What is enough for $\forall$ expression proofs?

Question

If a question is asking to prove that $(∀R)(∀S)((R ⊕ S) ⊕ R = S)$. Then, is it enough to prove using some propositional logic statements (using commutative, associative properties, etc..) to show how simplifying LHS equals the RHS, or do I have to prove by cases for all possible $4$ cases ($x ∈ R$ and $x ∈ S$, $x ∈ R$ and $x ∉ S$ , $x ∉ R$ and $x ∈ S$, $x ∉ R$ and $x ∉ S$) ?

Answer 1

Unless otherwise specified, an equation such as, say $$\lnot(R\land S)=(\lnot R)\lor(\lnot S)$$ is implicitly true for all values of $R$ and $S$ and is called an identity.

You can prove it by checking with all possible combinations of the logical variables and you obtain a theorem which is forever true.

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Now if we replace $R$ by $\lnot R$ and $S$ by $\lnot S$, we obtain another identity (as the first theorem is always true):

$$\lnot((\lnot R)\land(\lnot S))=R\lor S$$

and taking the negation of both members,

$$(\lnot R)\land(\lnot S)=\lnot(R\lor S),$$

another interesting identity, obtained without checking cases one by one.