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Proof step 80 from proof byte-mult-256 (previous step)

i0 + i = ~1, i1 < 0, ~256 ≤ i1, i < 0, 0 ≤ i0 ⊦ i = 0, i2b(i1) = i2b(i1 + 256)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 81:
⊦ i2b(i) = i2b(j) ↔ b2i(i2b(i)) = b2i(i2b(j))
Proof step 82:
i0 + i = ~1, i1 < 0, ~256 ≤ i1, i < 0, 0 ≤ i0 ⊦ i = 0, b2i(i2b(i1)) = b2i(i2b(i1 + 256))

Rule argument

name: i2b-equal rotate: F subst: [j, i] → [i1 + 256, i1]

Additional information

Used terms: i1 + 256, i1