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Proof step 18 from proof bint-mod (previous step)

¬(j ≠ 0 ∧ abs(j) ≤ 2147483647 + 1), 2147483647 < i % j, ~2147483648 ≤ j, j ≤ 2147483647, 0 < i % j, j < i % j ⊦ j = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: numlslsdel-01: i0 ≠ 0, i0 < i ⊦ (* i) < j → (* i0) < j; from specification int-basic1%abs-06: ⊦ 0 < abs(i) ↔ i ≠ 0; from specification int-abs%forward-pos: ⊦ i < j ∧ 0 < i → 0 < j ∧ 1 < j; (forward) from specification int-basic1%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs%lels-01: ⊦ ¬ i ≤ j ↔ j < i; from specification int-abs