grammar & semantics

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Reply Tue 8 Sep, 2009 01:12 pm
Hello there,

I certainly don't have the ability, as of yet, to contribute meaningfully at this forum in many ways, but I hope my history of posts, while seeking help, do show that I am willing to do the work needed. Without that history it might seem that right now I am fishing for answers, and an easy way to complete some homework.

That said, I have five questions that I am lost on...

Let "language L" be simply to describe quantifier language.
Let a,b,c... be variables, and
Let #, % be formulas...

1. Find a formula # of L such that # is not a sentence but contains a part that is a sentence...

2. Find a formula # of L such that # contains occurences of a variable a that are bound and at least one occurence of a that is free.
#=
a=

3. Find a sentence ! of L such that ! contains no proper part that is a sentnce but does contain at least one binary sentential connective.

!=

4. If #= [Fx->[[upsidedownAxFx->Fx]->Fx]] and a='x' and b=y, then #[a/b] equals...

5. Find a sentence ! of L such that ! contains each of the logical signs in L at least once....

Is there anyone that can provide me some reference material on grammer and semantics, some sort of tutorial, perhaps some guidance on these question. As I am always, I am thankful for the help I receive here.


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Folks, I have been working through these 5 questions for more than the expected time, and while I have received a lot of help from people here and elsewhere, I am at the point where I could use either some validation of these answers, some request for further clarification, or some future help. THanks as always.

1. Find a formula ψ such that ψ is not a sentence but contains a part that is a sentence...


AaFa v Gb

This formula is not a sentence because "b" is a free occurrence, but it contains part that is a sentence (AaFa) because the variable is bound in both cases. There's no free occurrence, and so this works for #1.


Is my understanding sound?





2. Find a formula ψ such that ψ contains occurrences of a variable α that are bound and at least one occurrence of a that is free.


This was offered as a slight modification of my answer: Ax[Ey[Hy v Gz] --> Jxy]


There does seem to be the free occurrence of "z," but aren't there also two bound occurrences of variable "y", and one occurrence of variable "y" that is free? In which case this formula would work?

3. Find a sentence ? such that ? contains no proper part that is a sentence but does contain at least one binary sentential connective.

Aa(Fa v Ga)


4. If ψ= [Fx->[[AxFx->Fx]->Fx]] and a='x' and b=y, then ψ[a/b] equals...


I haven't come across the sign in "a/b," but we seem to think it means that all free occurrences of "a" will be replaced by variable "b." Based on #2, my understanding of bound and free is shaky, but aren't all these variables bound, in which case I wouldn't do anything?

5. Find a sentence ? that ? contains each of the logical signs (if/then, or, and, if and only if, not, universal quantiifer, existential quantifier...) at least once....


I don't know.



 
Horace phil
 
Reply Mon 14 Sep, 2009 03:25 pm
@Horace phil,
Ok, I've been doing a lot of reading, and these are my conclusions for the first two. Can someone tell me if I am "getting" them. As for the other three I am still working on them. Any further explanation is appreciated.

1. Ok. So I need to find a formula that is not a sentence, but contains a part that is a sentence.

A "sentence" is basically a formula that contains no free occurence of a variable.

So this has been offered:

AaFa v Gb

It is not a sentence because B is a free occurence, but it contains part that is a sentence (AaFa) because the variable is bound in both cases. There's no free occurence, and so this works for #1. Is my understanding sound?

2. Now I need to find a formula which contains occurences of a variable that are bound, and at least one that is free. So something might have been offered, but how bout this one? upside downAx[backwardsEy[Hy or Ga] therefore Jxy]
 
 

 
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