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Proof step 117 from proof bxor-flip-one-bit (
previous step
)
⊦ # bits = # bits0 ∧ 1 < # bits → bitsxor(bits, bits0) = (¬(bits.first ↔ bits0.first))' + bitsxor(bits.rest, bits0.rest)
Lemma
bitsxor-first