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Proof step 6 from proof odd-add (previous step)

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

User applied rule case distinction

And obtained the following new goals

Rule argument

pos: right 2

Additional information

Used simplifier rules: coef-04: ⊦ m * n + m0 * n = m1 * n ↔ ¬ ¬(m + m0 = m1 ∨ n = 0); from specification nat-mult%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-basic2%c: ⊦ m + n = n + m; from specification nat-basic2