Propositional logic proofs: Prove X ∧ (Y v Z) = (X ∧ Y) v (X ∧ Z)

Question

The question asks me to Prove X ∧ (Y ∨ Z) = (X ∧ Y) ∨ (X ∧ Z). Can somebody walk me through this proof?

The axioms we are allowed to use are:

• ∨ Identity
• ∨ null
• ∨ commutative
• ∨ associative
• ∨ distributive
• implication
• ∨ DeMorgan
• ∨ idempotent
• self implication
• double negation

Answer 1

$$ \begin{align} X \land (Y \lor Z) & = (X \land (Y \lor Z))'' \\ & = (X' \lor (Y \lor Z)')' \\ & = (X' \lor (Y' \land Z'))' \\ & = ((X' \lor Y') \land (X' \lor Z'))' \\ & = (X' \lor Y')' \lor (X' \lor Z')' \\ & = (X'' \land Y'') \lor (X'' \land Z'') \\ & = (X \land Y) \lor (X \land Z) \end{align} $$

I used $\, '$ to denote negation (rather than $\lnot$) - easier on my fingers.

Can you tell which axiom from your list was used at each step? I believe I wrote the proof in such a way that only one axiom is used at each step (but perhaps the same axiom may be used more than once in the same step).