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Predicate Logic help please

  • Thread starter yankes2k
  • Start date
11
0
Hey everyone,

Just need some help on some truth values of a predicate logic interpretation. Here is the info. I will be using A(capital variable) to mean universal quantifier and EEX(capital variable) to mean existential quantifier.

UD: Set of positive integers
Bx: x is an even number
Gxy: x is greater than y.
Exy: x equals y.
Mxyz: x minus y equals z
a: 1
c: 3

a) (AX)(AY)(~Exy<->Gxy)
b) (AX)(Eax v Gxa)
c) (AX)(EEXY)(Gyx & (EEXZ)Mzxy)
d) (EEXX)((Bx & Gxc) & ~(EEXY)mxcy)

One more also...symbolize the following sentences of PL using the key.

UD:Mammals
Cxy: x is chasing y.
Lx: x is a lion
Ax: x is a formidable animal
Fx: x is a ferocius
Tx: x is a tiger
Bx: x is best avoided
b: Bruce willis
d: Danny Devito

1) Danny devito and ferocious lions and tigers are all best avoided.
2)If any tiger is ferocious, all formidable animals that chase danny devito are.
3)Not all things that are chased by lions are ferocious.

I am really having trouble please advise...I will be up until 1am Easter US time...thanks guys for the help.

-TOny
 

honestrosewater

Gold Member
2,071
5
What do you need to do? With the first group, you're trying to find a proof or counterexample?
With the second, you just need to formalize them?

Edit: Oh, I'm blind. I'll take a look at the second group.
 
Last edited:

honestrosewater

Gold Member
2,071
5
yankes2k said:
UD:Mammals
Cxy: x is chasing y.
Lx: x is a lion
Ax: x is a formidable animal
Fx: x is a ferocius
Tx: x is a tiger
Bx: x is best avoided
b: Bruce willis
d: Danny Devito

1) Danny devito and ferocious lions and tigers are all best avoided.
I haven't actually done this much, but here goes. You can take one piece at a time at first and combine them afterward. So
"Danny Devito is best avoided" is easy: Bd.
It isn't clear whether the ferocious property applies to only lions or both lions and tigers! I'll assume it applies to both.
"Ferocious lions are best avoided". You're attributing the property B to all ferocious lions, so what quantifier do you need? A.
AX(If x is a lion and x is ferocious, then x is best avoided)
Can you handle that?
 

honestrosewater

Gold Member
2,071
5
Okay, well, time is up. Just a suggestion: If you want help with your problems, a few hours is unrealistic. You have a much better chance of getting help if you post them a day in advance, so people have a chance to see them in time. And the more time you give us, the better. :smile:
But in case you check this in the morning...
yankes2k said:
UD: Set of positive integers
Bx: x is an even number
Gxy: x is greater than y.
Exy: x equals y.
Mxyz: x minus y equals z
a: 1
c: 3

a) (AX)(AY)(~Exy<->Gxy)
So obviously Gxy -> ~Exy. But ~Exy -> Gxy doesn't hold. The thing you need to know here is that for all x and y, exactly one of the following is true: Exy or Gxy or Lxy. (Lxy means x is less than y.) More pointedly, AXAY(Gxy -> ~Lxy) and AXAY(Lxy <-> Gyx), so AXAY(Gxy -> ~Gyx).
b) (AX)(Eax v Gxa)
IOW, is there any positive integer less than 1?
c) (AX)(EEXY)(Gyx & (EEXZ)Mzxy)
For any x, is there a y that's greater? The set of positive integers is infinite. (z - x = y) <=> (z = y + x). If x and y are positive integers, does their sum exist, and is it also a positive integer?
d) (EEXX)((Bx & Gxc) & ~(EEXY)mxcy)
So the first part: Does there exist a positive even integer greater than 3? Second part: And for which there is no y such that x - 3 = y? So you need x - 3 to not be a positive integer; There are two ways that can happen: x - 3 = 0 or x - 3 < 0. You can rule out x - 3 = 0 because x > 3. If x - 3 < 0, then x < 3 (add 3 to both sides). So you can rule out x - 3 < 0. So there exists no such x.

Not sure if this is really what you were looking for, but it should at least get you started.
 
11
0
that's great. With the first set I only had to assign true values to them based on the information listed in the interpretation. Next time I will definetly post in advance. Again, thanks for the help. Based on what you have said I'm going to conclude the following for a-d.

a)False
b)True
c)False
d)False

What do you think does this sound right?
 

honestrosewater

Gold Member
2,071
5
yankes2k said:
a)False
b)True
c)False
d)False

What do you think does this sound right?
c is claiming that, in the set of positive integers, for every x that you choose, you can find a y that is greater than x (this is correct so far) and the sum of x and y is also a positive integer. (z - x = y) <=> (z = y + x). So if you add two positive integers together, will you get another positive integer? Yes. So c is true. The rest are correct, I believe.
 
11
0
Actually,

I think if z-x=y and y+x=Z is true then (c) must be true.
 
11
0
up sorry posted before refreshing the page. Well at least I figured it out lol.

Thanks again so much.
 

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