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Proof step 14 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 2

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