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Proof step 2 from proof min-l-and-r (previous step)

min(m, n) < m, min(n, m) < n ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
⊦ min(n, m) = (n < m ⊃ n;m)
Proof step 4:
(m < n ⊃ m;n) < m, (n < m ⊃ n;m) < n ⊦

Rule argument

Additional information

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