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3 3 votes
Everyone has exactly one best friend

Are all three below same?

Let B(x, y) to be the statement “y is the best friend of x"

$ ∀x∃y(B(x, y) ∧ ∀z((z = y)→¬B(x, z))) $

$ ∀x ∃!y (B(x, y) $

$ ∀x ∃y (B(x, y) ∧ ∀z (B(x, z) → (y = z))) $

2 Answers

Best answer
7 7 votes
B is saying For all x there is exactly one y such that he is bestfriend of x. So yes here.
C is saying For all x there exist one y such that he is friend of x and if for all other person are best friends then he is same as y. So yes same as B.
A is saying some thing else For all x there exist one y such that he is friend of x and for all z if y is same as z then they cant be friends.

use Y$\neq$ Z then it saying same.
• edited by
0 0 votes

a) Statement 2 is true without any doubt since we have used the symbol for "exactly one" phrase.

b) Now coming to statement for validity of this statement 

let us have B(x,y) be true indicating that y is the best friend of x and in addition to that let us concentrate on the part:

∀z(B(x,z)→(y=z)))

For the above predicate logic to be true we have two cases :

Case 1 :

B(x,z) is false then the implication is trivially true and which also indicates only y is the best friend of x.

Case 2:

If B(x,z) is true for the whole implication to be true , then y = z must also be true and hence we are sure that if z is the best friend of x , then that "z" is nothing but y.

Hence the given statement 2 predicate logic also implies the same thing in statement 3.

c) Arguing in the similar manner , 

no problem if LHS i.e. (z=y) is false

But problem arises when it is true .

Then for implication to hold ¬B(x,z) should be true meaning B(x,z) should be false , thus suggesting that z is not friend of x meaning y is not friend of x(since z = y is true)..

But the first part of the entire logic says B(x,y) meaning y is the best friend of x which hence contradicts with the 2nd part.

Hence statement 1 is not correct according to the meaning which is required to be conveyed.

Hence statement 2 and 3 are correct only.

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