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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. $((R\vee Q)\wedge(P\vee \neg Q)\wedge (\neg R\vee P))$

### 1 comment

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

*

why QP remains here =RP+R~Q+PQR+~P~QR+QP

edited
sorry for that i did wrong ,ans will be (2) and thank you

((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

### 1 comment

I think there is some mistake here,

((R ∨ Q) ∧ (P ∨ ¬Q))

(R ∨ Q)

(P ∨ ¬Q)

cancel the  Q and ¬Q

till this point it is correct but (R ∨ Q) ^ (P ∨ ¬Q) these two are in AND with each other so result should be R ^ P. same in above case too.

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.

### 1 comment

What will be exact answer of this question
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.