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

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 7:
⊦ i < 0 → abs(i) = ~ i
Proof step 8:
(abs(i) / abs(j)) / abs(j0) = abs(i) / (abs(j) * abs(j0)) ⊦ (i / j) / j0 = i / (j * j0), j = 0, j0 = 0, i < 0
Proof step 17:
(~ i / abs(j)) / abs(j0) = ~ i / (abs(j) * abs(j0)), i < 0 ⊦ (i / j) / j0 = i / (j * j0), j = 0, j0 = 0

Rule argument

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

Additional information

Used terms: i