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Proof step 4 from proof even-add (previous step)

2 * m0 = m + n ⊦ (∃ m0. 2 * m0 = m) ∧ (∃ m0. 2 * m0 = n), 2 * m1 ≠ m ∧ 2 * m2 ≠ n

User applied rule case distinction

And obtained the following new goals

Proof step 5:
2 * m0 = m + n ⊦ ∃ m0. 2 * m0 = m, 2 * m1 ≠ m ∧ 2 * m2 ≠ n
Proof step 14:
2 * m0 = m + n ⊦ ∃ m0. 2 * m0 = n, 2 * m1 ≠ m ∧ 2 * m2 ≠ n

Rule argument

pos: right 1

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