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Proof step 3 from proof num-i2b-equal (previous step)

i2b(i) = i2b(j), ~128 ≤ i, i ≤ 127, ~128 ≤ j, j ≤ 127 ⊦ i = j

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ b2i(byi) = b2i(byj) ↔ byi = byj

Rule argument

name: byte-equal rotate: T subst: [byj, byi] → [i2b(j), i2b(i)]

Additional information

Used simplifier rules: bib-in-01: ⊦ ~128 ≤ i ∧ i ≤ 127 → b2i(i2b(i)) = i; Used terms: i2b(j), i2b(i)