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

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

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;% ⊦ m < n ↔ ¬ n ≤ m; (ss)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat% ⊦ m < n → n ≠ 0; (s)%trans-01: ⊦ n < n0 ∧ n0 < n1 → n +1 < n1; (forward) from specification nat-basic%trans: ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward) from specification nat-basic Used formulas: n < n0