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Proof step 22 from proof int2bytes-leading-zeros (previous step)

b2i(i2b(i1)) = 0, ~256 ≤ i1, ~129 < i1, i1 < 0, ∀ bytes, i0. i→n(i0) < i→n(i1 + 256) → int2bytes(i0) = i2b(0) ' + i2b(0) ' + bytes ∧ 0 ≤ i0 → false ⊦

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 23:
⊦ ~256 < i ∧ i < ~128 → b2i(i2b(i)) = i + 256
Proof step 24:
¬(~256 < i1 ∧ i1 < ~128), b2i(i2b(i1)) = 0, ~256 ≤ i1, ~129 < i1, i1 < 0, ∀ bytes, i0. i→n(i0) < i→n(i1 + 256) → int2bytes(i0) = i2b(0) ' + i2b(0) ' + bytes ∧ 0 ≤ i0 → false ⊦

Rule argument

seq: ⊦ ~256 < i ∧ i < ~128 → b2i(i2b(i)) = i + 256 subst: [i] → [i1] rotate: F spec: byte inst: name: bib-lower

Additional information

Used simplifier rules: ⊦ (* j) = i0 + (* j0) ↔ j - j0 = i0;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + j = j + i;%i0 ≠ 0, i0 < i ⊦ (* i) < j → (* i0) < j; Used terms: i1