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Proof step 2 from proof mod-lsz (previous step)

j + i % j < 0, 0 < j ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
i % j < 0, i % j < ~1, 0 < j, j + i % j < 0, i % j + 1 < j ⊦

Rule argument

Additional information

Used simplifier rules: mod-less-01: ⊦ 0 < j → i % j < j;%numlsf: ⊦ i < (* j) ∧ j < 0 → i < 0; (forward) from specification int-basic1%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2% ⊦ i + (* j) < (* j0) ↔ i < j0 - j; (s)% ⊦ i + j < j ↔ i < 0; (s)% ⊦ i +1 = i + 1; (s)%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