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Homework Help: Basic Propositional Calc

  1. Nov 12, 2007 #1
    This is mostly some basic stuff, but I just want to make sure I am doing these right. I have a hard time understanding what the questions are saying sometimes.

    1. The problem statement, all variables and given/known data
    1)Explain how we may conclude that if f is a decreasing function, then f(5) < f(3)
    Make reference to a logical principle. [Note- By defintion, a function g is decreasing i.f.f the conditional "if x<y, then g(x)>g(y)" is true for any real numbers x and y]

    2)Suppose A and B are sets and w is an object, for which it is known that A[tex]\subseteq[/tex]B and w[tex]\in[/tex]A. Explain how we may conclude w[tex]\in[/tex](A[tex]\cap[/tex]B) Make reference to a logical principle.

    3)Explain why it is true that if a real number x satisfies |x|>a, but it is not the case that x > a, then x < -a must hold. Make reference to a logical principle.

    2. Relevant equations
    (p[tex]\wedge[/tex]q) -> p Law of Simplification
    p -> (p[tex]\vee[/tex]q) Law of Addition
    [p[tex]\wedge[/tex](p ->q)] -> q Modus Ponens
    [(p[tex]\vee[/tex]q)[tex]\wedge[/tex] ~q] -> p Modus Tollendo Ponens
    [(p -> q)[tex]\wedge[/tex]~q] -> ~p Modus Tollens
    (p -> r) -> [(p[tex]\wedge[/tex]q) -> r]
    [~p -> (q[tex]\wedge[/tex]~q] -> ~p Law of Contradiction


    3. The attempt at a solution
    1) The definition of a decreasing function states, "if x<y, then g(x)>g(y)" is true for any real numbers x and y. By hypothsis we know f is a decreasing function, so that the preticular case of the definition, "if 5 < 3, then g(5) > g(3)" is known to be true. --That's as far as I can get using an example form the book as a guideline. I'm not sure where to go from here and which logical principle applies to this argument.--

    2) --I know how to explain it but I don't know which logical principle to use.--
    A[tex]\subseteq[/tex] B includes all the elements in A that are also elements of B. Since it is given w[tex]\in[/tex] A, by this definiton w is also an element of B. A[tex]\cap[/tex]B includes only the elements that A and B share in common. Since w is both an element of A and B, it is one of the elements that A and B share in common, making it an element of A[tex]\cap[/tex]B.

    3)If a real number x satisfies |x|>a, that means x < -a or x > a. It is also given that it is not the case that x > a. Therefore by Modus Tollendo Ponens we can conclude that x < -a.


    If anyone could help me with finding which logical principles apply to one and two, that'd help a lot.
     
    Last edited: Nov 12, 2007
  2. jcsd
  3. Nov 12, 2007 #2
    Anybody out there who could help?
     
  4. Nov 12, 2007 #3
    I could help you with number 2, you could make use of the "intersection of a subset rule." But you've got the basic idea nailed.

    Edit: I think that you could use Universal Modus Ponens for #1.
     
    Last edited: Nov 12, 2007
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