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Proof step 18 from proof ls-even-01 (previous step)

~ i % 2 = 0 ∧ j % 2 = 0 ∧ 0 ≤ ~ i ∧ 0 ≤ j → (~ i / 2 < j / 2 ↔ ~ i < j), i % 2 = 0, j % 2 = 0, i ≤ ~2, i < 0 ⊦ i / 2 < j / 2 ↔ ¬ j ≤ i

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 19:
i % 2 = 0, j % 2 = 0, i ≤ ~2, i < 0, 0 ≤ j → (0 < j / 2 + i / 2 ↔ 0 < j + i) ⊦ i / 2 < j / 2 ↔ ¬ j ≤ i

Rule argument

Additional information

Used simplifier rules: mod-neg-01: j ≠ 0 ⊦ ~ i % j = ~(i % j);%neg-01: j ≠ 0 ⊦ ~ i / j = ~(i / j);%neg-13: ⊦ ~ i < j ↔ 0 < j + i; from specification int-abs%zero-gels-03: ⊦ i < 0 ∧ 0 ≤ j → i ≤ j; from specification int-abs%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-zero-rev: ⊦ 0 + i = i; from specification int-basic2%negle-07: ⊦ i ≤ ~ j ↔ i + j ≤ 0; from specification int-abs%neg-10: ⊦ ~ i = 0 ↔ i = 0; from specification int-abs