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Proof step 24 from proof bytes2int-rec (previous step)

bitsinvert(bytes2bits(bytes)) ≠ [], bytes2bits(bytes).first ⊦ bitsinvert(bytes2bits(bytes)) + bitsinvert(byte2bits(by)) = [], bytes2bits(bytes) + byte2bits(by) = [], bytes = [], (bits2uint(bitsinvert(bytes2bits(bytes))) + 1) * 256 = (bits2uint(bitsinvert(bytes2bits(bytes)) + bitsinvert(byte2bits(by))) + band(b2i(by), 255)) + 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 25:
bytes2bits(bytes).first ⊦ bitsinvert(bytes2bits(bytes)) + bitsinvert(byte2bits(by)) = [], bytes2bits(bytes) + byte2bits(by) = [], bytes = [], (bits2uint(bitsinvert(bytes2bits(bytes))) + 1) * 256 = (bits2uint(bitsinvert(bytes2bits(bytes)) + bitsinvert(byte2bits(by))) + band(b2i(by), 255)) + 1

Rule argument

Additional information

Used simplifier rules: bytes2bits-empty-01: ⊦ bytes2bits(bytes) = [] ↔ bytes = [];% ⊦ bitsinvert(bits) = [] ↔ bits = [];