Q.Show, using a truth table, that $(p \wedge q) \rightarrow p$ is a tautology.

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

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 $F \rightarrow F = T$. …

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 $\sim p \rightarrow (p \rightarrow \sim q)$.

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

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).

…