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Proving an or statement

  1. Dec 31, 2009 #1
    Proving an "or" statement

    What is the general procedure when proving a statement like "A implies B or C"? Is it more common to assume A and then split into cases (i.e. case 1: A implies B, case 2: A implies C, therefore A implies B or C)? Or is it more common to assume A and that one of B or C is false and then prove that A implies C or B (depending on which is assumed false)?
  2. jcsd
  3. Dec 31, 2009 #2
    Re: Proving an "or" statement

    Both are common. In my experience the most common is your second approach, i.e. assume A is true and B is false, and then conclude that C is true (interchange B and C if you want to). It's not uncommon either to do contradiction and assume A is true, B is false, C is false and then reach a contradiction, or possibly contraposition where you assume B and C are false and then conclude A is false. Really the approach to use depends on the problem and there is no best general way.
  4. Dec 31, 2009 #3
    Re: Proving an "or" statement

    If you did it the first way, you would have proven A implies B *and* C.
  5. Dec 31, 2009 #4
    Re: Proving an "or" statement

    I took his statement to mean:
    Assume A
    Show that either P or Q is true.
    Case 1: Assume P is true. ... then B is true.
    Case 2: Assume Q is true. ... then C is true.

    which is valid.
  6. Dec 31, 2009 #5
    Re: Proving an "or" statement

    Where did P and Q come from?
  7. Dec 31, 2009 #6
    Re: Proving an "or" statement

    I introduced them to make the point more easily. When he wrote:
    Case 1: A implies B
    Case 2: A implies C
    I'm reading it as a way of writing that in one case which I call P we prove that A implies B (we don't assume it), and in another case which I called Q we prove that A implies C. That is the only way I see for the argument to make sense.
  8. Dec 31, 2009 #7
    Re: Proving an "or" statement

    No, you still get A -> (B /\ C) with that approach.
  9. Dec 31, 2009 #8
    Re: Proving an "or" statement

    You must misunderstand me because I'm pretty sure my argument form is correct (whether altcmdesc meant it or not). For instance let:
    A be "n is an integer square m^2".
    B be "[itex]n \equiv 0 \pmod 3[/itex]".
    C be "[itex]n \equiv 1 \pmod 3[/itex]"
    Assume [itex]n=m^2[/itex] for some integer m.
    Case 1 ([itex]m \equiv 1 \pmod 3[/itex] or [itex]m \equiv 2 \pmod 3[/itex]): Then [itex]n=m^2 \equiv 1 \pmod 3[/itex] which proves C in this case.
    Case 2 ([itex]m \equiv 0 \pmod 3[/itex]): Then [itex]n=m^2 \equiv 0 \pmod 3[/itex] which proves B in this case.

    This doesn't prove that every square is congruent to 0 AND 1 modulo 3, simply to one of them.

    Here the conditions in parentheses are what I called P and Q and I think altcmdesc simply omitted them for simplicity. I think we're just understanding altcmdesc in different ways, so I guess we have to wait for him to clear up exactly what he meant.
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