Hello everyone Originally Posted by

**XDemonoid** Ok, I need to convert an English sentence to predicate logic.

The domain is C , the set of Olympic competitors.

f(x,y) = x is faster than y

s(x,y) = x is the same person as y.

b(x,y) = x beats y

w(x) = x is the winner.

Translate the following English statements into precise symbolic notation. **Only use the predicates and domain defined above.**

1. There is exactly one fastest competitor.

2. If every competitor beats exactly one other competitor, then there is no winner.

Thank you guys

First of all, you don't need to keep saying$\displaystyle x \in C$, $\displaystyle y \in C$, etc

because $\displaystyle C$ is the domain, the universe from which all elements are chosen. (Which at least makes things a bit easier.)

Then for number 1 (and this is complicated, so bear with me) I think you need to say:There is a competitor, $\displaystyle x$, that is faster than all others ($\displaystyle y$), and whenever we find a competitor, $\displaystyle z$, that is faster than all others, then $\displaystyle x$ and $\displaystyle z$ are the same person.

The second half is there to ensure that there is only *one* $\displaystyle x$ that is faster than all others.

Now to begin translating this. The first phrase 'There is an $\displaystyle x$' is obviously $\displaystyle \exists x$. But the phrase 'all others ($\displaystyle y$)' is a bit tricky. You can't just say $\displaystyle \forall y$, because that will include $\displaystyle x$, and I think we must assume that $\displaystyle x$ cannot be faster than itself. So we must include somewhere that $\displaystyle y$ and $\displaystyle x$ are not the same person.

So the first section, that states the existence of a fastest competitor is:$\displaystyle \exists x\forall y[\neg s(x,y) \Rightarrow f(x,y)]$

Then the bit that makes $\displaystyle z$ a fastest competitor is, in the same way:$\displaystyle \forall y[\neg s(z,y) \Rightarrow f(z,y)]$

We now need to say that whenever this second statement is true (in other words, for all $\displaystyle z$), then $\displaystyle x$ and $\displaystyle z$ are the same person. This will look like this:$\displaystyle \forall z \forall y[ \neg s(z,y) \Rightarrow f(z,y)]\Rightarrow s(x,z)$

We now join this to the first phrase with a logical 'and':$\displaystyle \exists x\Big[\Big(\forall y[\neg s(x,y) \Rightarrow f(x,y)] \Big)\land \Big(\forall z \forall y[ \neg s(z,y) \Rightarrow f(z,y)]\Rightarrow s(x,z)\Big)\Big]$

I don't know whether all the brackets are strictly necessary, or whether there's a simpler expression, but I think that works.

If I have time, I'll have a look at number 2.

Grandad