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Proof step 5 from proof odd-add (previous step)

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

User applied rule case distinction

And obtained the following new goals

Proof step 6:
2 * m1 = m ⊦ ∃ m0. 2 * m0 = m, ∃ m0. 2 * m0 = m + n
Proof step 8:
⊦ ∃ m0. 2 * m0 = m, ∃ m0. 2 * m0 = m + n, ∃ m0. 2 * m0 = n

Rule argument

pos: right 3

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