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Proof step 9 from proof mod-rec-01 (previous step)

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

Heuristic elimination applied rule insert elim lemma

And closed the goal!

Rule argument

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

Additional information

Used simplifier rules: mod-neg-01: j ≠ 0 ⊦ ~ i % j = ~(i % j);% ⊦ i - j + i = ~ j; (s)%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2% ⊦ i ≤ i + j ↔ 0 ≤ j; (s)%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%lf-03: ⊦ i < j ∧ i = i0 → i0 < j; (forward) from specification int-basic1% ⊦ ~ i + i = 0; (s)% ⊦ i0 = i - j ↔ i = i0 + j; (elim) Used terms: j, i, i0