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Proof step 7 from proof exdivmod (previous step)

0 < j, 0 ≤ j0, 0 ≤ i0, 0 ≤ j, j0 < j ⊦ ∃ i1, j1. i0 * j + j0 = i1 * j + j1 ∧ abs(j1) < j

Heuristic quantifier closing applied rule exists right

And closed the goal!

Rule argument

pos: right 1 terms: i0, j0

Additional information

Used simplifier rules: absle: ⊦ 0 ≤ i → abs(i) = i; from specification int-abs Used terms: i0, j0