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Proof step 22 from proof int2bytes-first-neg (previous step)

int2bytes(i1 / 256) = by ' + bytes.butlast, int2bytes((i1 + 255) % 256).last = bytes.last, 0 ≤ b2i(by), i1 < ~383, i1 < 0, ∀ bytes, by, i0. i→n(i0) < i→n(i1 + 255) → int2bytes(i0) = by ' + bytes ∧ by ∈ int2bytes(i0) ∧ 0 ≤ b2i(by) ∧ i0 < 0 → false ⊦ bytes = []

User applied rule insert elim lemma

And obtained the following new goals

Proof step 23:
bytes = bytes0 + by0 ', int2bytes(i1 / 256) = by ' + bytes0, int2bytes((i1 + 255) % 256).last = by0, 0 ≤ b2i(by), i1 < ~383, i1 < 0, ∀ bytes, by, i0. i→n(i0) < i→n(i1 + 255) → int2bytes(i0) = by ' + bytes ∧ by ∈ int2bytes(i0) ∧ 0 ≤ b2i(by) ∧ i0 < 0 → false ⊦ bytes = []

Rule argument

seq: bytes ≠ [] ⊦ by = bytes.last ∧ bytes0 = bytes.butlast ↔ bytes = bytes0 + by ' subst: [bytes, by, bytes0] → [bytes, by0, bytes0] rotate: T spec: list-last inst: bytelist name: last-elim

Additional information

Used simplifier rules: bytes ≠ [] ⊦ by = bytes.last ∧ bytes0 = bytes.butlast ↔ bytes = bytes0 + by '; (elim) Used terms: bytes, by0, bytes0