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Proof step 8 from proof odd-add (previous step)

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

User applied rule case distinction

And obtained the following new goals

Proof step 9:
⊦ ∃ m0. 2 * m0 = m, ∃ m0. 2 * m0 = n, ∃ m0. 2 * m0 = m + n
Proof step 22:
⊦ 2 * m1 ≠ n, ∃ m0. 2 * m0 = n, ∃ m0. 2 * m0 = m + n

Rule argument

pos: right 3

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