Use logical Equivalence and rules of inference to prove the proposition If $\{w\Rightarrow x,(w\Rightarrow y)\Rightarrow(z\wedge x),\neg z\}$, then conclusion is $x$.
(Can you show what rules you are using to solve this problem?)
 A: Hint: 


*

*from the premiss $\neg z$ infer $\neg(z \land x)$

*from $\neg(z \land x)$ and the premiss $(w \to y) \to (z \land x)$ infer $\neg(w \to y)$.

*from $\neg(w \to y)$ infer $w$.

*from $w$ and the premiss $w \to x$ infer the desired conclusion $x$.
The rules you use for each step will depend on your particular implementation of a deduction system for propositional logic.
[Alternatively you could proceed by reductio. 


*

*suppose $\neg x$ for the sake of argument.

*from that and the premiss $w \to x$ infer $\neg w$.

*from $\neg w$ infer $w \to z$.

*from $w \to z$ and the premiss $(w \to y) \to (z \land x)$ infer $(z \land x)$ and hence $z$.

*that contradicts the premiss $\neg z$, so the initial assumption at line 1 is false.]
A: Looking at the shape of the premises, it looks like the third one can be used to simplify the second one.  So let's try that:
\begin{align}
& (w \Rightarrow y) \Rightarrow (z \land x) & \text{-- second premise}\\
\equiv & \;\;\;\;\;\text{"use third premise $\;\lnot z\;$"} \\
& (w \Rightarrow y) \Rightarrow (\text{false} \land x) \\
\equiv & \;\;\;\;\;\text{"simplify"} \\
& (w \Rightarrow y) \Rightarrow \text{false} \\
\equiv & \;\;\;\;\;\text{"write $\;p \Rightarrow q\;$ as $\;\lnot p \lor q\;$, twice"} \\
& \lnot (\lnot w \lor y) \lor \text{false} \\
\equiv & \;\;\;\;\;\text{"DeMorgan on left hand part; simplify"} \\
& w \land \lnot y \\
\end{align}
Now comparing this last line to our goal $\;x\;$, we see that the first premise immediately gets us there:
\begin{align}
& w \land \lnot y \\
\Rightarrow & \;\;\;\;\;\text{"weaken -- to leave out the irrelevant part"} \\
& w \\
\Rightarrow & \;\;\;\;\;\text{"using first premise $\;w \Rightarrow x\;$"} \\
& x \\
\end{align}
That completes the proof.
