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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
n < n0, n ≤ m ⊦ n = min(min(m, n), n0), n0 = 0

Rule argument

Additional information

Used simplifier rules: Min-left: ⊦ m < n → min(m, n) = m;%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1