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Proof step 2 from proof bytes2bytes-zero (previous step)

b2i(bytes.first) = 0, b2i(bytes.rest.first) < 0 ⊦ int2bytes(bytes2int(bytes)) = bytes, # bytes = 1, bytes = []

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 3:
bytes = by + bytes0, b2i(by) = 0, b2i(bytes0.first) < 0 ⊦ int2bytes(bytes2int(bytes)) = bytes, # bytes = 1, bytes = []

Rule argument

seq: bytes ≠ [] ⊦ by = bytes.first ∧ bytes0 = bytes.rest ↔ bytes = by + bytes0 subst: [bytes, by, bytes0] → [bytes, by, bytes0] rotate: T spec: list-data inst: bytelist name: elim-list-c

Additional information

Used simplifier rules: bytes ≠ [] ⊦ by = bytes.first ∧ bytes0 = bytes.rest ↔ bytes = by + bytes0; (elim) Used terms: bytes, by, bytes0