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Proof step 3 from proof min-02 (previous step)

j0 < j, j0 < i, min(i, j) < j0, min(i, j) < j, min(i, j) < i, min(i, j) +1 < j, min(i, j) +1 < i ⊦

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 → ¬ j < i; (s) Used formulas: i < j