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Proof step 25 from proof odd-add (previous step)

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

User applied rule case distinction

And obtained the following new goals

Proof step 26:
¬(∃ m0. 2 * m0 = m) ∧ 2 * m1 = n, 2 * m0 = m + n ⊦
Proof step 33:
2 * m2 = m ∧ ¬(∃ m0. 2 * m0 = n), 2 * m0 = m + n ⊦

Rule argument

pos: left 2

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