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Proof step 6 from proof max-a (previous step)

n < n0, m < n0 ⊦ max(m, n) < n0, n0 = max(max(m, n), n0), n0 = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
n < max(m, n), n0 < max(m, n), m < n0, n < n0, m + 1 < max(m, n), m < max(m, n), n + 1 < max(m, n) ⊦ max(m, n) = 0, n0 = max(m, n), n0 = 0

Rule argument

Additional information

Used simplifier rules: Max-right: ⊦ ¬ m < n → max(m, n) = m;%succ: ⊦ m +1 = m + 1; from specification nat-basic2%trans-1: ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward) from specification nat-basic1%lf: ⊦ m < n ∧ n < n0 → m +1 < n0; (forward) from specification nat-basic1%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%neq-01: ⊦ m ≠ n → (¬ m < n ↔ n < m); from specification nat-basic1