Is the answer b??

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1 vote

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

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

1 vote

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

*

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.

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.