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Proof step 5 from proof fits-law-div-stronger (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 6:
⊦ ~128 ≤ i ∧ i ≤ 127 → b2i(i2b(i)) = i
Proof step 7:
¬(~128 ≤ b2i(by1) / i ∧ b2i(by1) / i ≤ 127), b2i(by1) / i ≤ abs(b2i(by1)), 0 ≤ abs(b2i(by1)) + b2i(by1) / i ⊦ i = ~1, i = 0, b2i(by1) / i = b2i(i2b(b2i(by1) / i))

Rule argument

name: bib-in-01 rotate: F subst: [i] → [b2i(by1) / i]

Additional information

Used terms: b2i(by1) / i