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Proof step 8 from proof mod-rec-01 (previous step)

i ≤ j, i < 0, j < 0 ⊦ (j - i) % j + (i - j) % j = 0

Heuristic elimination applied rule insert elim lemma

And closed the goal!

Rule argument

spec: int-abs inst: name: elim-sub seq: ⊦ i0 = i - j ↔ i = i0 + j subst: [i, j, i0] → [j, i, i0] flags:

Additional information

Used simplifier rules: mod-neg-01: j ≠ 0 ⊦ ~ i % j = ~(i % j);%sub-add-09: ⊦ i - j + i = ~ j; from specification int-abs%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%lep: ⊦ i ≤ i + j ↔ 0 ≤ j; from specification int-abs%forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs%lf-02: ⊦ i < j ∧ j = i0 → i < i0; (forward) from specification int-basic1%forward-lsle: ⊦ i < j ∧ j ≤ j0 → i < j0; (forward) from specification int-abs%forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs%ls-01: ⊦ i + j < i + i0 ↔ j < i0; from specification int-basic2%irref: ⊦ ¬ i < i; from specification int-basic1%neg-03: ⊦ ~ i + i = 0; from specification int-abs% ⊦ i0 = i - j ↔ i = i0 + j; (elim) Used terms: j, i, i0