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Proof step 13 from proof int2bytes-first-neg-01 (previous step)

b2i(i2b(i)) = 0, ~128 ≤ i, i < 0, ∀ bytes, i0. i→n(i0) < i→n(i) → int2bytes(i0) = i2b(0) ' + bytes ∧ i0 < 0 ∧ i2b(0) ∈ int2bytes(i0) → false ⊦

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 14:
⊦ ~128 ≤ i ∧ i ≤ 127 → b2i(i2b(i)) = i

Rule argument

seq: ⊦ ~128 ≤ i ∧ i ≤ 127 → b2i(i2b(i)) = i subst: [i] → [i] rotate: F spec: byte inst: name: bib-in-01

Additional information

Used simplifier rules: i0 ≠ 0, i0 ≤ i ⊦ (* i) ≤ j → (* i0) ≤ j;% ⊦ ¬ i ≤ j ↔ j < i;% ⊦ i < j ∧ ¬ i0 < j → ¬ i0 < i; Used terms: i