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

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

User applied rule case distinction

And obtained the following new goals

Proof step 5:
⊦ ∃ m0. 2 * m0 = m + n, ∃ m0. 2 * m0 = m, 2 * m1 ≠ m ∧ (∃ m0. 2 * m0 = n)
Proof step 21:
2 * m0 = n ⊦ ∃ m0. 2 * m0 = m + n, 2 * m1 ≠ m ∧ (∃ m0. 2 * m0 = n)

Rule argument

pos: right 2

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