Discrete Mathematics | Propositional Logic | Test 1 | Question: 1

Question

A Proposition is a written or uttered declarative sentence used in such a way that it is true or false, but not both. Now, Consider the following statement:
$S:$ ‘If George is a duck then Ralph is a dog and Dusty is a horse’.
 Which one of the following statements is true?
 

• A. $S$ is not a proposition.     
      
• B. $S$ is ambiguous and there are two ways to parse it.     
       
• C. $S$ is unambiguous and there is only one way to parse it.    
       
• D. $S$ is ambiguous and there are three ways to parse it.

Comments

Let, George is a duck = G, Ralph is a dog = R, and Dusty is a horse = D

the given statement can be written as,

• $(G\rightarrow R)\wedge D$

• $G\rightarrow( R\wedge D)$

because there are 2 ways to represent single statement that’s why we can say that the given statement is ambiguous.

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Sir @Deepak Poonia according to precedence order of logical connectives... And ^ having more priority than implications -> So it is unambiguous and parse with 1 way only...plj correct if i m wrong

 

 

 

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we consider precedence order when we are given propositional formula. here we only have simple statement which can be interpreted in 2 ways,and that's why it's ambiguous. Above both propositional formula represents same statement.

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@Muaaz 

Precedence of operators comes into picture when we are dealing with a “Propositional Logic Expression”, Not when we have an English statement.

In English statements, we don’t have precedence of “words”. In propositional logic expression, we have precedence of “operators”.

Answer 1

Given, statement $S$: “If George is a duck then Ralph is a dog and Dusty is a horse.”

Let assume, “George is a duck”: $G_{d}$; “Ralph is a dog”: $R_{d}$; “Dusty is a horse”: $D_{h}$

One way to represent:

If George is a duck then Ralph is a dog and Dusty is a horse $\equiv$ $(G_{d} → R_{d})$ $\wedge$ $D_{h}$

Another way to represent:

If George is a duck then Ralph is a dog and Dusty is a horse $\equiv$ $G_{d} → (R_{d} \wedge D_{h})$

So, $S$ seems ambiguous and we can represent it in two ways also.
$Ans: B$

Comments

for ambiguity we will check precedence order??

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@geeks_god Hi Thanks for asking your doubt.

For ambiguity there is no precedence checking.

I am giving you example how ambiguity coming:

“If the fish is cooked then dinner is ready and I am hungry.”

These sentence we can interpret like this:

$f → (r \wedge h)$;  $(f → r) \wedge h$

But why this is happening ? The actual statement is ambiguous; if there is a comma inserted in the sentence or ordered in a perfect manner then we may got one logical statement and it will also unambiguous.

See this, “If the fish is cooked, then dinner is ready and I am hungry.”

Unambiguous !! fish is cooked so dinner is ready and I am also hungry.

logical statement $f → (r \wedge h)$

I hope it helps !!

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@geeks_god

Precedence of operators comes into picture when we are dealing with a “Propositional Logic Expression”, Not when we have an English statement.

In English statements, we don’t have precedence of “words”. In propositional logic expression, we have precedence of “operators”.