Let $F(x,y)$ denote the statement $x+y = 0$, where the domain of discourse for both the variables is $\mathbb{R}$. Write the truth values of the following with justification.

  1. $\exists\, y: \forall x\ F(x,y)$
  2. $\forall x: \exists\, y\ F(x,y)$

The verbal forms of the above expression are

  1. "There is some value of $y$, which can be used for all values of $x$ in order to satisfy the given relation."
  2. "For every $x$ there is some $y$ satisfying given relation."

The first statement is false and the second statement is true.

I am not getting the actual difference between the statements. What importance does the order of nested quantifiers have? Are the truth values right?


Let's take a real-life example. $x$ and $y$ represent people; $F(x, y)$ is true if and only if $x$ and $y$ are friends. Now let's examine the statements:

  1. $\exists\,y : \forall x\ F(x, y)$: This statement states that there is a person $y$ such that for all people $x$, $F(x, y)$ is true or in other words, $x$ and $y$ are friends. In normal English we would say that $y$ is friends with everyone.
  2. $\forall x: \exists\,y\ F(x, y)$: This statement states that for all people $x$, there is a person $y$ such that $F(x, y)$ is true or in other words, $x$ and $y$ are friends. In normal English we would say that everybody has a friend.

Clearly you can see that the two statements are different. In this example, we cannot say anything about their truth values but with the original problem, I think the distinction should be clear. To make sense of the original problem, replace 'person' with 'number' and 'friend' with 'additive inverse' in all the explanation above.

The truth values you mention are certainly correct (can you now see why?), and the order of quantifiers does indeed make a lot of difference.

  • $\begingroup$ Brilliant answer. Not only i got the answer but also the example made my concepts strong. $\endgroup$ – Kavita Sahu May 24 '16 at 9:30
  • $\begingroup$ @KavitaSahu Thanks! $\endgroup$ – shardulc May 24 '16 at 9:35

The difference is that in the first statement, you have to fix $y$ once and for all. Once you have chosen $y$, you need it to work for every $x$, without changing $y$.

In the second one, you have to use an arbitrary $x$, and then you can choose the $y$ you want, which can (and will !) depend on $x$.

Examples :

In the first one, if you choose and $y$, may be a $x$ will work, but then $x+1$ surely won't.

In the second one, for an arbitrary $x$, you can choose $y=-x$ and it will work.

  1. Exists at least one $y$ that, for all possible values of $x$, $F(x,y)$ always holds.

You are trying to answer the following question: Is there any value $y$ that summed with any $x \in \mathbb{R}$ will always result in zero? You are asking for a specific $y$ that summed with any value $x \in \mathbb{R}$ will be equal to zero.

  1. For all possible values of $x$ there is always a value for $y$ that $F(x,y)$ holds.

You are answering the question: For each $x \in \mathbb{R}$, can I always find a value $y$ where $F(x,y)$ holds? Is it always possible for all $x \in \mathbb{R}$ to find a $y$ that summed with that specific $x$ will result in zero? In this case, you can always choose an $y$ such as $y = -x$.


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