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Having trouble proving this. I know how to prove the first conjunct of the conclusion, but not the second one. Picture shown is the attempt proof of the second conjunct (rules haven't been added yet). I have a feeling that we need to use ∀Intro for the second conjunct of conclusion as well as ∀z (Cube(z) → (z = x v z = y)) part of premise.

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I have a feeling that we need to use ∀Intro for the second conjunct of conclusion as well as ∀z (Cube(z) → (z = x v z = y)) part of premise.

Yes, you need to apply Universal Elimination thrice, once to each arbitrary term.

You can then derive three disjunctions through Conditional Eliminations.

Up next is a nested disjunction elimination, where you use equality eliminations and disjunction introduction; you are aiming to obtain a=b v a=c v b=c at the end of each sub-proof.

Recall, a=d and b=d entails a=b by Equality Elimination.

And so on.

See also this

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