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Proof step 15 from proof mod-nls-04 (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: int-abs name: absls-02 seq: ⊦ i ≤ 0 ∧ j ≤ 0 → (abs(i) < abs(j) ↔ j < i) subst: [j, i] → [i0, i % j] flags: <all paths>

Additional information

Used simplifier rules: mod-less-01: ⊦ 0 < j → i % j < j;% ⊦ i < 0 ∧ 0 ≤ j → i ≤ j; (s)% ⊦ i < 0 → abs(i) = ~ i; (s)% ⊦ i ≤ ~ j ↔ i + j ≤ 0; (s)%add-c: ⊦ i + j = j + i; from specification int-basic2% ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j; (s)% ⊦ i < j → ¬ j < i; (s)% ⊦ i < ~ j ↔ i + j < 0; (s)% ⊦ i ≤ j ↔ ¬ j < i; (ss)% ⊦ 0 < j → abs(j) = j; (s)%forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs%forward-lels: ⊦ i ≤ j ∧ j < j0 → i < j0; (forward) from specification int-abs%forward-pos: ⊦ i < j ∧ 0 < i → 0 < j ∧ 1 < j; (forward) from specification int-basic1%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs Used terms: i0, i % j