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Proof step 15 from proof bytes2bytes-zero (previous step)

bytes = [], b2i(by) = 0, b2i(by0) < 0, ∀ by0, by, bytes0. # bytes0 < # bytes → b2i(by) = 0 ∧ b2i(by0) < 0 → bits2bytes(sgn(byte2bits(by) + byte2bits(by0) + bytes2bits(bytes0))) = by ' + by0 ' + bytes0 ⊦ bits2bytes(sgn(byte2bits(by) + byte2bits(by0) + bytes2bits(bytes))) = by ' + by0 ' + bytes

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 16:
b2i(by) = 0, b2i(by0) < 0, ∀ by0, by, bytes0. # bytes0 < # [] → b2i(by) = 0 ∧ b2i(by0) < 0 → bits2bytes(sgn(byte2bits(by) + byte2bits(by0) + bytes2bits(bytes0))) = by ' + by0 ' + bytes0 ⊦ bits2bytes(sgn(byte2bits(by) + byte2bits(by0))) = by ' + by0 '

Rule argument

Additional information

Used simplifier rules: bytes2bits-empty: ⊦ bytes2bits([]) = [];% ⊦ (bytes + bytes0) + bytes1 = bytes + bytes0 + bytes1;% ⊦ bytes + [] = bytes;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ bits + [] = bits;