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Proof step 25 from proof le-abs-numerator (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 26:
0 < j, abs(i) + abs(i) / j < 0, abs(i) / j * j ≤ abs(i) ⊦ i = 0, j = 0

Rule argument

Additional information

Used simplifier rules: forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs% ⊦ 0 ≤ abs(i); (s)% ⊦ 0 < abs(i) ↔ i ≠ 0; (s)% ⊦ ~ i < 0 ↔ 0 < i; (s)% ⊦ 0 < ~ i ↔ i < 0; (s)%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2% ⊦ ~ i + ~ j = ~(i + j); (s)