In a room there are only two types of people, namely Type 1 and Type 2. Type 1 people always tell the truth and Type 2 people always lie. You give a fair coin to a person in that room, without knowing which type he is from and tell him to toss it and hide the result from you till you ask for it. Upon asking the person replies the following “The result of the toss is head if and only if I am telling the truth” Which of the following options is correct?

A. The result is head

B. The result is tail

C. If the person is of Type 2, then the result is tail

D. If the person is of Type 1, then the result is tail

Case 1: if the person is Type 1 person then

The result of the toss is head if and only if I am telling the truth

denote it by (P↔Q)

P : The result of the toss is head

Q: I am telling the truth

“I am telling the truth” is true for him. (yes, he always tells the truth)

his statements are always true, so (P↔Q) must be true.

So P is also true.

P↔Q = TRUE—(1)

Q = TRUE—(2)

P↔TRUE = TRUE ( 2 is substituted in 1)

**P is TRUE**

Case 2: if the person is Type 2 person then

P↔Q is false (yes, he is always false)

and more over Q is false as it is also said by him.

P↔Q = FALSE—(1)

Q = FALSE—(2)

——————————-

P↔FALSE = FALSE ( 2 is substituted in 1)

**P is TRUE**

if some thing is proved for an element in the domain with out assuming any particular property of that element then it will be applicable for all.

but how did you know it was not first order logic question? In the question it is mentioned that there are people(domain) and that there are different types of people.why didnt you use statmnt like forall(x)(if x is type 1 then toss is head ^ if toss is head then x is type 1…)