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

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

User applied rule case distinction

And obtained the following new goals

Proof step 5:
⊦ 2 * m0 ≠ m, ∃ m0. 2 * m0 = m + n, (∃ m0. 2 * m0 = m) ∧ 2 * m1 ≠ n
Proof step 8:
⊦ ∃ m0. 2 * m0 = n, ∃ m0. 2 * m0 = m + n, (∃ m0. 2 * m0 = m) ∧ 2 * m1 ≠ n

Rule argument

pos: right 2

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