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Thread: Predicate Logic Question Help

  1. #1
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    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)

    Thanks in advance!
    Last edited by richardmark; Nov 7th 2016 at 09:51 AM. Reason: Missing info
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  2. #2
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    Re: Predicate Logic Question Help

    Quote 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.
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    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
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  4. #4
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    Re: Predicate Logic Question Help

    Quote Originally Posted by richardmark View Post
    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.
    $\left( {\forall x} \right)\left( {\forall y} \right)\left( {\exists z} \right)\left[ {(L(x,y) \vee L(y,x)) \vee \left( {L(z,x) \wedge L(z,y)} \right)} \right]$
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  5. #5
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    Re: Predicate Logic Question Help

    Quote Originally Posted by richardmark View Post
    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)".

    Thanks in advance!
    Last edited by HallsofIvy; Nov 8th 2016 at 03:46 AM.
    Thanks from richardmark
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    Re: Predicate Logic Question Help

    Quote Originally Posted by richardmark View Post
    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]
    Quote Originally Posted by HallsofIvy View Post
    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 $\neg L(X,Y) \vee \neg L(Y,X)$.
    Thanks from richardmark
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  7. #7
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    Re: Predicate Logic Question Help

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