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Proof step 2 from proof odd-add (previous step)

⊦ ¬(∃ m0. 2 * m0 = m + n) ↔ ¬(∃ m0. 2 * m0 = m) ∧ (∃ m0. 2 * m0 = n) ∨ (∃ m0. 2 * m0 = m) ∧ ¬(∃ m0. 2 * m0 = n)

User applied rule case distinction

And obtained the following new goals

Proof step 3:
⊦ ∃ m0. 2 * m0 = m + n, ¬(∃ m0. 2 * m0 = m) ∧ (∃ m0. 2 * m0 = n), (∃ m0. 2 * m0 = m) ∧ ¬(∃ m0. 2 * m0 = n)
Proof step 24:
∃ m0. 2 * m0 = m + n, ¬(∃ m0. 2 * m0 = m) ∧ (∃ m0. 2 * m0 = n) ∨ (∃ m0. 2 * m0 = m) ∧ ¬(∃ m0. 2 * m0 = n) ⊦

Rule argument

pos: right 1

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