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Proof step 3 from proof mod-nls-04 (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: int-abs inst: name: Abs-pos seq: ⊦ ¬ j < 0 → abs(j) = j subst: [j] → [j] flags: rewrite

Additional information

Used simplifier rules: abs-08: ⊦ i < 0 → abs(i) = ~ i; from specification int-abs%neg-07: ⊦ i < ~ j ↔ i + j < 0; from specification int-abs%neg-15: ⊦ 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%sub-le: ⊦ i - j ≤ i0 ↔ i ≤ j + i0; from specification int-abs%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%succ-02: ⊦ i < j + 1 ↔ i ≤ j; from specification int-abs%forward-lels: ⊦ i ≤ j ∧ j < j0 → i < j0; (forward) from specification int-abs%sub-ls: ⊦ i - j < i0 ↔ i < j + i0; from specification int-abs%add-zero-rev: ⊦ 0 + i = i; from specification int-basic2%lenls: ⊦ i ≤ j → ¬ j < i; from specification int-abs Used terms: j