About the logic associative and commutative laws, does this apply?

Question

Is this valid?

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

And if it is, would this equivalency be valid?

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

Answer 1

No , it doesn't hold:

Take p=False, q=False, r=True

Then (False $ ∧ $ (False ∨ True)) ≡ (False ∧ True)≡ False

But, ((False $ ∧ $ False) ∨ True) ≡ ( False$ ∨ $ True) ≡ True

Therefore, they do not have the same truth value under every truth assignment, so no they're not logically equivalent. There is actually a famous identity here, called the distributivity law which is the actual logical equivalence to that Well-Formed Formula. I'll let you think about what a distributivity law should look like here!

Answer 2

The logical connectives $\vee$ and $\wedge$ distribute over each other: $$\begin{align*} x\vee(y\wedge z) &\equiv (x\vee y)\wedge (x\vee z) x\wedge(y\vee z) &\equiv (x\wedge y)\vee (x\wedge z) \end{align*}$$ To see this: for $x\vee (y\wedge z)$ to be true you need either $x$ true (in which case both $x\vee y$ and $x\vee z$ are true, so the right hand side is true); or for $y$ and $z$ to both be true, in which case both $x\vee y$ and $x\vee z$ are true. Conversely, if both $x\vee y$ and $x\vee z$ are true, then ither $x$ is true, or else both $y$ and $z$ are true, giving the lef thand side. The second equivalence can be verified similarly.

They do not satisfy the equivalence you write; for instance, $x\wedge(y\vee z)$ will be false when $x$ is false, even if $y$ and $z$ are true; but if $x$ is false and $y$ Is true, then $(x\wedge y)\vee z$ will be true.

What $\vee$ and $\wedge$ do satisfy are two "absorption" laws: $$x\wedge(x\vee y)\equiv x\qquad\text{and}\qquad x\vee(x\wedge y)\equiv x.$$ Indeed, $x\wedge(x\vee y)$ is true when $x$ is true; and false when $x$ is false, regardless of the truth value of $y$. And $x\vee(x\wedge y)$ will be false if $x$ is false, and true if $x$ is true, regardless of $y$.