Can you see if i made a mistake in my check for tautology? I need to prove that the following is a tautology:
[(P -> Q) ∧ ¬ Q ] -> ¬P

so my solution was:
[(¬P V Q) ∧ ¬Q] -> ¬P]
[¬((¬P V Q) ∧ ¬Q)] V ¬P]
[(¬¬P ∧ ¬Q V ¬¬ Q) V ¬P]
[(P ∧ ¬Q V Q) V ¬P]

The way i see the final line is that: ¬Q V Q will always hold true and if P is not true, then (P ∧ ¬Q V Q) is not true. But if P is not satisfied, then ¬P is satisfied so it's always true. Is my logic correct?
 A: $$\begin{array}{c|c|c|c|c|c|c}
P &     Q&  \lnot P &P\rightarrow Q&  \lnot Q  & ( P\rightarrow Q)  \wedge \lnot Q  & [(P\rightarrow Q) \wedge \lnot Q ]\rightarrow \lnot P \\
\hline
T  &    T&  F   & T  &    F  &      F   &                  T\\
T   &   F&  F &   F  &    T  &      F   &                  T\\
F    &  T&  F &   T  &    F  &      F   &                  T\\
F     & F&  T &   T  &    T  &      T   &                  T
\end{array}$$
So in all cases you got T = true so it is a tautology !
Look at the last column it contains only T's.
A: Might be easier to do:
$[(P \rightarrow Q) \wedge \lnot Q] \rightarrow \lnot P \\
[(\lnot P \vee Q) \wedge \lnot Q] \rightarrow \lnot P \\
[(\lnot P \wedge \lnot Q) \vee (Q \wedge \lnot Q)] \rightarrow \lnot P \\
(\lnot P \wedge \lnot Q) \rightarrow \lnot P \\
\lnot (\lnot P \wedge \lnot Q) \vee \lnot P \\
(P \vee Q) \vee \lnot P \\
P \vee\lnot P \vee Q \\
T \vee Q \\
T
$
A: Due to the simple structure of the formula :

$[(P \rightarrow Q) \land \lnot Q ] \rightarrow \lnot P$

you can try with a "menmonic" truth-table :

Case 1) : if $P$ is False, then $\lnot P$ is True, so in this case, by property of truth-functional implication, the formula is True
Case 2) : $P$ is True
Case 2a) : if $Q$ is True, then $\lnot Q$ is False, so the LHS conjunction is False and the complete formula is $False \rightarrow False$, so is again True
Case 2b) : if $Q$ is False, then (because $P$ is True) $(P \rightarrow Q)$ is False, so that the LHS conjunction is False. Again, the complete formula is $False \rightarrow False$, i.e. True

All cases give us True, so the formula is a tautology.
