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Proof step 9 from proof min-a (previous step)

min(i, j) < j0, min(i, j) < j, min(i, j) < i ⊦ j0 = min(i, j), j < j0, i < j0

Heuristic weak cut applied rule cut formula

And closed the goal!

Rule argument

fma: i < j

Additional information

Used simplifier rules: Min-left: ⊦ j < i → min(j, i) = j;%Min-right: ⊦ ¬ j < i → min(j, i) = i;% ⊦ ¬ i < j → (i = j ↔ ¬ j < i); (ss)% ⊦ i < j → ¬ j < i; (s) Used formulas: i < j