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Proof step 6 from proof ex-two (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: two-succ-05: ⊦ 2 * m ≠ 2 * n + 1;%mult-a: ⊦ m0 * m * n = (m0 * m) * n;%mult-c: ⊦ m * n = n * m;%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2