translogi wrote: > herbzet wrote: > > Jack Higgs wrote:
> > > Hey guys,
> > > Any help you could provide with this is greatly appreciated.
> > > F = (AX)(p(s(X)) -> p(X)) and G = (EX)p(X)
> > > Where A = for all and E = exists
> > > Use as domain the set N = {0,1,2,...} of natural numbers.
> > > Having trouble with the following:
> > > 1. Describe an interpretation where F ^ ¬G is true or argue that such an > > > interpretation does not exist.
> > F ^ ~G = Ax(Psx -> Px) ^ ~ExPx transcription
> > = Ax(Psx -> Px) ^ Ax~Px
> > = Ax[(Psx -> Px) ^ ~Px]
> > = Ax[(~Psx v Px) ^ ~Px]
> > = Ax[(~Psx ^ ~Px) v (Px ^ ~Px)] distributive law
> > = Ax(~Psx ^ ~Px)
> > = Ax~(Psx v Px)
> > Let sx be x + 1. > > Let Px be x < 0.
> > Tricky: P must be a predicate that is true of no natural number.
> > > 2. Describe an interpretation where F ^ G and F ^ ¬G is true or argue > > > that such an interpretation does not exist.
> > (F ^ G) ^ (F ^ ~G) = F ^ (G ^ ~G) conjunction distributes over > > itself. > > = G ^ ~G
> > > 3. Describe an interpretation where F ^ G ^ (EX)¬p(X) is true or argue > > > that such an interpretation does not exist.
> > F ^ G ^ Ex~Px = Ax(Psx -> Px) ^ ExPx ^ Ex~Px
> > Let sx be x + 1. > > Let Px be x < 10
> Agree
> (IOnly realised later with s(x) Jack probably ment the successor > function. sx= x+1)
I think any interpretation of s(x) as a one-place function is OK for this problem, although it does suggest the successor function.
In your suggestion of s(x) = x (so s(x) is interpreted as the identity function) we have that F = Ax(Psx -> Px) comes out as meaning Ax(Px -> Px) which of course is true in every interpretation. Which is fun.
The OP's notation of P(x) is slightly ambiguous to me. It could mean that P is a one-place predicate symbol, or it could mean that P is a formula with x as a free variable.
It doesn't make any difference for this problem, I guess.
I would use 'F' for a random predicate symbol (and 'G' if I needed two), and for functions I'd use 'f' (and 'g', etc). I would save up 'P' to stand for a sentence (proposition). The OP does just the opposite, uses 'F' for a sentence and 'P' for a predicate.
> > > = Ax[(~Psx ^ ~Px) v (Px ^ ~Px)] distributive law
> > > = Ax(~Psx ^ ~Px)
> > > = Ax~(Psx v Px)
> > > Let sx be x + 1. > > > Let Px be x < 0.
> > > Tricky: P must be a predicate that is true of no natural number.
> > > > 2. Describe an interpretation where F ^ G and F ^ ¬G is true or argue > > > > that such an interpretation does not exist.
> > > (F ^ G) ^ (F ^ ~G) = F ^ (G ^ ~G) conjunction distributes over > > > itself. > > > = G ^ ~G
> > > > 3. Describe an interpretation where F ^ G ^ (EX)¬p(X) is true or argue > > > > that such an interpretation does not exist.
> > > F ^ G ^ Ex~Px = Ax(Psx -> Px) ^ ExPx ^ Ex~Px
> > > Let sx be x + 1. > > > Let Px be x < 10
> > Agree
> > (IOnly realised later with s(x) Jack probably ment the successor > > function. sx= x+1)
> I think any interpretation of s(x) as a one-place function is > OK for this problem, although it does suggest the successor > function.
> In your suggestion of s(x) = x (so s(x) is interpreted as the > identity function) we have that F = Ax(Psx -> Px) comes out > as meaning Ax(Px -> Px) which of course is true in every > interpretation. Which is fun.
> The OP's notation of P(x) is slightly ambiguous to me. It > could mean that P is a one-place predicate symbol, or it > could mean that P is a formula with x as a free variable.
> It doesn't make any difference for this problem, I guess.
> I would use 'F' for a random predicate symbol (and 'G' if I needed > two), and for functions I'd use 'f' (and 'g', etc). I would save > up 'P' to stand for a sentence (proposition). The OP does just the > opposite, uses 'F' for a sentence and 'P' for a predicate.
> -- > hz- Hide quoted text -
> - Show quoted text -
I agree again Wish i couls write it as eloquently as you.
but i guess Jack did mean the successor function. Anybody else studying at Bath University UK.
