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Proof step 21 from proof bytes2int-rec (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 22:
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 = [];%i < 0 ⊦ (* i) + j = j - ~ 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);