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Proof step 12 from proof abs-ls-same-01 (previous step)

j = abs(j) ⊦ abs(i) * abs(j) < i ↔ ¬ i ≤ 0 ∧ ¬ 0 < abs(j) ∨ ¬ 0 ≤ i ∧ ¬ ~1 ≤ abs(j)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs%numleledel-01: i0 ≠ 0, i0 ≤ i ⊦ (* i) ≤ j → (* i0) ≤ j; from specification int-abs%absle: ⊦ 0 ≤ i → abs(i) = i; from specification int-abs%abs-18: ⊦ abs(i) = i ↔ 0 ≤ i; from specification int-abs