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Proof step 3 from proof min-a (previous step)

m < min(n, n0) ⊦ m = min(min(m, n), n0), min(n, n0) = 0

Heuristic weak cut applied rule cut formula

And closed the goal!

Rule argument

fma: n < n0

Additional information

Used simplifier rules: Min-left: ⊦ m < n → min(m, n) = m;%Min-right: ⊦ ¬ m < n → min(m, n) = n;%Min-left: ⊦ m < n → min(m, n) = m;%lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat%lf: ⊦ m < n ∧ n < n0 → m +1 < n0; (forward) from specification nat-basic1%trans-1: ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward) from specification nat-basic1 Used formulas: n < n0