Determine whether {¬q∧(p→q)}→¬p is tautology

Question

Determine whether $\{¬q∧(p→q)\}→¬p$ is tautology .

this my solution : \begin{align} \{¬q∧(p→q)\}→¬p & ≡¬\{¬q∧(¬p∨q)\}∨¬p \\ &≡q∨(p∧¬q)∨¬p≡(q∨p)∧(¬q∨¬p) \\ &≡(q∨¬q)∧(p∨¬p) ≡T∧ T \\ &≡T \end{align} Is it correct ?

Answer 1

You can list all posible values for $p$ and $q$ and see what it does. \begin{array}{|c|c|c|c|c|c|c|} \hline p& q & \neg q & p \to q & \neg q \wedge (p \to q) & \neg p & \neg q \wedge (p \to q) \to \neg p\\ \hline 0& 0& 1&1&1&1&1\\ \hline 0& 1& 0&1&0&1&1\\ \hline 1& 0& 1&0&0&0&1\\ \hline 1& 1& 0&1&0&0&1\\\hline \end{array}

That means, no matter of truth value of $p$ or $q$, the stetement $\neg q \wedge (p \to q) \to \neg p$ is always true, hence its tautology.

Answer 2

$\left<(\neg q\wedge(p\rightarrow q))\rightarrow\neg p\right>\Leftrightarrow$ $\left<1+(1+q)(1+p+pq)+(1+q)(1+p+pq)(1+p)\right>$, since

$\begin{cases} \neg p\Leftrightarrow (1+p)\\ (p\rightarrow q)\Leftrightarrow(1+p+pq) \end{cases}$

where $+$ is XOR and $\cdot$ is AND. The expression can be simplified:

$1+(1+p+pq)+(q+qp+qpq)+(1+p+pq+q+qp+qpq)+(p+pp+pqp+qp+qpp+qpqp)=$$1+1+p+pq+q+qp+qpq+1+p+pq+q+qp+qpq+p+pp+pqp+qp+qpp+qpqp$= $=1+1+1+p+p+p+p+q+q+pq+pq+pq+pq+pq+pq+pq+pq+pq+pq=1$, because of associativity, commutativity, idempotency and the annihilation law $x+x=0$.

That is, the expression is a tautology.