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Math Help - Predicate Logic with quantifiers

  1. #1
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    Predicate Logic with quantifiers

    For each of the following:
    i. Write/expand each of the following predicates using disjunctions, conjunctions, and negations.
    ii. Determine the truth value of the following statements.

    Given:
    P(x,y): x < y^3
    Q(x,y): x –1 > y^2
    R(x,y,z): x + y^3 > z

    Universe of discourse for variable x is N (natural numbers)
    Universe of discourse for variable y is Z+ (positive integers)
    Universe of discourse for variable z is R+ (positive real numbers)


    a. \forall x \exists y   P(x,y)




    b. \exists y \forall x   P(x,y)




    c. \forall y \forall x   Q(x,y)




    d. \exists y \exists x   Q(x,y)



    I understand the problem except that in the discrete book the problems always give a set of particular numbers to use the conjunctions, disjunctions, and negations on.
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  2. #2
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    This problem is also confusing me, can anyone shed some light on at least how to expand one of them?
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  3. #3
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    Quote Originally Posted by inferno View Post
    This problem is also confusing me, can anyone shed some light on at least how to expand one of them?
    1. For each natural number x there exists a positive integer y such that x is less than y cubed.

    i.e. x = 1, y = 2 (1 < 8)
    x = 2, y = 2 (2 < 8)
    x = 3, y = 2 (3 < 8)
    ...
    x = 8, y = 3 (8 < 27)
    ....

    I think that works.....
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  4. #4
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    Sorry, you wanted it expanded using conjunctions, disjunctions etc.
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  5. #5
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    I do believe I'm in your class

    Here's what I'm going to do. Since the universe of discourse is infinite, it is impossible to exhaustively write/expand the predicates. I chose to write the first few in the sequence, then use ellipses to infer the existence of the rest.

    If you have any other ideas, let me know
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  6. #6
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    Quote Originally Posted by gbenga View Post
    Here's what I'm going to do. Since the universe of discourse is infinite, it is impossible to exhaustively write/expand the predicates. I chose to write the first few in the sequence, then use ellipses to infer the existence of the rest.

    If you have any other ideas, let me know
    So, you'd go

    (1 < 2 ^ 3) & (2 < 2 ^3) & (3 < 2 ^ 3) & (4 < 2 ^ 3) & (5 < 2 ^ 3) & (6 < 2 ^ 3) & (7 < 2 ^ 3) & (8 < 3 ^3).....and so on for
    VxEy(x < y ^ 3) where x is a natural number and y is an integer greater than zero?

    & = conjunction, < = less than, ^ = power.

    I wish I knew how to formulate < and ^ in predicate logic instead of using mathematical symbols....

    I'm not in your class. I'm just an interested bystander.
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  7. #7
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    Same here

    Quote Originally Posted by gbenga View Post
    Here's what I'm going to do. Since the universe of discourse is infinite, it is impossible to exhaustively write/expand the predicates. I chose to write the first few in the sequence, then use ellipses to infer the existence of the rest.
    I was thinking the same.


    For the expansion, I was thinking more of something along the lines of

    (P(1,1) V P(1,2) V P(1,3) V …) ^ (P(2,1) V P(2,2) V P(2,3) V …) ^ (P(3,1) V P(3,2) V P(3,3) V …) ^ ...


    Quote Originally Posted by Skids View Post
    (1 < 2 ^ 3) & (2 < 2 ^3) & (3 < 2 ^ 3) & (4 < 2 ^ 3) & (5 < 2 ^ 3) & (6 < 2 ^ 3) & (7 < 2 ^ 3) & (8 < 3 ^3)...
    I would use this for part ii, to show how I arrived at the predicate being true.


    What do you guys think?
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  8. #8
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    That's exactly how I did mine, except in a matrix form:

    [P(0,1) v P(0,2) v P(0,3)] ^ ...
    [P(1,1) v P(1,2) v P(1,3)] ^ ...
    [P(2,1) v P(2,2) v P(2,3)] ^ ...
    .
    .
    .

    Remember that x's domain is natural numbers, and y's is positive integers (z is not bound, so I disregarded it).
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  9. #9
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    Quote Originally Posted by gbenga View Post
    That's exactly how I did mine, except in a matrix form:

    [P(0,1) v P(0,2) v P(0,3)] ^ ...
    [P(1,1) v P(1,2) v P(1,3)] ^ ...
    [P(2,1) v P(2,2) v P(2,3)] ^ ...
    .
    .
    .

    Remember that x's domain is natural numbers, and y's is positive integers (z is not bound, so I disregarded it).
    That makes sense. I was thinking the natural numbers start from 1, and was wondering why distinguish positive integers from natural numbers if they amount to the same thing.
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  10. #10
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    Quote Originally Posted by gbenga View Post
    Remember that x's domain is natural numbers, and y's is positive integers (z is not bound, so I disregarded it).

    I wasn't sure whether to include 0 as a natural number. I looked it up and most sites said that it isn't included.
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