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Proof step 14 from proof mod-rec-any (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

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

Additional information

Used simplifier rules: mod-mult-04: ⊦ j ≠ 0 → ~ j |%| j = 0;%sub-ls: ⊦ i - j < i0 ↔ i < j + i0; from specification int-abs%ax-04: ⊦ i + 0 = i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%neg-06: ⊦ 0 - i = ~ i; from specification int-abs%minus-sub: ⊦ ~(i - j) = j - i; from specification int-abs%sub-ls-01: ⊦ i < j - i0 ↔ i + i0 < j; from specification int-abs%add-mult-two: ⊦ i + i = 2 * i; from specification int-mult%sub-le-01: ⊦ i ≤ j - i0 ↔ i + i0 ≤ j; from specification int-abs%lep-01: ⊦ i + j ≤ i ↔ j ≤ 0; from specification int-abs%anti: ⊦ i ≤ j → (j ≤ i ↔ i = j); from specification int-abs%lel: ⊦ i < i0 ∧ i = i1 → i1 < i0; (forward) from specification oelem[intlista]%Abs-pos: ⊦ ¬ j < 0 → abs(j) = j; from specification int-abs%ls: ⊦ i + j < j ↔ i < 0; from specification int-basic2%ls-01: ⊦ i < j → ¬ j < i; from specification int-basic1 Used terms: i - j