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Math Help - Logic

  1. #1
    kit
    kit is offline
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    Logic

    Hi,

    I don't know if this the place to post this question, but I guess I can always try.

    My questions are regarding Logic.

    1) I'm trying to prove 'proof by cases' - X v Y, X -> Z, Y -> Z |- Z
    I'm trying to prove it using the natural rules, but am having problems!

    2) Using proof by cases, I want to prove the implication rule: A -> B |- A v B
    Either that, or by using natural rules.

    This is really starting to annoy me - as I have spent a long time working on this!

    Any help / advice would be helpful!

    Thanks.
    Kit
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  2. #2
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    I think that 'proof by cases' refers to the process of writing down all the possible assignments of truth values to X,Y,Z,... which make all the statements preceding the |- true; then you check that for every such assignment the statement folliwing the |- is true as well. So for A -> B |- A v B you find that the truth values (A,B) = (T,T), (F,T), (F,F) make A->B true and for every one of these assignments A v B is true as well.
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  3. #3
    kit
    kit is offline
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    Question

    Hi,

    Thanks for that - I actually want to show this without using truth tables.

    So if X v Y, X -> Z, Y -> Z |- Z

    Would it be ok to say the following:

    1) X assume
    2) X-> Z ->I, 1, and X->Z
    3) Z -> E, 2
    4) Y assyne
    5) Y -> Z ->I, 4, Y->Z
    6) Z -> E, 5
    7) Z 3, 6.

    Would this work?

    For the implication rule - I'm really stuck on how to do that one...
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  4. #4
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    Quote Originally Posted by kit
    Hi,

    I don't know if this the place to post this question, but I guess I can always try.

    My questions are regarding Logic.

    1) I'm trying to prove 'proof by cases' - X v Y, X -> Z, Y -> Z |- Z
    I'm trying to prove it using the natural rules, but am having problems!

    2) Using proof by cases, I want to prove the implication rule: A -> B |- A v B
    Either that, or by using natural rules.

    This is really starting to annoy me - as I have spent a long time working on this!

    Any help / advice would be helpful!

    Thanks.
    Kit
    to your second question you can use this process:
    1)A->B
    2)A
    3)B 1,2 M.P
    4)Bv~A 3,Add.

    about your first question:
    1) XvY
    2)X->Z
    3)Y->Z
    4)~X->Y 1,m.c
    5)~X->Z 4,3 h.s
    6)~Z->~X 2,trans
    7)~Z->Z 6,5 h.s
    8)ZvZ 7,m.c
    9)Z 8, idemp.

    that's all there is to it.
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