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Proof step 18 from proof bytes2bytes-mone (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 19:
b2i(by0) = ~1, 0 ≤ b2i(by), ∀ bytes1. # bytes1 < #((by0 + []) + by ') → b2i(bytes1.first) = ~1 ∧ 0 ≤ b2i(bytes1.rest.first) → bits2bytes(sgn(bytes2bits(bytes1))) = bytes1 ∨ # bytes1 = 1 ∨ bytes1 = [] ⊦ bits2bytes(sgn(byte2bits(by0) + byte2bits(by))) = by0 ' + by ', true ∈ byte2bits(by0) ∧ false ∈ byte2bits(by0)

Rule argument

Additional information

Used simplifier rules: bytes2bits-one: ⊦ bytes2bits(by ') = byte2bits(by);% ⊦ by ' ≠ [];% ⊦ (bytes + bytes0) + bytes1 = bytes + bytes0 + bytes1;% ⊦ bytes + [] = bytes;% ⊦ by + bytes = by ' + bytes;