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Proof step 2 from proof min-ls-right (previous step)

n1 < n0 ⊦ min(n0, n1) = n1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

Used simplifier rules: min: ⊦ min(n, m) = (n < m ⊃ n;m);% ⊦ m < n → ¬ n < m; (s)