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Proof step 12 from proof div-div (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 13:
⊦ i < 0 → abs(i) = ~ i

Rule argument

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

Additional information

Used simplifier rules: Div-right: ⊦ j ≠ 0 → i / ~ j = ~(i / j);%div-eq-same-01: j ≠ 0 ⊦ i / j = i ↔ i = 0 ∨ j = 1;%neg: j ≠ 0 ⊦ i / ~ j = ~(i / j);% ⊦ ~ i * j = ~(i * j); (s)%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult% ⊦ i * j = 0 ↔ i = 0 ∨ j = 0; (ws)% ⊦ i * 0 = 0; (s)% ⊦ ~ i = ~ j ↔ i = j; (s)% ⊦ ¬ j < 0 → abs(j) = j; (s) Used terms: j0