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Logical Notation

  1. Dec 2, 2007 #1
    I need help on converting these to the correct logical notation form using only the propositional connectives and the quantifiers. I'd appreciate if there are some explanation for the reasons of converting in such ways.


    This is what I got so far:
    (a) Problem: A ⊆ B ∩ C
    What I tried getting: ∀x(x ∈ A → x ∈ B ∧ x ∈ C)

    (b) Problem: A ⊆ {x, y}
    What I tried getting (incomplete... don't understand the {x, y} part): ∀x(x ∈ A → ... )

    (c) Problem: A = ø
    What I tried getting (don't understand this one, either): ∃x(x ¬∈ A)

    (d) Problem: A = {x}
    What I tried getting (incomplete): ∀x(x ∈ A ⇔ ...)


    These are the examples given:
    (a) Problem: A ⊆ B
    Converted to: ∀x(x ∈ A → x ∈ B)

    (b) Problem: A ∩ B ≠ ø
    Converted to: ∃x(x ∈ A ∧ x ∈ B)


    Thanks, your help would be greatly appreciated. :)
     
    Last edited: Dec 2, 2007
  2. jcsd
  3. Dec 2, 2007 #2

    cristo

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    (b) {x,y} is just the set consisting of the elements x and y.

    (c) your statement doesn't describe "A is the empty set." Try again.

    (d) you need to say that every element of A is an element of {x}; that is, x is the only element of A.
     
  4. Dec 2, 2007 #3
    I'm pretty new to this... so I don't really know the symbols for representing everything. But I'll give it another try. :)

    (b) ∀x(x ∈ A → x ^ y)

    ^ I don't know the symbol for "set," do I just leave it as it is "{x, y}?"

    (c) ∀x(x ¬∈ A) or ¬∃x(x ∈ A)

    (d) ∀x(x ∈ A)
     
    Last edited: Dec 2, 2007
  5. Dec 4, 2007 #4

    cristo

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    Yes; so you will have [itex]\forall x (x\in A\Rightarrow x \in \{x,y\})[/itex]

    I don't know whether you're allowed to use "not in". An alternative expression is [itex]\forall x(x\in A^{c})[/itex]
    Try using a different letter. What can you say for every y that is in A?
     
  6. Dec 6, 2007 #5
    (d) ∀x(z ∈ A ⇔ z = x)
     
  7. Dec 6, 2007 #6

    CRGreathouse

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    I'm really uncomfortable with the use of x for a particular value as well as one of your bound values. I'd prefer
    [tex]\forall z (z\in A\Rightarrow z\in\{x,y\})[/tex]
    or
    [tex]\forall z (z\in A\Rightarrow(z=x\vee z=y)[/tex].
     
  8. Dec 6, 2007 #7

    CRGreathouse

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    Close. This could also describe the empty set; you should rule this out.
     
  9. Dec 6, 2007 #8
    for the one A equals the empty set, use the example AB equals the empty set and make the necessary substitution.
     
  10. Dec 8, 2007 #9
    Here we go. I don't know how to do the coding, so I'll try to spell everything out and then give formulae that look like what should be written.

    (a) A is a subset of the intersection of B and C. Therefore all the members of A are in both B and C.
    (Ax)[(x E A) --> ((x E B) & (x E C))]
    'For all X, if X is an element of A then both X is an element of B and X is an element of C.'

    (b) A is a subset of {x,y}. All elements of A are either x or y.
    (Az)[(z E A)-->(z=x) v (z=y)]
    'For all Z, if Z is an element of A then Z is either identical to X or identical to Y.'

    (c) A is the null set.
    (Ax)~[x E A]
    For all X, it is not the case that x is an element of A.

    (d) A is a set consisting of one element: x.
    (Ex)(Ay)[(x E A) & ((y E A) --> (y=x))]
    There exists an X, and for all Y, X is an element of A, and if Y is an element of A then y is identical to X.

    Hope this helps.
     
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