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Proof step 15 from proof odd-add (previous step)

¬ even(i) ∧ even(j) ∨ even(i) ∧ ¬ even(j) ⊦ (¬ even(i) ∨ ¬ even(j)) ∧ (even(i) ∨ even(j))

User applied rule case distinction

And closed the goal!

Rule argument

pos: left 1

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