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Proof step 1 from proof bitsxor-append-same-size (conclusion)

⊦ # bits0 = # bits2 ∧ bits ≠ [] ∧ bits0 ≠ [] ∧ bits1 ≠ [] → bitsxor(bits + bits0, bits1 + bits2) = bitsxor(bits, bits1) + bitsxor(bits0, bits2)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information