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Proof step 14 from proof min-a (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
min(m, n) < n0, min(m, n) < m, min(m, n) < n, n0 ≤ m, n0 ≤ n ⊦ n0 = min(m, n)

Rule argument

Additional information

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