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Proof step 3 from proof even-add (previous step)

(i + j) % 2 = 0 ⊦ i % 2 ≠ 0 ∧ j % 2 ≠ 0, i % 2 = 0 ∧ j % 2 = 0

User applied rule case distinction

And obtained the following new goals

Proof step 4:
i % 2 = 0, (i + j) % 2 = 0 ⊦ i % 2 = 0 ∧ j % 2 = 0
Proof step 13:
j % 2 = 0, (i + j) % 2 = 0 ⊦ i % 2 = 0 ∧ j % 2 = 0

Rule argument

pos: right 1

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