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**kompik** I think that your solution is ok. You may notice that the last line is in fact $\displaystyle p\Rightarrow q$.

You can also use truth table to verify that your answer is correct.

For proposition which have the form of implication there is a shorter way:

Clearly $\displaystyle p\Rightarrow q$ if false only in one case: p=1, q=0.

Let us try to the same thing with the first expression

$\displaystyle [p\land \neg(p\Rightarrow q)] \Rightarrow q$.

This implication is false only if [tex][p\land \neg(p\Rightarrow q)]=1[/math and q=0.