Normalization Lectures

DM-Test1-Logic-Q15

+2 votes

S_1: \forall _x(P(x)\rightarrow A) \equiv \exists_x P(x) \rightarrow A

S_2: \exists _x(P(x)\rightarrow A) \equiv \forall_x P(x) \rightarrow A

Which of the following statements is true?

(A) Only S1

(B) Only S2

(C) Both S1 and S2

(D) Neither S1 not S2

asked Jun 28, 2019 by gbeditor-2 (72,810 points)
reshown Jun 30, 2019 by developer

3 Answers

+2 votes
C is correct one
answered Jun 30, 2019 by (34,820 points)
edited Jun 30, 2019 by
could someone tell me when the rank list for this test will be released ?
commented Jul 1, 2019 by (550 points)
don't know man many of the answers are wrong
commented Jul 1, 2019 by (870 points)
@tsvikasthamke2095 in couple of days
commented Jul 1, 2019 by (1,750 points)
I think 11 wrong other than 11 all are now correct.
commented Jul 1, 2019 by (34,820 points)
Or topper list may be released tomorrow before next test.
commented Jul 1, 2019 by (34,820 points)
Is all the questions of DM uploaded? I am able to see only 15th question.
commented Jul 1, 2019 by (16,750 points)
@tskushagra-guptacse , yes all DM questions are uploaded.
commented Jul 1, 2019 by (550 points)
Yes. I found it. They should upload in the recent section.
commented Jul 1, 2019 by (16,750 points)
Click on discreet mathematics
commented Jul 1, 2019 by (34,820 points)
Now all questions are updated
commented Jul 1, 2019 by (34,820 points)
+16 votes

\\ \forall x(P(x)\rightarrow A)=\exists xP(x)\rightarrow A\\ (P1\rightarrow A).(P2\rightarrow A)=(P1+P2)\rightarrow A\\ (P1'+A).(P2'+A)=(P1+P2)'+A\\ P1'P2'+A=P1'P2'+A\\ \\ \exists x(P(x)\rightarrow A )=\forall x P(x)\rightarrow A\\ (P1\rightarrow A)+(P2\rightarrow A)=P1P2\rightarrow A\\ (P1'+A)+(P2'+A)=(P1P2)'+A\\ P1'+P2'+A=P1'+P2'+A\\

Therefore ans should be C

 

answered Jun 30, 2019 by (16,750 points)
you are correct bhai ,I have made mistake in this question
commented Jun 30, 2019 by (3,730 points)
You r correct bro.
commented Jun 30, 2019 by (34,820 points)
In the first one, shouldn't there be a term (P1' + P2')A? Where did that go?
commented Jun 30, 2019 by (5,270 points)
I have given the final result.
Do like this: P1'P2'+P1'A+P2'A+A  
                      P1'P2'+P1'A+(P2'+1)A
                      P1'P2'+P1'A+A
                      P1'P2'+(P1'+1)A
                      P1'P2'+A
commented Jun 30, 2019 by (16,750 points)
+1 vote
C is the answer
answered Jul 1, 2019 by (870 points)
For problem 11
commented Jul 1, 2019 by (34,820 points)
@tsnikhilsharmagate2018 none of these both are satisfiable.
commented Jul 1, 2019 by (870 points)
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