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

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

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: int-abs name: Abs-neg seq: ⊦ j < 0 → abs(j) = 0 - j subst: [j] → [i0] flags: <all paths>

Additional information

Used simplifier rules: ⊦ 0 - i = ~ i; (s)% ⊦ ~ i = j ↔ i + j = 0; (s)% ⊦ (* j) = i0 + (* j0) ↔ j - j0 = i0; (s)%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs%i < 0 ⊦ (* i) + j = j - ~ i; (s)% ⊦ i - j = i0 ↔ i = i0 + j; (s)% ⊦ ¬ j < 0 → abs(j) = j; (s)% ⊦ i < j ↔ ¬ j ≤ i; (ss)%forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs Used terms: i0