Predicates, English to Symbolic Notation

Printable View

February 3rd 2010, 09:21 PM
Alterah

Predicates, English to Symbolic Notation

Hello, I am wanting assistance/confirmation that I am understanding how to the do the following correctly. I need to turn a statement into symbolic notation. I have:

S(x): "x is a spy novel"
M(x): "x is a mystery"
N(x): "x is a nonfiction book"
B(x,y): "x is better than y"

Statement: Some mysteries are better than all spy novels and all nonfiction books.

My attempt:∃x∀y[M(x) ^ N(y) -> B(x,y)] ^ ∃x∀z[M(x) ^ N(z) -> B(x,z)]

I realize this might be more bulky than necessary, but it makes sense to me. Did I do this correctly? Thanks!
February 3rd 2010, 11:12 PM
bmp05

My guess is that you need something like:

$ (\exists x)(\forall y)[B(x, y) | x = M(x) \wedge y = S(y) \wedge N(y)] $

"There exists at least one mystery book that is better than all spy and non-fiction books."
February 4th 2010, 12:19 AM
Grandad

Hello Alterah

Quote:

Originally Posted by Alterah

Hello, I am wanting assistance/confirmation that I am understanding how to the do the following correctly. I need to turn a statement into symbolic notation. I have:

S(x): "x is a spy novel"
M(x): "x is a mystery"
N(x): "x is a nonfiction book"
B(x,y): "x is better than y"

Statement: Some mysteries are better than all spy novels and all nonfiction books.

My attempt:∃x∀y[M(x) ^ N(y) -> B(x,y)] ^ ∃x∀z[M(x) ^ N(z) -> B(x,z)]

I realize this might be more bulky than necessary, but it makes sense to me. Did I do this correctly? Thanks!

This answer is close (although I think you intended your first N to be an S), but the problem is that it means:
Some mysteries are better than all spy novels, and some mysteries are better than all nonfiction books.

and these might not be the same mysteries. Whereas the original statement means that the same mysteries are better than both of the other types of book.

So you just need to combine these into a single expression, something like:
$\exists x\;\forall y, z \Big[M(x)\land\big(S(y) \Rightarrow B(x,y)\big) \land \big(N(z) \Rightarrow B(x,z)\big)\big]$

Grandad

PS See my amended version in posting #7
February 4th 2010, 12:38 AM
bmp05

Hi Grandad, this is fascinating- can you give us a walk-through of what this means?

$ (\exists x)(\forall y)[M(x) \wedge S(y) \Rightarrow B(x, y)] \equiv (\exists x)(\forall y)[M(x) \wedge (S(y) \Rightarrow B(x, y))] $

So, if M(x) and S(y) then B(x, y) is the same as M(x) and, if S(y) then B(x, y)? (without the quantors)

And what I wrote above isn't a statement it's an intensional representation of the set B(x, y)?
February 4th 2010, 02:50 AM
Grandad

Hello bmp05

Quote:

Originally Posted by bmp05

Hi Grandad, this is fascinating- can you give us a walk-through of what this means?

$ (\exists x)(\forall y)[M(x) \wedge S(y) \Rightarrow B(x, y)] \equiv (\exists x)(\forall y)[M(x) \wedge (S(y) \Rightarrow B(x, y))] $

So, if M(x) and S(y) then B(x, y) is the same as M(x) and, if S(y) then B(x, y)? (without the quantors)

And what I wrote above isn't a statement it's an intensional representation of the set B(x, y)?

First, let's establish that, given that $\land$ has precedence over $\Rightarrow$ :
$p \land q \Rightarrow r$
means:
$(p \land q) \Rightarrow r$
Then I think you're asking whether:
$(p \land q) \Rightarrow r$
is logically equivalent to:
$p \land (q \Rightarrow r)$

If so, the answer is: no. A truth table will show you immediately that the two are not equivalent.

If you're asking what the expression in my first post means, then it's this:
There's an $x$ such that for all $y$ and $z$ , ( $x$ is a mystery) and (whenever $y$ is a spy novel $x$ is better than $y$ ) and (whenever $z$ is a nonfiction book $x$ is better than $z$ ).

I think that means the same as "Some mysteries are better than all spy novels and all nonfiction", doesn't it?

Grandad
February 4th 2010, 03:34 AM
bmp05

Thanks Grandad, I guess, what I was wondering was how B(x, y) effects the proposition?
So does B(x, y) return True for 'x' and False for 'y'? I can't put my finger on it, but it seems strange that you can have a proposition, like $S(n) \Rightarrow B(x, y)$ ? S(n) implies there is 'something' better than it!

$S(n) \Rightarrow B(x, y) \Leftrightarrow \neg S(n) \vee B(x, y)$
Would you say this as "something (x) is better than Spy novels."
February 4th 2010, 05:36 AM
Grandad

Hello bmp05

Quote:

Originally Posted by bmp05

Thanks Grandad, I guess, what I was wondering was how B(x, y) effects the proposition?
So does B(x, y) return True for 'x' and False for 'y'? I can't put my finger on it, but it seems strange that you can have a proposition, like $S(n) \Rightarrow B(x, y)$ ? S(n) implies there is 'something' better than it!

$S(n) \Rightarrow B(x, y) \Leftrightarrow \neg S(n) \vee B(x, y)$
Would you say this as "something (x) is better than Spy novels."

$B(x,y)$ is an example of a two-place predicate, since it takes two parameter values, $x$ and $y$ . But it simply has a single Boolean value, True or False, like any other predicate, once you supply it with the correct number of parameters.

Another example of a two-place predicate might be:
$L(x, y)$ meaning $x$ loves $y$ .

So, for example, if $x =$ Grandad and $y =$ Curry, then $L(x,y)$ has the value True. But if $x$ = Grandad and $y$ = Pop Music, then $L(x,y)$ has the value False. (Sad, aren't I?)

Incidentally, I have realised that we don't need the $z$ in my expression. It works with:
$\exists x\;\forall y \Big[M(x)\land\big(S(y) \Rightarrow B(x,y)\big) \land \big(N(y) \Rightarrow B(x,y)\big)\Big]$

which can be written in English as:
There's at least one book, $x$ , such that whenever we choose another book, $y$ , $x$ is a mystery, and if $y$ a spy novel then $x$ is better than $y$ , and if $y$ is a nonfiction book then $x$ is better than $y$ .

In other words:
There's at least one book, $x$ , such that whenever we choose another book, $y$ , $x$ is a mystery, and if $y$ a spy novel or a nonfiction book then $x$ is better than $y$ .

which can be symbolised more simply as:
$\exists x\;\forall y \Big[M(x)\land \Big(\big(S(y) \lor N(y)\big) \Rightarrow B(x,y)\Big)\Big]$

Grandad