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Proof step 12 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 obtained the following new goals

Proof step 13:
(¬ even(i) ∨ ¬ even(j)) ∧ (even(i) ∨ even(j)) ⊦ ¬ even(i) ∧ even(j) ∨ even(i) ∧ ¬ even(j)
Proof step 15:
¬ even(i) ∧ even(j) ∨ even(i) ∧ ¬ even(j) ⊦ (¬ even(i) ∨ ¬ even(j)) ∧ (even(i) ∨ even(j))

Rule argument

pos: right 1

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