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Confused between Nested Quantifiers

I asked the question about two sentences. interpreting mixed quantifier

But, I don't know the meaning difference between $$∀x∃y(\text{Cube}(x) → (\text{Tet}(y) ∧ \text{LeftOf}(x, y))),$$ and $$∃y∀x(\text{Cube}(x) → (\text{Tet}(y) ∧ \text{LeftOf}(x, y))),$$

"Every cube is to the left of a tetrahedron"
"There is a tetrahedron that is to the right of every cube"

I think these sentences have same meaning.

Is it wrong? please give me your opinion.

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marked as duplicate by Qiaochu Yuan May 16 '12 at 3:31

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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The difference between "for every x there is a y such that..." and "there is a y such that for every x ..." is the difference between "Everyone is loved by someone" and "Someone loves everyone." –  Arturo Magidin May 16 '12 at 3:17
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A better translation of the first sentence is "Every cube is to the left of some tetrahedron", to make it clear that it is not necessarily the same tetrahedron that works for every cube; whereas in the second sentence, the same tetrahedron must act as witness for every cube. –  Arturo Magidin May 16 '12 at 3:22
    
Related: this question. –  Arturo Magidin May 16 '12 at 3:30

2 Answers 2

The two sentences in your question do not have the same meaning: consider an infinite sequence $$\text{Cube}\qquad\text{Tetrahedron}\qquad\text{Cube}\qquad\text{Tetrahedron}\qquad\cdots$$ Every cube is to the left of a tetrahedron, but there is no tetrahedron that is to the right of every cube.

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I was about to write almost exactly the same answer. Ha! –  Alex Kruckman May 16 '12 at 3:16
    
@Alex: $\text{ }$:) –  Zev Chonoles May 16 '12 at 3:21
    
..I constructed your example in Tarski's world. But it says that the both sentences are true..... thehjh.com/tarski.png –  Jonghwan Hyeon May 16 '12 at 4:42

Let’s look at a simpler example, where we assume that the variables range over real numbers: $\forall y\exists x(x+y=0)$ and $\exists x\forall y(x+y=0)$. The first says that every real number has an additive inverse, which is true. The second says that there is some particular real number $-$ call it $z$, say $-$ such that $z+y=0$ no matter what $y$ is; that’s clearly false.

In general you cannot reverse $\forall$ and $\exists$.

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