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Proof step 43 from proof bytes2bytes-other (previous step)

false ∈ bytes2bits(bytes0), true ∈ bytes2bits(bytes0), ∀ bytes1. # bytes1 < #(bytes0 + by ') → bytes2bits(bytes1) ≠ [] → bytes1 = [] ∨ b2i(bytes1.first) = 0 ∨ b2i(bytes1.first) = ~1 ∨ bits2bytes(sgn(bytes2bits(bytes1))) = bytes1 ⊦ bits2bytes(sgn(bytes2bits(bytes0))) + i2b(bits2int(lastn(8, sgn(bytes2bits(bytes0)) + byte2bits(by)))) ' = bytes0 + by ', bytes0 = [], b2i(bytes0.first) = ~1, b2i(bytes0.first) = 0

Heuristic var elimination applied rule insert elim lemma

And obtained the following new goals

Rule argument

seq: bytes ≠ [] ⊦ by = bytes.first ∧ bytes0 = bytes.rest ↔ bytes = by + bytes0 subst: [bytes, by, bytes0] → [bytes0, by0, bytes] 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: bytes0, by0, bytes