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DeMorgan’s first theorem shows the equivalence of:

  1. OR gate and Exclusive OR gate.
  2. NOR gate and Bubbled AND gate.
  3. NOR gate and NAND gate.
  4. NAND gate and NOT gate
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Nor is (A+B)'.

(A + B)' = A' . B'  (By DeMorgan's first thm)

A'.B' is AND of A' and B' i.e. bubbled AND.

So, option B.
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