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Proof step 78 from proof bytes2bytes-mone (previous step)

bytes0 = by0 + bytes, bytes = [], b2i(by0) = ~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 = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 79:
b2i(by0) = ~1, 8 = # fillsgn(8, int2bits(b2i(by))), b2i([].first) < 0, 0 ≤ b2i(by), true ∈ byte2bits(by0), false ∈ byte2bits(by0), ∀ 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))) = 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;