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Proof step 6 from proof min-less-second (previous step)

n1 < (n0 < min(n1, n2) ⊃ n0;min(n1, n2)) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
n1 < (n0 < min(n1, n2) ⊃ n0;min(n1, n2)) ⊦ n0 = 0

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ n < min(n, m); (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic