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 307 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.