GO Classes DPP | GATE CS | Discrete Mathematics| First Order Logic

Answer 1

Option A: ∃x(H(x)∧∀y(H(y)→x=y)

Could be rewritten as : - ∃x∀y(H(x)∧H(y)→x=y)

It forces us to take a single x from the domain and check for all y and if the condition is True we have exactly one person in the domain who is happy, hence True.

For two happy person or no happy person the condition would be False.

Another way to quickly get the intuition behind this option would be:-

P : ∃x(H(x)) tells us about atleast one happy person in the domain.

Q: ∀y(H(y)→x=y) tells us about atmost one happy person in the domain.

P∧Q = atleast one happy AND atmost one happy = exactly one happy person.

 

Option B: ∃x∃y(H(x)∧H(y)∧x!=y)

Even if we get a pair satisfying this condition it’s not guaranteed to be a True for exactly one person, hence False.

 

Option C: ∀x∀y(H(x)∧H(y)→x=y)

This option will be true even if we have no happy person in the domain, hence False.

 

Option D: ∃xH(x)∧∃yH(y)∧x!=y

Not even a preposition as it contains free variables (x and y) which are free to take any values from the domain, hence we can’t satisfy the condition. 

 

Answer 2

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Answer 3

Exactly one happy person = atleast one happy and atmost one happy 

Option A:  ∃x(H(x)∧∀y(H(y)→x=y))  
The blue coloured part says there exist some person x who is happy (This is atleast one happy) 
The Red Coloured part is not independent it goes together with atleast part so its read like there exist some person x who is happy and for all person y if y is also happy then x and y must be same 
So this is true 

Option B: It says there exist some person x and some person y who are happy and they are different so it means it says there exist atleast two happy person 

Option C: Looks correct but if we closely observe and assume there are no happy person in the set then also it is true, so this one is atmost one happy 

Option D: It is not a proposition it has free variable x and y