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Assume

$A(x) = x$ is an American.

$D(y) = y$ is a dream.

$H(x,y) = x$ has $y$.

Then, Convert below first order logic to English statements :

  • $∀x ∃y \left ( A(x)\rightarrow D(y) ∧ H(x,y) \right )$

I tried to translate this as "Every American has his own set of dreams".

  • $∀x ∃y \left ( A(x) ∧ D(y) \rightarrow H(x,y) \right )$

Not getting how is this pronounced ?

For this, I guess it is like "For all x if x is an American and there exists some y, such that y is a dream then x has y".

How much both of them are correct ?

3 Answers 3

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You're not quite right on the first one. Here's a transition from "predicate English" to "proper English":

  • For all $x$, there exists a $y$, such that "x is an American" implies ("y is a dream" and "x has y")
  • For all $x$, if $x$ is an American then there exists a $y$ such that $y$ is a dream and $x$ has $y$.
  • For all Americans $x$, there is a dream $y$ that $x$ has.
  • Every American has a dream.

As for the second, here's what it looks like at the start:

  • For all $x$, there exists a $y$, such that ("x is an American") and ("y is a dream") implies "x has y".
  • For all $x$, there exists a $y$, such that if $x$ is an American and $y$ is a dream, then $x$ has $y$.

This is where it gets weird. Because it's saying that for every $x$ there's a $y$ that makes the inner part true. But the inner part can be true if:

  • $x$ is not American.
  • $y$ is not a dream.
  • $x$ has $y$.

In particular, it says that for every American $x$, there's something $y$ that, if $y$ is a dream, then $x$ has. But it's fine if it's just not a dream, then it doesn't matter if $x$ has it or not. If we use the equivalence of $A \rightarrow B$ and $\lnot(A \land \lnot B)$, then we get something like:

  • For every American $x$, there is something $y$ that, if $x$ doesn't have it, isn't a dream.

And that's incredibly hard to put into "real" English. Something like:

  • For every American there's something that, if it's a dream, they have it.

And that's about as good as I can get.

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    Thanks for the clear explaination :) – 2017-02-02
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The first one is pretty close. I would say "Every American has a dream," because we only know that there exists one $y$.

The second one is weird. If there exists any y that is not a dream, then it is vacuously true. Literally it's something like "For every American there is a thing y such that if y is a dream, then the American has that dream." But y could just not be a dream.

Edit: I think the second one is logically equivalent to "Either there exists something that is not a dream, or every American has a dream." Formally, ($A(x) \wedge D(Y) \implies H(x,y)$) is equivalent to $(\neg(A(x) \wedge D(y)) \vee H(x,y))$, and subsequently $(\neg A(x) \vee \neg D(y) \vee H(x,y))$.

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    Yes, I agree with you. 2nd is kinda weirdo. It says like "Every American has something y(Where y could be a dream or anything else)". – 2017-02-02
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What ∀x∃y(A(x)∧D(y)→H(x,y)) literally means in English, according to your given information is that for all Americans, there exists a dream y such that if every American x has at least one dream y, then they have their own dreams. In other words, every American with a dream sets has their own set of dreams.

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    I've edited with some more info. Can you check it ? – 2017-02-02
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    I don't love that. The y is not automatically the American's dream until we know that it is a dream at all. – 2017-02-02
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    Hmm but we need to conditions in order to satisfy this statement. – 2017-02-02
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    @JonGarrick we are still discussing it. I agree with Nathaniel what I said doesn't come off as clear. – 2017-02-02
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    @Oliver821 Yah !! I can see that :-) – 2017-02-02
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    @JonGarrick Look at Nathaniel's approach I will write back when I come up with a revised version of an answer to part b :) – 2017-02-02
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    @Oliver821 Yes !! Thanks :) – 2017-02-02