Q.Examine whether the statement pattern $\sim(p \wedge q) \vee p$ is a tautology.

Build the truth table; the pattern is false in every row.

The last column is F in every row, so the statement pattern $(p\wedge q)\wedge(\sim p\vee\sim q)$ is a contradiction.

We tabulate the truth values of $(p \wedge \sim q) \rightarrow (\sim p \wedge \sim q)$ over all four combinations of $p$ and $q$. The final column is $T, F, T, T$ — neither all true nor all false — so the pattern is a contingency.

We evaluate step by step. Recall $A \rightarrow B$ is false only when $A$ is true and $B$ is false.

Row-by-row justification:

• $p=T, q=T$: $p \wedge \sim q = T \wedge F = F$, so $F \rightarrow F = T$.
• $p=T, q=F$: $p \wedge \sim q = T \wedge T = T$ but $\sim p \wedge \sim q = F \wedge T = F$, so $T \rightarrow F = F$.
• $p=F, q=T$: antecedent $= F$, so the implication is $T$.
• $p=F, q=F$: antecedent $= F$, so the implication is $T$.

The truth values in the last column are $T, F, T, T$ — some true and some false.

Construct the truth table of $\sim p \rightarrow (p \rightarrow \sim q)$. All four rows evaluate to $T$, hence the pattern is a tautology.

A statement pattern is a tautology if its final column is $T$ throughout, a contradiction if it is $F$ throughout, and a contingency otherwise. We evaluate the inner implication $p \rightarrow \sim q$ first, then the whole pattern $\sim p \rightarrow (p \rightarrow \sim q)$.

Row-by-row justification of the last column: whenever $\sim p$ is $F$ (rows 1–2), the implication $\sim p \rightarrow (\dots)$ is automatically $T$ (an implication with a false antecedent is true). In rows 3–4, $\sim p$ is $T$ and the consequent $p \rightarrow \sim q$ is also $T$ (because $p$ is $F$ there), so $T \rightarrow T = T$.

Since the final column contains $T$ in every row, the statement pattern is true regardless of the truth values of $p$ and $q$.

Build the truth table for all 8 combinations of $p,q,r$.

The final column has both T and F entries (neither all T nor all F), so the statement pattern is a contingency.

Build the truth table for $[(p\to q)\land q]\to p$; it has both T and F rows, so it is a contingency.

The final column contains three T's and one F (row $p=F,q=T$: $(p\to q)\land q = T\land T=T$, but $p=F$, so the conditional $T\to F$ is False).

Since the column is neither all T (not a tautology) nor all F (not a contradiction), the statement pattern is a contingency.