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Proof step 4 from proof div-twoodd-01 (previous step)

i % 2 = 0, i < 0, i - 1 < 0 ⊦ 2 = 0, ~(~(i - 1) / 2) = i / 2

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 5:
⊦ ~(i - 1) = ~ i + 1
Proof step 6:
i % 2 = 0, i - 1 < 0, i < 0 ⊦ 2 = 0, ~((~ i + 1) / 2) = i / 2

Rule argument

spec: int-abs name: minus-02 seq: ⊦ ~(i - 1) = ~ i + 1 subst: [i] → [i] flags: <all paths>

Additional information

Used terms: i