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Proof step 24 from proof le-abs-numerator (previous step)

0 ≤ abs(i) ∧ 0 < j → abs(i) / j * j ≤ abs(i), j ≠ 0, 0 < j, 0 < ~(abs(i) / j) + ~ abs(i), ~ abs(i) < 0 ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 25:
0 < j, abs(i) / j * j ≤ abs(i), abs(i) + abs(i) / j < 0 ⊦ j = 0, i = 0

Rule argument

Additional information

Used simplifier rules: forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs%abs-03: ⊦ 0 ≤ abs(i); from specification int-abs%abs-06: ⊦ 0 < abs(i) ↔ i ≠ 0; from specification int-abs%neg-05: ⊦ ~ i < 0 ↔ 0 < i; from specification int-abs%neg: ⊦ 0 < ~ i ↔ i < 0; 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%neg-11: ⊦ ~ i + ~ j = ~(i + j); from specification int-abs