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Proof step 15 from proof abs-ls-same-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 16:
j < 0 ⊦ abs(i) * j < i ↔ ¬ i ≤ 0 ∨ i ≠ 0 ∧ j ≠ ~1

Rule argument

Additional information

Used simplifier rules: forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs% ⊦ i < 0 → (~1 ≤ i ↔ i = ~1); (s)% ⊦ i ≤ j → (j ≤ i ↔ i = j); (s)% ⊦ i < j → ¬ j < i; (s)% ⊦ i ≤ j ↔ ¬ j < i; (ss)% ⊦ abs(i) = i ↔ 0 ≤ i; (s)