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Proof step 2 from proof fits-law-div-stronger (previous step)

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

User applied rule insert spec-lemma

And obtained the following new goals

Proof step 3:
j ≠ 0 ⊦ ~ abs(i) ≤ i / j ∧ i / j ≤ abs(i)
Proof step 4:
~ abs(b2i(by1)) ≤ b2i(by1) / i ∧ b2i(by1) / i ≤ abs(b2i(by1)) ⊦ i = ~1, i = 0, b2i(by1) / i = b2i(i2b(b2i(by1) / i))

Rule argument

seq: j ≠ 0 ⊦ ~ abs(i) ≤ i / j ∧ i / j ≤ abs(i) subst: [j, i] → [i, b2i(by1)] rotate: F spec: int-div inst: name: le-abs-numerator

Additional information

Used terms: i, b2i(by1)