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Proof step 12 from proof min-a (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 13:
n0 ≤ m, n0 ≤ n ⊦ n0 = min(min(m, n), n0)

Rule argument

Additional information

Used simplifier rules: Min-right: ⊦ ¬ m < n → min(m, n) = n;% ⊦ m < n ↔ ¬ n ≤ m; (ss)