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Proof step 1 from proof fits-law-bxor (conclusion)

⊦ bxor(b2i(by1), b2i(by2)) = b2i(i2b(bxor(b2i(by1), b2i(by2))))

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))
Proof step 3:
⊦ bor(band(b2i(by1), bcompl(b2i(by2))), band(bcompl(b2i(by1)), b2i(by2))) = b2i(i2b(bor(band(b2i(by1), bcompl(b2i(by2))), band(bcompl(b2i(by1)), b2i(by2)))))

Rule argument

seq: ⊦ bxor(i, j) = bor(band(i, bcompl(j)), band(bcompl(i), j)) subst: [j, i] → [b2i(by2), b2i(by1)] rotate: F spec: intops inst: name: bxor-def-logic

Additional information

Used terms: b2i(by2), b2i(by1)