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Proof step 2 from proof min-02 (previous step)

min(i, j) < j, min(i, j) < i, j0 < j, j0 < i, min(i, j) < j0, 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-a: ⊦ min(i, min(j, j0)) = min(min(i, j), j0);%min-c: ⊦ min(i, j) = min(j, i);%Min-right: ⊦ ¬ j < i → min(j, i) = i;%min-a: ⊦ min(i, min(j, j0)) = min(min(i, j), j0);%min-c: ⊦ min(i, j) = min(j, i);%irref: ⊦ ¬ i < i; from specification int-basic1 Used formulas: i < j