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Proof step 1 from proof bitsxor-rec-01-01 (conclusion)

⊦ bits ≠ [] ∧ bits0 = true ' → bitsxor(bits, bits0) = bitsinvert(bits)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
⊦ bitsxor(bits, true ') = bitsinvert(bits), bits = []

Rule argument

Additional information