q$ (Associative Law) $\equiv T \vee q$ (Law of Excluded Middle) $\equiv T$ (Domination Law) Since the expression is always True, it is a tautology.

4. Both (b) & (c): Since proposition (b) is not a tautology, this option is incorrect.

Therefore, only proposition (c) is a tautology.