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Proof step 5 from proof mods-nls (previous step)

0 < i, i * abs(j) < j ⊦ j < 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
0 < i, i * j < j, 0 ≤ j ⊦

Rule argument

Additional information

Used simplifier rules: forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs% ⊦ i < j ↔ ¬ j ≤ i; (ss)% ⊦ ¬ j < 0 → abs(j) = j; (s)