2 votes 2 votes DeMorgan’s first theorem shows the equivalence of: OR gate and Exclusive OR gate. NOR gate and Bubbled AND gate. NOR gate and NAND gate. NAND gate and NOT gate Digital Logic tbb-digital-logic-2 + – Bikram asked Nov 26, 2016 • edited Aug 19, 2019 by Counsellor Bikram 319 views answer comment Share Follow See 1 comment See all 1 1 comment reply Praveen Saini commented Dec 1, 2016 reply Follow Share Option B (A+B)'= (A'B')' 1 votes 1 votes Please log in or register to add a comment.
Best answer 2 votes 2 votes 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. shraddha priya answered Apr 22, 2017 • selected Apr 22, 2017 by Bikram shraddha priya comment Share Follow See all 2 Comments See all 2 2 Comments reply swetha12 commented Oct 24, 2019 reply Follow Share isn't B and C options same? bubbled AND and NAND are same? 0 votes 0 votes Lakshman Bhaiya commented Jan 7, 2020 reply Follow Share @swetha12 No Bubble AND $= \overline{A}\cdot\overline{B} = \overline{A+B} =$ NOR NAND $ = \overline{A\cdot B} = \overline{A} + \overline{B} =$ Bubble OR 0 votes 0 votes Please log in or register to add a comment.