# Thread: Predicate Logic Question Help

1. ## Predicate Logic Question Help

Hi everyone,

I have been trying to solve the following but not sure if I am on the right lines. Please can you help?

1) Logical statement L (c,d) stands for (c knows d) where the discourse universe is a set of people at Ryan's party. Translate the following into logical notations.

(So P stands for party)

a) Ryan knows everyone at the party - ∀xP(x) => L (x)
b) Everyone knows each other - ∀x∀y L(x,y) => L(x,y)
c) There are people who don't know each other - ∀x∃y P(x) => ⌜L(y,x)
d) For any two people who dont know each other, there is someone who knows them both - ∃x ⌜P(x) ^ ∃y L(y,x)

2. ## Re: Predicate Logic Question Help

Originally Posted by richardmark;909816
1) Logical statement L (c,d) stands for (c knows d) [B
where the discourse universe is a set of people at Ryan's party[/B]. Translate the following into logical notations.
(So P stands for party)
a) Ryan knows everyone at the party - ∀xP(x) => L (x)
b) Everyone knows each other - ∀x∀y L(x,y) => L(x,y)
c) There are people who don't know each other - ∀x∃y P(x) => ⌜L(y,x)
d) For any two people who dont know each other, there is someone who knows them both - ∃x ⌜P(x) ^ ∃y L(y,x)
Because the domain of discourse is a set of people at Ryan's party, we do not need the predicate P(x).
When we write $\forall x$ we already know that $x$ is a person at Ryan's party.
So part (a) becomes $\left( {\forall x} \right)\left[ {L(\text{Ryan},x)} \right]$___you may use R for Ryan.

c) is $\left( {\exists x} \right)\left( {\exists y} \right)\left[ {\neg L(x,y) \wedge \neg L(y,x)} \right]$

Now you post (b) & (d) so we can help.

3. ## Re: Predicate Logic Question Help

Thank you so much for the quick response. I've been comfortable with all the topics so far but this logic is killing me!

Would B become - ∀x∃y R(x,y) ^ R(y,x) ?
And D - ∃x ⌜R(x,y) => ∃y R(y,x) ?

Thanks again

4. ## Re: Predicate Logic Question Help

Originally Posted by richardmark
Would B become - ∀x∃y R(x,y) ^ R(y,x) ?NO
And D - ∃x ⌜R(x,y) => ∃y R(y,x) ?NO
b) Everyone knows each other - $\left( {\forall x} \right)\left( {\forall y} \right)\left[ {L(x,y) \wedge L(y,x)} \right]$

d) For any two people who don't know each other, there is someone who knows them both.
$\left( {\forall x} \right)\left( {\forall y} \right)\left( {\exists z} \right)\left[ {(L(x,y) \vee L(y,x)) \vee \left( {L(z,x) \wedge L(z,y)} \right)} \right]$

5. ## Re: Predicate Logic Question Help

Originally Posted by richardmark
Hi everyone,

I have been trying to solve the following but not sure if I am on the right lines. Please can you help?

1) Logical statement L (c,d) stands for (c knows d) where the discourse universe is a set of people at Ryan's party. Translate the following into logical notations.

(So P stands for party)

a) Ryan knows everyone at the party - ∀xP(x) => L (x)
You defined "L(x, y)" so that L must have two parameters- "L(x)" is meaningless. You want "∀xP(x)=> L(Ryan, x)".

b) Everyone knows each other - ∀x∀y L(x,y) => L(x,y)
Seriously? "L(x,y)=> L(x,y)"? "A=> A" is always true for any A! You want just "∀x∀y P(x)^P(y) => L(x,y)".

c) There are people who don't know each other - ∀x∃y P(x) => ⌜L(y,x)
No. This says "for everyone at the party, there is someone they don't know." That would be impossible if Ryan know everybody. You want "∃x∃y P(x)^P(y) ^ ⌜L(y,x).

d) For any two people who dont know each other, there is someone who knows them both - ∃x ⌜P(x) ^ ∃y L(y,x)
Do you not see that this involves three people? You want "∀x∀y P(x)^P(y)^ ⌜L(x, y)=> ∃z P(z) ^ L(x, z) ^ L(y, z)".

6. ## Re: Predicate Logic Question Help

Originally Posted by richardmark
1) Logical statement L (c,d) stands for (c knows d) where the [SIZE=3]discourse universe is a set of people at Ryan's party.[/SIZE]
Originally Posted by HallsofIvy
You defined "L(x, y)" so that L must have two parameters- "L(x)" is meaningless. You want "∀xP(x)=> L(Ryan, x)".
Seriously? "L(x,y)=> L(x,y)"? "A=> A" is always true for any A! You want just "∀x∀y P(x)^P(y) => L(x,y)".
No. This says "for everyone at the party, there is someone they don't know." That would be impossible if Ryan know everybody. You want "∃x∃y P(x)^P(y) ^ ⌜L(y,x).
Do you not see that this involves three people? You want "∀x∀y P(x)^P(y)^ ⌜L(x, y)=> ∃z P(z) ^ L(x, z) ^ L(y, z)".
Since the domain of discourse is is a set of people at Ryan's party, people in this question are the party. Thus $P(x)$ is not nesessary.
Also note that $L$ is not necessarily symmetric. To say that 'X & Y know each other one must be formalize as $L(X,Y) \wedge L(Y,X)$.
That is $L(X,Y)\not \Rightarrow L(Y,X)$. To say that 'X & Y do not know each other one must be formalize as $\neg L(X,Y) \vee \neg L(Y,X)$.

7. ## Re: Predicate Logic Question Help

Thank you both, it's helped me understand it a bit more