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

n < n0, n ≤ m ⊦ min(m, n) < n0, n = min(min(m, n), n0), n0 = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
n < min(m, n), n0 ≤ min(m, n), n ≤ m, n < n0 ⊦ n = n0, n0 = 0, min(m, n) = 0

Rule argument

Additional information

Used simplifier rules: Min-right: ⊦ ¬ m < n → min(m, n) = n;%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat%lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat