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Math Help - Prop Logic shenanigans

  1. #1
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    Prop Logic shenanigans

    A, B, and C are arbitrary wffs, how would I prove that:

    {A->B, B->C} |- (A->C)
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  2. #2
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    Quote Originally Posted by snowtone View Post
    A, B, and C are arbitrary wffs, how would I prove that:

    {A->B, B->C} |- (A->C)

    1) Α===>Β............................................ ......................given

    2) B====>C........................................... ........................given

    3) A................................................. ...........................assumption

    4) B................................................. ..........................1,3 M.Ponens

    5) C................................................. ...........................2,4 M.Ponens

    6) A=====>C.......................................... ..................BY using the deduction theorem or called differently the law of conditional proof
    from the lines 3 to 5
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  3. #3
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    Logic proofs

    Hi -

    Quote Originally Posted by snowtone View Post
    A, B, and C are arbitrary wffs, how would I prove that:

    {A->B, B->C} |- (A->C)
    As an alternative, set up a truth table for the proposition

    A \wedge((A \implies B) \wedge (B \implies C)) \implies C

    and show that this is a tautology.

    Grandad
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  4. #4
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    Why not for the proposition:

    [(A---->B)&(B---->C)]====>(A---->C)
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  5. #5
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    Logic proofs

    Hello archidi -

    Quote Originally Posted by archidi View Post
    Why not for the proposition:

    [(A---->B)&(B---->C)]====>(A---->C)
    No particular reason. This works just as well.

    Grandad


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