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Proof step 6 from proof div-one-one-neg (previous step)

abs(i0) = 1, i0 + i = 0 ⊦ i = 1, i = ~1

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] → [i0] flags: rewrite

Additional information

Used simplifier rules: neg-06: ⊦ 0 - i = ~ i; from specification int-abs%negeq: ⊦ ~ i = j ↔ i + j = 0; from specification int-abs%numadd-01: ⊦ (* j) = i0 + (* j0) ↔ j - j0 = i0; 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%lf-02: ⊦ i < j ∧ j = i0 → i < i0; (forward) from specification int-basic1%forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs%minus-add: i < 0 ⊦ (* i) + j = j - ~ i; from specification int-abs%sub-eq: ⊦ i - j = i0 ↔ i = i0 + j; from specification int-abs%Abs-pos: ⊦ ¬ j < 0 → abs(j) = j; from specification int-abs%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs Used terms: i0