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Proof step 3 from proof two-succ-06 (previous step)

∀ m, m0. ¬(2 * m = 2 * n0 + m0 ∧ m < n0), 2 * m = 2 * n0 +1 + m0, m < n0 +1 ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
2 * m = (2 * n0 + 2) + m0, m ≤ n0, ∀ m1, m2. 2 * m1 ≠ 2 * n0 + m2 ∨ ¬ m1 < n0 ⊦

Rule argument

Additional information

Used simplifier rules: dist: ⊦ m0 * (m + n) = m0 * m + m0 * n;%mult-a: ⊦ m0 * m * n = (m0 * m) * n;%mult-c: ⊦ m * n = n * m;%lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat%s: ⊦ m < n + 1 ↔ ¬ n < m; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%succ: ⊦ m +1 = m + 1; from specification nat-basic2