Showing tautology without a truth table.

Question

Show that the conditional statement is a tautology without using a truth table.

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

My suggestion would be getting rid of the implication first, so

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

How should I continue hereafter?

Answer 1

$$ \begin{align}(p \wedge q) \to p &\equiv \neg(p \wedge q) \vee p \\&\equiv (\neg p \vee \neg q) \vee p \equiv \neg p \vee (\neg q \vee p) \\&\equiv (\neg p \vee p)\vee \neg q \equiv T \vee \neg q \equiv T \end{align}$$

Next step should be to use De Morgan's Law.

Answer 2

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

Answer 3

Well you are on the right track on getting rid of the implication.

Now you should see that for $\neg(p \wedge q) \vee p$

If $p$ is true and then we have $\neg(p \wedge q) \vee true$ which is true

Now if $p$ is false then we have $\neg(false \wedge q) \vee false$ which is $\neg(false) \vee false$ which is $true \vee false$ which is true.

Now you see that we don't even care about $q$ and that's why we don't need a truth table !