Q.Construct the truth table for the statement pattern $(p \wedge q) \rightarrow \sim p$ and hence state whether it is a tautology, a contradiction or a contingency.

Add columns for $p \wedge q$, for $\sim p$, and finally for the conditional, then read off the last column.

The conditional is false only in the single row where $p \wedge q$ is true while $\sim p$ is false.

With two statements there are $2^2 = 4$ rows. Build the intermediate columns $p \wedge q$ and $\sim p$, then the conditional.

Reading the rows. Row 1: $p \wedge q = T$ and $\sim p = F$, so $T \rightarrow F = F$. Rows 2-4: the antecedent $p \wedge q$ is $F$, and a conditional with a false antecedent is $T$.

The final column is $F, T, T, T$ — it contains both $T$ and $F$, so the pattern is a contingency.

Verification. The only way $(p \wedge q) \rightarrow \sim p$ can be false is $p \wedge q$ true (so $p = T$ and $q = T$) with $\sim p$ false (so $p = T$) — exactly row 1, and no other row. One false row rules out a tautology, and three true rows rule out a contradiction, confirming a contingency.

Students often forget that a conditional is true whenever its antecedent is false, and wrongly mark rows 2-4 as false. Always evaluate p -> q as false only in the single antecedent-true, consequent-false case.