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Proof step 2 from proof bytes2bits-append (previous step)

∀ bytes1. bytes2bits(bytes0 + bytes1) = bytes2bits(bytes0) + bytes2bits(bytes1) ⊦ bytes2bits((by + bytes0) + bytes1) = bytes2bits(by + bytes0) + bytes2bits(bytes1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: bytes2bits-rec: ⊦ bytes2bits(by ' + bytes) = byte2bits(by) + bytes2bits(bytes);% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ bits + bits0 = bits + bits1 ↔ bits0 = bits1;% ⊦ (bytes + bytes0) + bytes1 = bytes + bytes0 + bytes1;% ⊦ by + bytes = by ' + bytes;