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Proof step 58 from proof bxor-def-logic (previous step)

int2bits(i) ≠ [] ∧ int2bits(j) ≠ [] ⊦ bitsand(bitsinvert(int2bits(i)), int2bits(j)) = [], bitsand(bitsinvert(int2bits(j)), int2bits(i)) = [], bitsand(sgn(bitsinvert(int2bits(i))), int2bits(j)) = [], bits2int(bitsor(bitsand(bitsinvert(int2bits(i)), int2bits(j)), bitsand(bitsinvert(int2bits(j)), int2bits(i)))) = bits2int(bitsor(bitsand(bitsinvert(int2bits(j)), int2bits(i)), bitsand(bitsinvert(int2bits(i)), int2bits(j))))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 59:
⊦ bitsand(bitsinvert(int2bits(i)), int2bits(j)) = [], bitsand(bitsinvert(int2bits(j)), int2bits(i)) = [], bitsand(sgn(bitsinvert(int2bits(i))), int2bits(j)) = [], bits2int(bitsor(bitsand(bitsinvert(int2bits(i)), int2bits(j)), bitsand(bitsinvert(int2bits(j)), int2bits(i)))) = bits2int(bitsor(bitsand(bitsinvert(int2bits(j)), int2bits(i)), bitsand(bitsinvert(int2bits(i)), int2bits(j))))

Rule argument

Additional information

Used simplifier rules: ⊦ int2bits(j) ≠ [];