Can I use De Morgan's law in the third step as shown below to solve this problem?

Question

$(p \rightarrow q) \wedge (\neg p \rightarrow q)$

$\equiv(p \rightarrow q) \wedge (\neg p \rightarrow q)$

$\equiv(\neg p \vee q) \wedge (p \vee q)$

$\equiv \neg(\neg (\neg p \vee q) \vee \neg(p \vee q))$

$\equiv \neg((p \vee \neg q) \vee (\neg p \vee \neg q))$

$\equiv \neg((p \vee \neg p) \vee (\neg q \vee \neg q))$

$\equiv \neg(T \vee \neg q)$

$\equiv F \vee q$

$\equiv q$

Answer 1

Yes, the step is valid. I don't know if you are told to solve in a certain way, but there are better ways to solve this problem.

One would be a truth table: since there are only two variables, you can write out the result of the expression for each pair $(p,q)$, namely $(T,T), (T,F), (F,T), (F,F)$ and realize that it is equivalent to $q$. It is also not difficult to use your logic (no pun intended) to see that this expression is equivalent to $q$.

Answer 2

here is another way to do this.

$ (p \to q) \wedge (\neg p \to q) \equiv(q\vee\neg p)\wedge (q \vee p) \equiv q \vee (\neg p\wedge p) \equiv q\vee F \equiv q$