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Proof step 19 from proof bytes2int-append (previous step)

bytes2bits(bytes).first ∧ bytes2bits(bytes) ≠ [], bytes2bits(bytes).first ⊦ bytes2bits(bytes) + bytes2bits(bytes0) = [], bytes0 = [], bytes = [], bits2uint((bitsinvert(bytes2bits(bytes)) + bitsinvert(bytes2bits(bytes0)))+1) + ~ bits2uint(bitsinvert(bytes2bits(bytes)) +1) * 256 ^ n→i(# bytes0) + band(bits2int(bytes2bits(bytes0)), 256 ^ n→i(# bytes0) - 1) = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 20:
bytes2bits(bytes).first ⊦ bytes2bits(bytes) + bytes2bits(bytes0) = [], bytes0 = [], bytes = [], bits2uint((bitsinvert(bytes2bits(bytes)) + bitsinvert(bytes2bits(bytes0)))+1) + band(bits2int(bytes2bits(bytes0)), 256 ^ n→i(# bytes0) - 1) = bits2uint(bitsinvert(bytes2bits(bytes)) +1) * 256 ^ n→i(# bytes0)

Rule argument

Additional information

Used simplifier rules: bytes2bits-empty-01: ⊦ bytes2bits(bytes) = [] ↔ bytes = [];% ⊦ i + 0 = i;% ⊦ i - j = i0 ↔ i = i0 + j;% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + (j - i0) = i + j - i0;% ⊦ i + ~ j = i - j;% ⊦ i * j = j * i;% ⊦ i * j * j0 = (i * j) * j0;% ⊦ ~ i * j = ~(i * j);