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Proof step 11 from proof div-abs-02 (previous step)

¬ j < 0, i * i0 = j ⊦ ∃ i0. i * i0 = j

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
0 ≤ i * i0 ⊦ ∃ i1. i * i1 = i * i0

Rule argument

Additional information

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