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

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

Heuristic weak cut applied rule cut formula

And closed the goal!

Rule argument

fma: j < j0

Additional information

Used simplifier rules: Min-left: ⊦ j < i → min(j, i) = j;%Min-right: ⊦ ¬ j < i → min(j, i) = i;%Min-left: ⊦ j < i → min(j, i) = j;%lf-07: ⊦ ¬ i < i0 ∧ j < i0 → j < i; (forward) from specification int-basic1%lf-01: ⊦ i < j ∧ j < i0 → i +1 < i0; (forward) from specification int-basic1%trans: ⊦ i < j ∧ j < j0 → i < j0; (forward) from specification int-basic1 Used formulas: j < j0