Because according to this book it does:


Page 64 - > Quantifiers and scope - paragraph 4.

I personally think that the former translates into, 'there exists an x, such that if Ax, then Bx', and the latter saying 'if an x is Ax then Bx'.

In this specific example, the symbolization key is:

UD: people

Gx: x can play guitar

Rx: x is a rock star

l: Lenny

So here, it's:

∃x(Gx -> Gl)

∀xGx -> Gl

So for me the former is stating 'There exists someone such that if that someone plays guitar, then so can Lenny'. Where the latter is stating 'if everyone can play the guitar, then so can Lenny'.

So, the former seems to be picking out a specific member of a set, not just any member, but some particular person. The latter seems to be referring to every member of a set.

  • 1
    The equivalence is proved in predicate logic. See this post. Mar 2 '16 at 19:36
  • Obviously, the for-all case implies the there-exists case. The other way around is not true. Can you make the philosophical problem you're trying to solve more clear? Mar 2 '16 at 19:45
  • @JamesKingsbery - as per Jo's answer below, the title of the question is WRONG. Mar 2 '16 at 20:30
  • This is related to what Raymond Smullyan calls the Drinker's Principle (sometimes called the Drinker's Paradox). "There is somebody in the pub such that if that person is drinking then everyone is drinking." It is true if and only if the pub is non-empty. It holds in standard predicate logic, but not in free logics.
    – Bumble
    Mar 2 '16 at 20:57
  • Where in the paragraph you are linking to is there a variable x on both the sides of the implication? Mar 3 '16 at 9:03

You're right, this is extremely counterintuitive. However, remember, that statements are equivalent in the case that they have the same truth conditions.

∃x(Gx -> Gl)  

is true if there exists someone who is not a guitarist or if Lemmy is a guitarist. It is false just in the case that Lemmy is not a guitarist and all other people are.

∀xGx -> Gl 

is true in the same cases and false in the same cases.

I have to admit, it took me a long time to think through this one. It might be easier to remember that A -> B is equivalent to !A V B (with ! for NOT)

So one statement is

 ∃x(!Gx V Gl)  

One statement is

!∀xGx V Gl

In a world where Lemmy is a guitarist, both statements are true. In a world where Lemmy is not a guitarist, the truth value of the first reduces to ∃x!Gx (there exists someone who is not a guitarist) and the second to !∀xGx (not all people are guitarists). Those are equivalent.

IMPORTANT As Jo mentioned, ∃x(!Gx -> Gl) and ∃x(!Gx -> Bx) are quite different. In one, the consequent is a constant, in the other it is a variable. That DOES make a difference.


We can prove it intuitively considering the possible cases (of course, I'll refer to the correct example of Magnus' text: page 64, changing l with a for readibility).

(A) ∀xGx → Ga is true.

We ave two possible sub-cases:

(Ai) ∀xGx is true; thus (truth-table for conditional) also Ga is true.

If Ga is true, then also the conditional Gx → Ga is, and thus also ∃x(Gx → Ga) is true.

(Aii) ∀xGx is false; thus, for some value c for x, Gc is false.

Thus Gc → Ga is true (truth-table for conditional) and so ∃x(Gx → Ga) is true.

(B) ∀xGx → Ga is false.

This can happens only when ∀xGx is true and Ga is false. But if ∀xGx is true, then Gc is true for any value c for x; thus, Gc → Ga is false (truth-table for conditional) for any value c, and so ∃x(Gx → Ga) is false.

Conclusion: the two formuale are equivalent.

Of course, the equivalence holds also with any other predicate letter in place of the G in the consequent (i.e.Ga).

The key fact is:

∀xGx → R is equivalent to ∃x(Gx → R) provided that x is not free in R.


Note that the quoted passage of the textbook does not deal with the formulas from your quote.

It does not speak about "Bx" but about "Gl" with l being a fixed individual (Lemmy) not a variable.

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