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Proof step 23 from proof mod-ls-same-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: 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%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%numadd-01: ⊦ (* j) = i0 + (* j0) ↔ j - j0 = i0; from specification int-abs%negeq: ⊦ ~ i = j ↔ i + j = 0; from specification int-abs%neg-07: ⊦ i < ~ j ↔ i + j < 0; from specification int-abs%abs-08: ⊦ i < 0 → abs(i) = ~ i; from specification int-abs%lst-01: ⊦ ¬ i < j → (i ≠ j ↔ j < i); from specification int-basic1%lels-02: ⊦ i ≠ j → (i ≤ j ↔ i < j); from specification int-abs