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

⊦ ¬(∃ m0. 2 * m0 = m * 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)
Proof step 9:
∃ m0. 2 * m0 = m * n ⊦ ∃ m0. 2 * m0 = n, ∃ m0. 2 * m0 = m

Rule argument

pos: right 1

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