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Proof step 18 from proof fits-law-div-stronger (previous step)

b2i(by1) < 0, b2i(by1) / i + b2i(by1) ≤ 0, b2i(by1) < ~127, 0 < b2i(by1) / i, 127 < b2i(by1) / i ⊦ b2i(by1) = 0, b2i(by1) / i = b2i(i2b(b2i(by1) / i)), i = 0, i = ~1

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 19:
⊦ i < j ↔ i + 1 ≤ j
Proof step 20:
b2i(by1) < 0, b2i(by1) / i + b2i(by1) ≤ 0, b2i(by1) + 1 ≤ ~127, 0 < b2i(by1) / i, 127 < b2i(by1) / i ⊦ b2i(by1) = 0, b2i(by1) / i = b2i(i2b(b2i(by1) / i)), i = 0, i = ~1

Rule argument

seq: ⊦ i < j ↔ i + 1 ≤ j subst: [j, i] → [~127, b2i(by1)] rotate: F spec: int-abs inst: name: ls-rew-succ

Additional information

Used terms: ~127, b2i(by1)