> > > > > 3. Describe an interpretation where F ^ G ^ (EX)¬p(X) is true or argue > > > > > that such an interpretation does not exist.
> > > > F ^ G ^ Ex~Px = Ax(Psx -> Px) ^ ExPx ^ Ex~Px
> > > > Let sx be x + 1. > > > > Let Px be x < 10
> > > Agree
> > > (IOnly realised later with s(x) Jack probably ment the successor > > > function. sx= x+1)
> > I think any interpretation of s(x) as a one-place function is > > OK for this problem, although it does suggest the successor > > function.
> > In your suggestion of s(x) = x (so s(x) is interpreted as the > > identity function) we have that F = Ax(Psx -> Px) comes out > > as meaning Ax(Px -> Px) which of course is true in every > > interpretation. Which is fun.
> > The OP's notation of P(x) is slightly ambiguous to me. It > > could mean that P is a one-place predicate symbol, or it > > could mean that P is a formula with x as a free variable.
> > It doesn't make any difference for this problem, I guess.
> > I would use 'F' for a random predicate symbol (and 'G' if I needed > > two), and for functions I'd use 'f' (and 'g', etc). I would save > > up 'P' to stand for a sentence (proposition). The OP does just the > > opposite, uses 'F' for a sentence and 'P' for a predicate.
> I agree again > Wish i couls write it as eloquently as you.
I'm guessing that's due to me have much more experience writing *in english* than you do!
It's nice to receive a compliment -- thank you. The more I post, the better my writing gets, I think. I make an effort to be clear, but there's room for improvement.
> but i guess Jack did mean the successor function.
I don't know, maybe.
> Anybody else studying at Bath University UK.
> (Wondering if we are doing his home work)
Probably. Jack, you only get one bite. Next time, only hints!
> Any help you could provide with this is greatly appreciated.
> F = (AX)(p(s(X)) -> p(X)) and G = (EX)p(X)
> Where A = for all and E = exists
> Use as domain the set N = {0,1,2,...} of natural numbers.
> Having trouble with the following:
> 1. Describe an interpretation where F ^ ¬G is true or argue that such an > interpretation does not exist.
> 2. Describe an interpretation where F ^ G and F ^ ¬G is true or argue > that such an interpretation does not exist.
> 3. Describe an interpretation where F ^ G ^ (EX)¬p(X) is true or argue > that such an interpretation does not exist.
> Cheers
> Jack
It's surprising that the domain was actually specified. After all, selection of the domian of interpretation is a part of the interpretation.
And the key in the selection process, is to think finite -- small finite. Like singletons, doubletons. Certainly not infinite.
The small domains provide an opportunity to connect (explicitly) conjunction with universal quantification, and disjunction with existential quantification. They provide an opportunity to learn more about the meaning of the phrases, functions-in-extension and predicates-in-extension.
In 1), nothing more than a singleton is needed. Then simply give extensions of the function and predicate.
In 3), a singleton is not sufficient, but a doubleton will work.
In 2), it's irrelevant, since the set is clearly inconsistent. (There's no interpretation under which those formulas are simultaneously true.)
> > Any help you could provide with this is greatly appreciated.
> > F = (AX)(p(s(X)) -> p(X)) and G = (EX)p(X)
> > Where A = for all and E = exists
> > Use as domain the set N = {0,1,2,...} of natural numbers.
> > Having trouble with the following:
> > 1. Describe an interpretation where F ^ ¬G is true or argue that such an > > interpretation does not exist.
> > 2. Describe an interpretation where F ^ G and F ^ ¬G is true or argue > > that such an interpretation does not exist.
> > 3. Describe an interpretation where F ^ G ^ (EX)¬p(X) is true or argue > > that such an interpretation does not exist.
> > Cheers
> > Jack
> It's surprising that the domain was actually specified. > After all, selection of the domian of interpretation is a part of the > interpretation.
> And the key in the selection process, is to think finite -- small > finite. > Like singletons, doubletons. Certainly not infinite.
> The small domains provide an opportunity to connect (explicitly) > conjunction with universal quantification, > and disjunction with existential quantification. They provide an > opportunity to learn more about the meaning > of the phrases, functions-in-extension and predicates-in-extension.
> In 1), nothing more than a singleton is needed. > Then simply give extensions of the function and predicate.
> In 3), a singleton is not sufficient, but a doubleton will work.
> In 2), it's irrelevant, since the set is clearly inconsistent. > (There's no interpretation under which those formulas are > simultaneously true.)
