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

m0 + m1 < n ⊦ min(n, m0) = m0

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 + m0 < n → ¬ n + n0 < m; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m +1 = m + 1; (s)%trans: ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward) from specification nat-basic%lf: ⊦ m < n ∧ n < n0 → m +1 < n0; (forward) from specification nat-basic% ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic