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Proof step 5 from proof bytes2bytes-identity (previous step)

# bytes = # int2bytes(bytes2int(bytes)) ⊦ int2bytes(bytes2int(bytes)) = bytes, bytes = [], b2i(bytes.first) ≠ ~1 ∧ b2i(bytes.first) ≠ 0

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 6:
bytes = by + bytes0, # bytes = # int2bytes(bytes2int(bytes)) ⊦ int2bytes(bytes2int(bytes)) = bytes, bytes = [], b2i(by) ≠ ~1 ∧ b2i(by) ≠ 0

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