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Proof step 10 from proof bytes2int-append (previous step)

¬(¬ (bytes2bits(bytes) + bytes2bits(bytes0)).first ∧ bytes2bits(bytes) + bytes2bits(bytes0) ≠ []) ⊦ bytes0 = [], bytes = [], bits2int(bytes2bits(bytes) + bytes2bits(bytes0)) = bits2int(bytes2bits(bytes)) * 256 ^ n→i(# bytes0) + band(bits2int(bytes2bits(bytes0)), 256 ^ n→i(# bytes0) - 1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
(bytes2bits(bytes) + bytes2bits(bytes0)).first ⊦ bytes0 = [], bytes = [], bits2int(bytes2bits(bytes) + bytes2bits(bytes0)) = bits2int(bytes2bits(bytes)) * 256 ^ n→i(# bytes0) + band(bits2int(bytes2bits(bytes0)), 256 ^ n→i(# bytes0) - 1)

Rule argument

Additional information

Used simplifier rules: bytes2bits-empty-01: ⊦ bytes2bits(bytes) = [] ↔ bytes = [];% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ ¬ bits + bits0 ≠ [] ↔ bits = [] ∧ bits0 = [];