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Math Help - Predicate logic (2 Questions)

  1. #1
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    Predicate logic (2 Questions)

    Hi,
    I'm having test tomorrow, but I have problem with these questions:

    1. How to build an anti example that overturns equivalence

    \forall x (P(x)  \oplus Q(x)) = \forall x P(x)  \oplus \forall x Q(x)<br />

    2. Given predicate  L (a, b) = "a>b"
    Write predicate expressions:

    a)  "a< b < c"

    b)  "a \neq b "

    c) "At least two of the numbers a, b, c are the same"

    a, b and c are real numbers
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  2. #2
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    Is this plus sign should be an "or" connective?

    Anyway, the second question is:
    2) 1.L(a,b)&L(b,c)
    2. L(a,b)&L(b,a)
    3.ExEyAz(~L(x,y)&~L(y,x)&((L(z,x)vL(x,z)v(~L(z,x)& ~L(x,z)))&(L(z,y)vL(y,z)v(~L(z,y)&~L(y,z)))

    the three was a bit long, but I hope you got it.
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  3. #3
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    Thank you for your reply.

    With \oplus I meant "exclusive or" (XOR)
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  4. #4
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    Quote Originally Posted by Bernice View Post
    Hi,
    I'm having test tomorrow, but I have problem with these questions:

    1. How to build an anti example that overturns equivalence

    \forall x (P(x) \oplus Q(x)) = \forall x P(x) \oplus \forall x Q(x)<br />

    2. Given predicate  L (a, b) = "a>b"
    Write predicate expressions:

    a)  "a< b < c"

    b)  "a \neq b "

    c) "At least two of the numbers a, b, c are the same"

    a, b and c are real numbers
    For (1) let:

    P(x) be the expression : a student x is tall,and

    Q(x) be the expression : a student x is plond.

    Then the formula \forall x(P(x) \oplus Q(x)) is in words : All the student (in a class ) are tall,xor,blond.


    In this case if we examine the students in a class there are some tall and some blond.


    But the formula \forall x(P(x)) \oplus \forall x(Q(x)) in words is:

    All the students are tall ,xor, All the students are blond.


    In that case if we examine the students in the class they will be All tall,

    or they will be All blond.

    Hence the two formulas are not equivalent.

    For (2):

    a) L(b,a)\wedge L(c,b).


    b) since the Nos a,b,c are real Nos then by the trichotomy law in real Nos ,

     a\neq b\rightarrow a>b\vee b>a.

    Hence  a\neq b\rightarrow L(a,b)\vee L(b,a)
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  5. #5
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    I finally understood.

    Quote Originally Posted by archidi View Post
    For (1) let:

    P(x) be the expression : a student x is tall,and

    Q(x) be the expression : a student x is plond.
    So anti example will also be:

    P(x) =" x=1"
    Q(x) ="x=2"
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  6. #6
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    Quote Originally Posted by Bernice View Post
    I finally understood.









    So anti example will also be:

    P(x) =" x=1"
    Q(x) ="x=2"
    First of all "=" (equality) is a two place predicate symbol,while x is tall or x is blond are one place predicate symbols

    In mathematics equality is a two place predicate symbol

    However the ,x=1 concept can be considered as a one place predicate symbol in the following interpretation:

    AN X box in a storehouse having one meter height.

    Again in this case the formula, \forall x( P \oplus Q)) is not equivalent to \forall x(P(x))\oplus\forall x(Q(x))

    In general predicate calculus gives the following theorems with respect to the v (=or) logical symbol:

    \forall x( P\vee Q) <===> \forall x P\vee Q if x is not free in Q,and

    \forall x( P\vee Q) <====>  P\vee\forall x Q if x is not free in P

    Where P and Q are well formed formulas ,like x=1,x=2 ,x is tall e.t.c,e.t.c


    For xor i could not find any
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