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For all x,y: (x ∨ y) ∧ x = x

This is my attempt, I am just getting back to the question:

(x ∨ y) ∧ x

= x ∧ (x ∨ y)

= (x ∧ x) ∨ (x ∧ y)

= x ∨ (x ∧ y)

= (x ∨ x) ∧ (x ∨ y)

= x ∧ (x ∨ y)

So where am I going wrong here?

weak_at_math
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    It is the absorption rule that must be applied. Only associativity, distributivity and/or commutativity will make run in loops only. – drhab Nov 18 '20 at 12:21

1 Answers1

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You are not "going wrong"; $x \land (x \lor y)$ follows. Its truth table has the same result as $x$.From line 4 above, we have $$\neg(\neg x) \lor (x \land y)$$

which is $\neg x \implies (x \land y)$. Applying the absorption rule as in the comment gives $$\neg x \implies (\neg x \land x \land y) $$

This is $x \lor F$ or $x$, as required.

DavidW
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