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Proof step 24 from proof mod-ls-same-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 25:
0 < i, j < 0 ⊦ i % j < i ↔ ¬ i + j < 0, ~1 = j, 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 < j → ¬ j < i; (s)% ⊦ i < ~ j ↔ i + j < 0; (s)%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2% ⊦ (* j) = i0 + (* j0) ↔ j - j0 = i0; (s)% ⊦ ~ i = j ↔ i + j = 0; (s)% ⊦ i < 0 → abs(i) = ~ i; (s)% ⊦ ¬ i < j → (i = j ↔ ¬ j < i); (ss)% ⊦ i ≠ j → (i ≤ j ↔ i < j); (s)