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

b2i(by) < 0 ⊦ bits2int(mklist(true, 8) + byte2bits(by) + bytes2bits(bytes)) = bits2int(byte2bits(by) + bytes2bits(bytes))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
b2i(by) < 0 ⊦ bits2int((true ' + true ' + true ' + true ' + true ' + true ' + true ' + true ') + byte2bits(by) + bytes2bits(bytes)) = bits2int(byte2bits(by) + bytes2bits(bytes))

Rule argument

Additional information

Used simplifier rules: ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ bits + [] = bits;% ⊦ mklist(boolvar, 0) = [];%n ≠ 0 ⊦ mklist(true, (* n)) = true ' + mklist(true, n -1);