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Proof step 4 from proof abs-eqall (previous step)

⊦ i < 0, abs(i * j) = abs(i) * abs(j)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
0 ≤ i ⊦ abs(i * j) = i * abs(j)

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%Abs-pos: ⊦ ¬ j < 0 → abs(j) = j; from specification int-abs