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Proof step 1 from proof bytes2bits-append (conclusion)

⊦ bytes2bits(bytes + bytes0) = bytes2bits(bytes) + bytes2bits(bytes0)

User applied rule structural induction

And obtained the following new goals

Proof step 2:
∀ bytes1. bytes2bits(bytes0 + bytes1) = bytes2bits(bytes0) + bytes2bits(bytes1) ⊦ bytes2bits((by + bytes0) + bytes1) = bytes2bits(by + bytes0) + bytes2bits(bytes1)

Rule argument

indvar: bytes

Additional information

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