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Proof step 77 from proof bytes2bytes-mone (previous step)

bytes0.rest = [], b2i(bytes0.first) = ~1, 8 = # fillsgn(8, int2bits(b2i(by))), 0 ≤ b2i(by), true ∈ bytes2bits(bytes0), false ∈ bytes2bits(bytes0), b2i([].first) < 0, ∀ bytes1. # bytes1 < #(bytes0 + by ') → b2i(bytes1.first) = ~1 ∧ 0 ≤ b2i(bytes1.rest.first) → bits2bytes(sgn(bytes2bits(bytes1))) = bytes1 ∨ # bytes1 = 1 ∨ bytes1 = [] ⊦ bits2bytes(sgn(bytes2bits(bytes0))) = bytes0, bytes0 = []

User 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