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Proof step 27 from proof le-abs-numerator (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 28:
0 < j, abs(i) / j * j ≤ abs(i), abs(i) + abs(i) / j < 0, 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%lels-01: ⊦ ¬ i ≤ j ↔ j < i; from specification int-abs%abs-03: ⊦ 0 ≤ abs(i); from specification int-abs