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Proof step 12 from proof mod-nls-04 (previous step)

abs(i % j) < j, i % j < i0, i0 + j ≤ 1, 0 < j, 0 ≤ i0 ⊦

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: int-abs inst: name: le-rec seq: ⊦ i ≠ j → (i ≤ j ↔ i ≤ j - 1) subst: [j, i] → [1, i0 + j] flags: rewrite

Additional information

Used simplifier rules: one-01: ⊦ i = 1 → j % i = 0;%forward-lels: ⊦ i ≤ j ∧ j < j0 → i < j0; (forward) from specification int-abs%forward-le-04: ⊦ j ≤ i ∧ i ≤ 0 → j ≤ 0; (forward) from specification int-abs%forward-lele: ⊦ j ≤ j0 ∧ j0 ≤ j1 → j ≤ j1; (forward) from specification int-abs%ls: ⊦ i + j < j ↔ i < 0; from specification int-basic2%lenls: ⊦ i ≤ j → ¬ j < i; 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%one-03: ⊦ 0 < i ∧ 0 ≤ j → (1 = i + j ↔ i = 1 ∧ j = 0); from specification int-abs%lf-02: ⊦ i < j ∧ j = i0 → i < i0; (forward) from specification int-basic1%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs Used terms: 1, i0 + j