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Proof step 22 from proof odd-add (previous step)

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

User applied rule case distinction

And obtained the following new goals

Rule argument

pos: right 2

Additional information

Used simplifier rules: ⊦ m * n + m0 * n = m1 * n ↔ m + m0 = m1 ∨ n = 0; (s)%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%mult-c: ⊦ m * n = n * m; from specification nat-mult%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic