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Proof step 12 from proof bxor-def-logic (previous step)

⊦ bits2int(bitsxor(int2bits(i), int2bits(j))) = bor(bits2int(bitsand(int2bits(i), int2bits(bits2int(bitsinvert(int2bits(j)))))), bits2int(bitsand(sgn(bitsinvert(int2bits(i))), int2bits(j))))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: bits ≠ [] ⊦ int2bits(bits2int(bits)) = sgn(bits);% ⊦ bitsinvert(bits) = [] ↔ bits = [];% ⊦ int2bits(j) ≠ [];