We start by applying De Morgan's law:
[tex]\neg (p \wedge q) \vee p \equiv [(\neg p) \vee (\neg q)] \vee p.[/tex]
We now use the fact that the disjunction is commutative:
[tex][(\neg p) \vee (\neg q)] \vee p \equiv p \vee [(\neg p) \vee (\neg q)].[/tex]
We now use the fact that the disjunction is distributive:
[tex]p \vee [(\neg p) \vee (\neg q)] \equiv [p \vee (\neg p)] \vee (\neg q).[/tex]
Using the law of excluded middle, we have [tex]p \vee (\neg p) \equiv \textrm{TRUE}[/tex], so that:
[tex][p \vee (\neg p)] \vee (\neg q) \equiv \textrm{TRUE} \vee (\neg q).[/tex]
We finally use the fact that TRUE is the absorbing element of the disjunction:
[tex]\textrm{TRUE} \vee (\neg q) \equiv \textrm{TRUE}.[/tex]
