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Proof step 11 from proof mod-ls-same-01 (previous step)

0 ≤ i ⊦ i = 0, i % j < i ↔ 0 < i ∧ (abs(j) = 1 ∨ abs(j) ≤ i), j = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
0 < i ⊦ j = 0, i = 0, i % j < i ↔ abs(j) = 1 ∨ abs(j) ≤ i

Rule argument

Additional information

Used simplifier rules: forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs%lels-02: ⊦ i ≠ j → (i ≤ j ↔ i < j); from specification int-abs