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Proof step 96 from proof byte-mult-256 (previous step)

i1 < 0, ~256 ≤ i1, 0 ≤ i0, i < 0, i0 + i < ~1, i0 + i < 0 ⊦ i0 + i = 0, i = 0, b2i(i2b(i1)) = b2i(i2b(i1 + 256))

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 97:
⊦ ~256 < i ∧ i < ~128 → b2i(i2b(i)) = i + 256
Proof step 98:
¬(~256 < i1 ∧ i1 < ~128), i1 < 0, ~256 ≤ i1, 0 ≤ i0, i < 0, i0 + i < ~1, i0 + i < 0 ⊦ i0 + i = 0, i = 0, b2i(i2b(i1)) = b2i(i2b(i1 + 256))
Proof step 102:
~256 < i1 ∧ i1 < ~128, i1 < 0, ~256 ≤ i1, 0 ≤ i0, i < 0, i0 + i < ~1, i0 + i < 0 ⊦ i0 + i = 0, i = 0, i1 + 256 = b2i(i2b(i1 + 256))

Rule argument

name: bib-lower rotate: F subst: [i] → [i1]

Additional information

Used terms: i1