Is Implying an Implication in Nested Quantifier Problem Correct?

[SamitC]

The problem statement:
Suppose P(x). And if I want to write "For exactly one x, P(x) then:
If we write ∃x [P(x) ∧ ∀y {P(y) → x = y}]. This is as per answer in the books.
Now, suppose P(y) is false. It will still mean x = y.
Shouldn't it be ∃x [P(x) ∧ ∀y {P(y) ↔ x = y}]

Same problem with ∃x [P(x) ∧ ∀y {x ≠ y → ¬ P(y)}]

Thanks

 

[FactChecker]

The problem statement:
Suppose P(x). And if I want to write "For exactly one x, P(x) then:
If we write ∃x [P(x) ∧ ∀y {P(y) → x = y}]. This is as per answer in the books.
Now, suppose P(y) is false. It will still mean x = y.
Shouldn't it be ∃x [P(x) ∧ ∀y {P(y) ↔ x = y}]

The backwards implication is redundant because you already have P(x). So if y=x, you already know P(y).

 

[SamitC]

The backwards implication is redundant because you already have P(x). So if y=x, you already know P(y).

Hi,
Thanks for your reply.
Now i am confused with another example:
Suppose L (x,y) means "x loves y". Now to write "There is someone who loves no one besides himself or herself". The answer in the book is:
∃x ∀y {L(x,y) ↔ x = y}]
Here what is the need for the Bi-conditional or backward implication?
Thanks

 

[FactChecker]

Hi,
Thanks for your reply.
Now i am confused with another example:
Suppose L (x,y) means "x loves y". Now to write "There is someone who loves no one besides himself or herself". The answer in the book is:
∃x ∀y {L(x,y) ↔ x = y}]
Here what is the need for the Bi-conditional or backward implication?
Thanks

Your earlier question modified answer was not wrong, just redundant. So the book answer was also right (without being redundant).
You can include the first part of the "and" condition, or you can use "↔", or you can do both and be redundant.
For this new example, it is fine to say:
∃x ∀y {L(x,y) ↔ x = y}
This second answer can also be redundant and say
∃x ∀y {L(x,x) ∧ L(x,y) ↔ x = y}
Or you could remove the redundancy another way by saying
∃x ∀y {L(x,x) ∧ L(x,y) → x = y}
All three versions are correct, one is redundant.

 

[sysprog]

The problem statement:
Suppose P(x). And if I want to write "For exactly one x, P(x) then:
If we write ∃x [P(x) ∧ ∀y {P(y) → x = y}]. This is as per answer in the books.

That answer is correct.

Now, suppose P(y) is false. It will still mean x = y.

If P(x) is true, then if P(y) is false, x cannot equal y --
(P(x) ∧ ¬ P(y)) → ¬ (x = y) can be derived via identity elimination.

Shouldn't it be ∃x [P(x) ∧ ∀y {P(y) ↔ x = y}]

From P(y) → x = y you can derive ¬ (x = y) → ¬ P(y) via modus tollens, but you cannot, merely from the fact that if x = y then everything true of x is also true of y, derive
(P(x) ∧ ¬ (x = y)) → ¬ P(y).

Same problem with ∃x [P(x) ∧ ∀y {x ≠ y → ¬ P(y)}]

This too is correct, and is equivalent to the book answer.
∃x [P(x) ∧ ∀y {P(y) → x = y}] ↔ ∃x [P(x) ∧ ∀y {x ≠ y → ¬ P(y)}] --
∃x [P(x) ∧ ∀y {P(y) → x = y}] → ∃x [P(x) ∧ ∀y {¬P(y) V x = y}] via material implication,
(¬ P(y) V x = y) → (x = y V ¬ P(y)) via commutation,
(x = y V ¬ P(y)) → (¬ (x = y) → ¬ P(y)) via conjunctive syllogism ...

The backwards implication is redundant because you already have P(x).

The biconditional being used instead of simple implication is an incorrect symbolization.

So if y=x, you already know P(y).

If P(x) and y=x, then P(y) is derivable via identity substitution, but it's not part of a correct symbolization.

Hi,
Thanks for your reply.
Now i am confused with another example:
Suppose L (x,y) means "x loves y". Now to write "There is someone who loves no one besides himself or herself". The answer in the book is:
∃x ∀y {L(x,y) ↔ x = y}]

Taking "besides" to mean "other than", this answer presumes that the someone loves himself or herself. If that is an enthymemic premise, it should be stated explicitly, before symbolization. The symbolic statement presented is a correct symbolization of "there is someone who loves himself or herself and loves no-one else".

Here what is the need for the Bi-conditional or backward implication?
Thanks

The biconditional is necessary if we mean to say that the someone loves himself or herself and no-one else. Simple implication would say there exists a someone who loves no-one other than himself or herself",.but would not say "... who loves himself or herself and who loves no-one ..." -- the inclusion of the self-love premise enthymemically and not explicitly renders the correctness of the symbolization open to question.

Your earlier question modified answer was not wrong, just redundant.

It was incorrect.

So the book answer was also right (without being redundant).

It was correct.

You can include the first part of the "and" condition, or you can use "↔", or you can do both and be redundant.

You can use the → symbol in a correct symbolization, and the ↔ symbol in an incorrect one.

For this new example, it is fine to say:
∃x ∀y {L(x,y) ↔ x = y}

Assuming the inclusion of the enthymemic self-love premise, this is correct.

This second answer can also be redundant and say
∃x ∀y {L(x,x) ∧ L(x,y) ↔ x = y}

This is an incorrect symbolization because of the superfluity and because the scopes of the binary operators are ambiguous. Resolving the ambiguity via distribution of the conjunction:
∃x ∀y {L(x,x)} ∧ ∃x ∀y {(x,y) ↔ x = y} The first term of the conjunction as distributed has a universal quantifier over y without predicating on y. Resolving the ambiguity via parenthesis grouping of the disjunction as the first term of the biconditional statement:
∃x ∀y {(L(x,x) ∧ L(x,y)) ↔ x = y} This would mean that there exists an x such that for all y, x loves x and x loves y, if and only if x is y.

Or you could remove the redundancy another way by saying
∃x ∀y {L(x,x) ∧ L(x,y) → x = y}

This formulation is also not well-formed.

All three versions are correct, one is redundant.

The first, and only the first, if and only if the enthymemic self-love premise is included, is correct.

 

[FactChecker]

@sysprog makes a good point. I stand corrected.
Even though the following two statements, taken as a whole, may indicate the same thing, they are not really equivalent statements:

This second answer can also be redundant and say
∃x ∀y {L(x,x) ∧ L(x,y) ↔ x = y}
Or you could remove the redundancy another way by saying
∃x ∀y {L(x,x) ∧ L(x,y) → x = y}

No part of the expression should imply more than that part merits. Just because one part of an expression makes an implication does not mean that the same implication can be added to other parts. (I am not sure that I have expressed this correctly, but I hope that the meaning is clear.)