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Proof step 12 from proof bint-mod (previous step)

¬(~2147483648 - 1 ≤ ~ abs(j) ∧ j ≠ 0), i % j < ~2147483648, ~2147483648 ≤ j, j ≤ 2147483647, i % j < j, i % j < 0 ⊦ 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%sub-le: ⊦ i - j ≤ i0 ↔ i ≤ j + i0; from specification int-abs%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%minus-add: i < 0 ⊦ (* i) + j = j - ~ i; from specification int-abs%negle-07: ⊦ i ≤ ~ j ↔ i + j ≤ 0; from specification int-abs