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Proof step 46 from proof bytes2bytes-mone (previous step)

b2i(by0) = ~1, 0 ≤ b2i(bytes.first), ∀ bytes1. # bytes1 < #((by0 + bytes) + by ') → b2i(bytes1.first) = ~1 ∧ 0 ≤ b2i(bytes1.rest.first) → bits2bytes(sgn(bytes2bits(bytes1))) = bytes1 ∨ # bytes1 = 1 ∨ bytes1 = [] ⊦ bytes = [], bits2bytes(sgn((byte2bits(by0) + bytes2bits(bytes)) + byte2bits(by))) = (by0 ' + bytes) + by ', (true ∈ byte2bits(by0) ∨ true ∈ bytes2bits(bytes)) ∧ (false ∈ byte2bits(by0) ∨ false ∈ bytes2bits(bytes))

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 47:
bytes = by1 + bytes0, b2i(by0) = ~1, 0 ≤ b2i(by1), ∀ bytes1. # bytes1 < #((by0 + bytes) + by ') → b2i(bytes1.first) = ~1 ∧ 0 ≤ b2i(bytes1.rest.first) → bits2bytes(sgn(bytes2bits(bytes1))) = bytes1 ∨ # bytes1 = 1 ∨ bytes1 = [] ⊦ bytes = [], bits2bytes(sgn((byte2bits(by0) + bytes2bits(bytes)) + byte2bits(by))) = (by0 ' + bytes) + by ', (true ∈ byte2bits(by0) ∨ true ∈ bytes2bits(bytes)) ∧ (false ∈ byte2bits(by0) ∨ false ∈ bytes2bits(bytes))

Rule argument

seq: bytes ≠ [] ⊦ by = bytes.first ∧ bytes0 = bytes.rest ↔ bytes = by + bytes0 subst: [bytes, by, bytes0] → [bytes, by1, 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, by1, bytes0