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Proof step 22 from proof div-div (previous step)

(i / j) / abs(j0) = i / (abs(j0) * j), i < j, i < 0, 0 < j, i + 1 < j ⊦ j0 = 0, (i / j) / j0 = i / (j * j0)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 23:
⊦ i < 0 → abs(i) = ~ i

Rule argument

spec: int-abs inst: name: abs-08 seq: ⊦ i < 0 → abs(i) = ~ i subst: [i] → [j0] flags: rewrite

Additional information

Used simplifier rules: Div-right: ⊦ j ≠ 0 → i / ~ j = ~(i / j);%neg: j ≠ 0 ⊦ i / ~ j = ~(i / j);%minus-left: ⊦ ~ i * j = ~(i * j); from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult%zero-neq-01: ⊦ 0 < i → (i * j = 0 ↔ j = 0); from specification int-mult%neg-12: ⊦ ~ i = ~ j ↔ i = j; from specification int-abs%Abs-pos: ⊦ ¬ j < 0 → abs(j) = j; from specification int-abs Used terms: j0