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The first order logic (FOL) statement $((R\vee Q)\wedge(P\vee \neg Q))$ is equivalent to which of the following?

  1. $((R\vee \neg Q)\wedge(P\vee \neg Q)\wedge (R\vee P))$
  2. $((R\vee Q)\wedge(P\vee \neg Q)\wedge (R\vee P))$
  3. $((R\vee Q)\wedge(P\vee \neg Q)\wedge(R\vee \neg P))$
  4. $((R\vee Q)\wedge(P\vee \neg Q)\wedge (\neg R\vee P))$
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6 Answers

1 votes
1 votes

i can write this like:

    (R+Q) (P+~Q)

    =RP+R~Q+PQ

now option (b)

(R+Q) (P+~Q) (R+P)

(PR+~QR+PQ)(R+P)

PR+~QR+PQR+PR+P~QR+PQ

(~QR+P~QR)+(PQR+PQ)+PR

~QR+PQ+PR

that is same as given

so (b) is answer


 *

1 votes
1 votes
i think answer is (2)

 ((R ∨ Q) ∧ (P ∨ ¬Q) ∧ (R ∨ P))

R ∨ Q

P ∨ ¬Q

R ∨ P

------------

cancel the Q and ¬Q only remain  (R ∨ P)  ∧ (R ∨ P)

(R ∨ P)  result 1

for

 ((R ∨ Q) ∧ (P ∨ ¬Q))

(R ∨ Q)

(P ∨ ¬Q)

-------------------

cancel the  Q and ¬Q

you will get R ∨ P which is similar to result 1
0 votes
0 votes
option b

((R∨Q)∧(P∨¬Q)∧(R∨P))((R∨Q)∧(P∨¬Q)∧(R∨P))

I have just drawn the truth tables of all, and found this match.
0 votes
0 votes
P is equivalent to Q means P is true, Q is true and P is false, Q is false.

If (R  $\vee$ Q) is true and (P $\vee$ ~Q) is true then we can conclude that (P v R) is also true.  (R $\vee$ Q) <==> (~R $\rightarrow$ Q) similarly (P $\vee$ ~Q) <==> (Q $\rightarrow$ P).

1. (~R $\rightarrow$ Q)
2. (Q $\rightarrow$ P)

$\therefore$ (~R $\rightarrow$ P) From 1 and 2 by applying Hypothetical syllogism.

(~R $\rightarrow$ P) <==> (R $\vee$ P).

So ((R∨Q)∧(P∨¬Q)∧(R∨P))((R∨Q)∧(P∨¬Q)∧(R∨P)) is true whenever  ((R∨Q)∧(P∨¬Q)) is true
Also ((R∨Q)∧(P∨¬Q)∧(R∨P))((R∨Q)∧(P∨¬Q)∧(R∨P)) is false whenever ((R∨Q)∧(P∨¬Q)) is false.
So option B.
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Answer:

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