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Proof step 2 from proof even-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 23:
(∃ m0. 2 * m0 = m) ∧ (∃ m0. 2 * m0 = n) ∨ ¬(∃ m0. 2 * m0 = m) ∧ ¬(∃ m0. 2 * m0 = n) ⊦ ∃ m0. 2 * m0 = m + n

Rule argument

pos: right 1

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