Why $p \lor (\lnot p \land q)$ is not equal to $(p \lor \lnot p) \land q$ ? I had trying making Truth Table but, still can't figure it out?

Question

Why $p \lor (\lnot p \land q)$ is not equal to $(p \lor \lnot p) \land q$ ? I had trying making Truth Table but, still can't figure it out ?

Answer 1

If $p$ is true then the first is automatically true. If $q$ is false then the second is automatically false.

I have no idea what you mean by $\sim$ but it doesn't even matter.

Answer 2

The term $(p\vee\neg p)\vee q$ is equivalent to $q$ since term in parenthesis is logically true.

The term $p\vee (\neg p\wedge q)$ has the truth value $1$ if $p=1$ independent of the value of $q$.

Answer 3

$(p\vee\sim p)\wedge q=1\wedge q=q\because p\vee\sim p=1$

$p∨(\sim p∧q)=p\vee q$ by the Absorption law.

Try $p=1,q=0$.

Answer 4

The statement $p\vee\neg p$ is a true statement in Boolean algebra.

Then:

$$p\vee(\neg p\wedge q)=(p\vee\neg p)\wedge(p\vee q)=p\vee q$$by distribution and:$$(p\vee\neg p)\wedge q=q$$

If you are approaching with truth tables then you encounter the difference for $q=0$ and $p=1$.