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Proof step 3 from proof mod-ls-same-01 (previous step)

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 4:
i % j < 0, i < 0, i % j < i, i % j < ~1 ⊦ j = 0

Rule argument

Additional information

Used simplifier rules: add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%numls-03: ⊦ i + (* j) < (* j0) ↔ i < j0 - j; from specification int-abs%succ-01: ⊦ i +1 = i + 1; from specification int-basic2%forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs%trans: ⊦ i < j ∧ j < j0 → i < j0; (forward) from specification int-basic1%lf-01: ⊦ i < j ∧ j < i0 → i +1 < i0; (forward) from specification int-basic1%ls-01: ⊦ i < j → ¬ j < i; from specification int-basic1