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Proof step 1 from proof min-abs-zero (conclusion)

⊦ min(0, abs(j)) = 0

Heuristic weak cut applied rule cut formula

And closed the goal!

Rule argument

fma: 0 < abs(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);%abs-07: ⊦ abs(i) = 0 ↔ i = 0; from specification int-abs%abs-06: ⊦ 0 < abs(i) ↔ i ≠ 0; from specification int-abs Used formulas: 0 < abs(j)