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Proof step 26 from proof bits2int-pos (previous step)

7 < # bits0, # bits0 % 8 ≤ 7, # bits0 % 8 < # bits0 ⊦ bits0 = [], (# bits0 + 1) % 8 = 0, 0 ≤ bits2int(bytes2bits(i2b(bits2int(boolvar ' + firstn(# bits0 % 8, bits0))) ' + bits2bytes(restn(# bits0 % 8, bits0)))) ↔ ¬ boolvar

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 27:
7 < # bits0, # bits0 % 8 ≤ 7, # bits0 % 8 < # bits0 ⊦ bits0 = [], (# bits0 + 1) % 8 = 0, 0 ≤ bits2int(byte2bits(i2b(bits2int(boolvar ' + firstn(# bits0 % 8, bits0)))) + bytes2bits(bits2bytes(restn(# bits0 % 8, bits0)))) ↔ ¬ boolvar

Rule argument

Additional information

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