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Proof step 10 from proof min-a (previous step)

n < n0, n ≤ m ⊦ n = min(min(m, n), n0), min(m, n) < n0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
n < n0, n < min(m, n), n ≤ m, n0 ≤ min(m, n) ⊦ n = n0

Rule argument

Additional information

Used simplifier rules: Min-right: ⊦ ¬ m < n → min(m, n) = n;% ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat% ⊦ m < n ↔ ¬ n ≤ m; (ss)