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Proof step 26 from proof exdivmod (previous step)

0 ≤ ~ i ∧ 0 < j → (∃ i0, j0. ~ i = i0 * j + j0 ∧ 0 ≤ i0 ∧ 0 ≤ j0 ∧ j0 < j), i < 0, 0 < j ⊦ 0 ≤ i, ∃ i1, j1. i = i1 * j + j1 ∧ abs(j1) < j

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: forward-lsle: ⊦ i < j ∧ j ≤ j0 → i < j0; (forward) from specification int-abs%forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs%ls-01: ⊦ i < j → ¬ j < i; from specification int-basic1%negeq: ⊦ ~ i = j ↔ i + j = 0; 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: ⊦ i + 0 = i; from specification int-basic2%negle-07: ⊦ i ≤ ~ j ↔ i + j ≤ 0; from specification int-abs%zero-gels-01: ⊦ i < 0 ∧ 0 ≤ j → i < j; from specification int-abs%lels-01: ⊦ ¬ i ≤ j ↔ j < i; from specification int-abs