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Proof step 11 from proof even-mult (previous step)

2 * m0 = m * n, ∃ m0. m = m0 + m0 + 1 ⊦ ∃ m0. 2 * m0 = m, ∃ m0. 2 * m0 = n

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
2 * m0 = n * 2 * m1 + n ⊦ ∃ m0. 2 * m0 = n

Rule argument

Additional information

Used simplifier rules: ⊦ 2 * m ≠ 2 * n + 1; (s)%mult-c: ⊦ m * n = n * m; from specification nat-mult%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult% ⊦ m * (n + 1) = m * n + m; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + m = 2 * m; (s)