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Proof step 57 from proof bxor-def-logic (
previous step
)
⊦ bits ≠ [] ∧ bits0 ≠ [] → bits2int(bitsxor(bits, bits0)) = bits2int(bitsor(bitsand(bitsinvert(bits), bits0), bitsand(bitsinvert(bits0), bits)))
Lemma
bitsxor-is-orand