Predicates and Models ... Goldrei

[Math Amateur]

I am reading Goldrei's book: "Propositional and Predicate Calculus: A Model of Argument", Chapter 4: Predicates and Models.

I need help in understanding Goldrei's answer to Exercise 4.5 (a) ...

Exercise 4.5 together with Goldrei's solution reads as follows:

Can someone please explain exactly how x < y can be represented by

$( \exists z \ (x * z) = y \land \neg x = y )$






Help will be much appreciated.


Goldrei mentions the language given above ... that reads as follows:

Peter

 

[Mark44]

Your scanned or photographed images are difficult to read, especially because some symbols don't show up.
In the image for ex. 4.5, it says "(involving *, e and , interpreted respectively by 1 (?), 0, and equality)." It looks like whatever was written after "and" doesn't show up. Possibly it was in a different color. Also, is the first symbol after "respectively by" the digit 1 or the symbol |? It looks like the latter, but I don't know what that means in this context.

Can someone please explain exactly how x < y can be represented by
$(\exists z (x * z) = y \wedge \neg x = y)$

The text in the second image is smaller, so more difficult to read. Several symbols in the explanation in this image also don't show up, so I still don't understand what is meant by *.

 

[TeethWhitener]

* represents addition (I agree it's very tough to read). In more standard mathematical language:
$$( \exists z \ (x * z) = y \land \neg x = y )$$
says that there's a $z$ such that $x+z=y$ and $x\neq y$. Alternatively, the other explanation
$$( \exists z \ (x * z) = y \land \neg z = \mathbf{e} )$$
says that there's a $z$ such that $x+z=y$ and $z\neq 0$.

 

[SSequence]

Alternatively, the other explanation
$$( \exists z \ (x * z) = y \land \neg z = \mathbf{e} )$$
says that there's a $z$ such that $x+z=y$ and $z\neq 0$.

For clarity of notation, I think it is probably better to have the $z$ variable to be outside all of the brackets.

===========


One thing to note about these formulas is that since the predicate is of the form $less(x,y)$, any variables other than $x$ and $y$ that are used must be bound. This is true in general regardless of the specific predicate so it is worth a mention [since checking for it can sometimes help identify a mistake in more complicated formulas].

Also, I agree that it is not difficult to understand the solution if you write it with more familiar symbols.
(1) $\exists z [x+z=y] \land (x\neq y)$
(2) $\exists z [(x+z=y) \land (z\neq 0)]$

In the first formula if you look separately at:
$\exists z [x+z=y]$
then this is just the representation of predicate $x \leq y$. So the first formula is just saying that $x<y$ iff we have (i) $x \leq y$ and (ii) $x \neq y$.

 

[TeethWhitener]

For clarity of notation, I think it is probably better to have the z variable to be outside all of the brackets.

I agree, but that’s how it was written in the book. Not sure why.

 

[SSequence]

I agree, but that’s how it was written in the book. Not sure why.

Given the first image posted in OP, the book does put $\exists z$ outside all of the brackets in the second formula!

Anyway, it also occurred to me that formula-(1) in post#4 also works [as in correctly describes "less than" predicate] even if the domain of quantification is $\mathbb{R}$ or $\mathbb{Q}$. However, while formula-(2) works for $\mathbb{N}$, it doesn't work for $\mathbb{R}$ or $\mathbb{Q}$.

 

[SSequence]

Anyway, it also occurred to me that formula-(1) in post#4 also works [as in correctly describes "less than" predicate] even if the domain of quantification is $\mathbb{R}$ or $\mathbb{Q}$.

Sorry for the mistake. Formula-(1) in post#4 doesn't represent the "less than" predicate either [if the domain of quantification is $\mathbb{R}$ or $\mathbb{Q}$]. It represents the "not equal" predicate in that case.

In fact, if we look at it:
(1) $\exists z [x+z=y] \land (x\neq y)$
Then basically it is just equivalent to the predicate $x\neq y$ if the domain of quantification is $\mathbb{R}$, $\mathbb{Q}$ or $\mathbb{Z}$.


And while we are at it, we might also look at the second formula:
(2) $\exists z [(x+z=y) \land (z\neq 0)]$
This formula is also equivalent to the predicate $x\neq y$ if the domain of quantification is $\mathbb{R}$, $\mathbb{Q}$ or $\mathbb{Z}$.