Determine whether {¬q∧(p→q)}→¬p is tautology 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 ?
 A: 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.
A: $\left<(\neg q\wedge(p\rightarrow q))\rightarrow\neg p\right>\Leftrightarrow$
$\left<1+(1+q)(1+p+pq)+(1+q)(1+p+pq)(1+p)\right>$, 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.
