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Proof step 1 from proof bxor-minus-one (conclusion)

⊦ bxor(i, ~1) = bcompl(i)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 2:
⊦ bxor(i, j) = bor(band(i, bcompl(j)), band(bcompl(i), j))

Rule argument

name: bxor-def-logic rotate: F subst: [j, i] → [~1, i]

Additional information

Used simplifier rules: bcompl-num: ⊦ bcompl((* i)) = bits2int(bitsinvert(int2bits(i)));%band-zero: ⊦ band(0, i) = 0;%band-associativity: ⊦ band(i, band(i0, j)) = band(band(i, i0), j);%band-commutativity: ⊦ band(i, j) = band(j, i);%band-minus-one: ⊦ band(~1, i) = i;%bor-zero: ⊦ bor(i, 0) = i;%bor-com: ⊦ bor(i, j) = bor(j, i);% ⊦ int2bits(~1) = true ';% ⊦ bitsinvert(true ') = false ';% ⊦ bits2int(false ') = 0; Used terms: ~1, i