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Use Boolean algebras to prove the following: $x ∨ ((x ∨ y) ∧ (x ∨ y)) = x ∨ ((x ∨ y) ∧ x) ∨ ((x ∨ y) ∧ y))$

I'm not sure but I'm starting with right-hand-side because I think it might be a little bit easier

$x ∧ x = x$ idempotent

$x ∨ ((x ∧ x) ∨ (x ∨ y)) ∨ ((x ∧ y) ∨ (y ∧ y))$ Substitution

$x ∨ (x ∨ (x ∧ y)) ∨ ((x ∧ y) ∨ y) $

I stuck here

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HINT

Use Idempotence: $\phi \land \phi = \phi$

And Absorption: $(\phi \lor \psi) \land \phi = \phi$

So:

$x \lor ((x \lor y) \land x) \lor ((x \lor y) \land y))=$ (Absorption x2)

$x\lor (x \lor y)=$ (Idempotence)

$x \lor ((x \lor y)\land (x \lor y))$

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  • $\begingroup$ Thank you very much! I got it now. $\endgroup$ – Haley_Q Apr 6 '17 at 2:49
  • $\begingroup$ @Haley_Q You're welcome! $\endgroup$ – Bram28 Apr 6 '17 at 2:50

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