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Proof step 4 from proof two-succ-06 (previous step)

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

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 3 terms: m, m0 + 2

Additional information

Used simplifier rules: a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat%anti: ⊦ m ≤ n → (n ≤ m ↔ m = n); from specification nat%eq: ⊦ m = m + n ↔ n = 0; from specification nat-basic2%add-num-03: n < n0 ⊦ (* n) ≠ m0 + (* n0); from specification nat Used terms: m, m0 + 2