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Proof step 1 from proof bor-notAndNot (conclusion)

⊦ bor(i, j) = bcompl(band(bcompl(i), bcompl(j)))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: bor-def: ⊦ bor(i, j) = bits2int(bitsor(int2bits(i), int2bits(j)));%bcompl-def: ⊦ bcompl(i) = bits2int(bitsinvert(int2bits(i)));%band-def: ⊦ band(i, j) = bits2int(bitsand(int2bits(i), int2bits(j)));%band-associativity: ⊦ band(i, band(i0, j)) = band(band(i, i0), j);%band-commutativity: ⊦ band(i, j) = band(j, i);%bits ≠ [] ⊦ int2bits(bits2int(bits)) = sgn(bits);% ⊦ bitsinvert(bits) = [] ↔ bits = [];% ⊦ int2bits(j) ≠ [];