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I have to convert the given predicates to English sentences. I have tried from my side but I am very confused (as there can be many forms to represent a statement and also that my English statement might be wrong). Here are the predicates and English statements. Please tell me if there can be a better answer for any of these?

  1. $\forall x\exists y R(x,y)$

    For every $x$ there is at least one $y$ such that $R(x,y)$

  2. $\exists x\forall y R(x,y)$

    For every $x$ there is a single $y$ such that $R(x,y)$


    There is an x $x$ for which there is a single $y$ such that $R(x,y)$

  3. $\forall x(\neg Qx)$

    For every $x$ $Q(x)$ is false; or $Q(x)$ is false for every $x$

  4. $\exists y(\neg P(y))$

    There is at least one $y$ for which $P(y)$ is false.

Please help me out. Thanks.

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All but the second one are correct. The second one says that "there is at least one $x$ with the property that for every $y$, $R(x,y)$ is true", which is very different from what you wrote. It is also important to keep the order right: "there is an $x$ such that for every $y$, $R(x,y)$ is true" says something different from "for every $y$, there is an $x$ such that $R(x,y)$ is true". After all, while everyone has a mother, there isn't anyone who is everybody's mother. – Arturo Magidin Sep 22 '11 at 16:28
some people can argue about your last statement – Ilya Sep 22 '11 at 16:30
@Gortaur: People can argue all they want. If they argue that there is someone who is everybody's mother, then they'd be wrong. – Arturo Magidin Sep 22 '11 at 16:41
@ArturoMagidin: Thanks. Will there only be a single y that satisfies all the values of x in 2 ? – Fahad Uddin Sep 22 '11 at 17:11
@Akito: No, you are misinterpreting it again. Number 2 says that there is at least one $x$ for which every $y$ satisfies $R(x,y)$. For example, if $R(x,y)$ meant "$x$ is less than or equal to $y$", then the statement would say "there is an $x$ which is less than or equal to each and every $y$". There is a single value of $x$ at issue. – Arturo Magidin Sep 22 '11 at 17:19
up vote 2 down vote accepted

You have only mistake in number 2 as Arturo has pointed. The right expression is: There exists at least one $x$ such that for all $y$ the formula $R(x,y)$ is true. What you wrote is encoded as $\forall x\exists yR(x,y)$.

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You may want to change "mistake in 1 number 2" to "one mistake, in number 2". It's confusing, otherwise... – Arturo Magidin Sep 22 '11 at 16:40
Thanks, just mentioned that the line is nr.2 rather then nr.1 and haven't deleted first version. This numeration is a bit confusing, so maybe it should be fixed in the question? – Ilya Sep 22 '11 at 16:45
I fixed the original post. The problem is that if you have a new paragraph and the lines are not indented, the "counter" restarts at 1. By adding some spaces, you can get the counter to not reset. – Arturo Magidin Sep 22 '11 at 16:46
@ArturoMagidin: And one more comment - your answer was first, so I would appreciate if you formulate it as an answer, then I can delete mine since they are quite the same. – Ilya Sep 22 '11 at 16:47
No need; your answer is good. – Arturo Magidin Sep 22 '11 at 16:53

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