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Math Help - Propositional Logic Proof

  1. #1
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    Propositional Logic Proof

    If someone could please confirm if this proof is correct:

    p -> q (conditional ->) line 1
    ~p -> r line 2
    r -> s line 3
    ~q ^ v line 4
    ---------
    s

    My proof:

    p -> q premise line 5
    r -> s premise line 6
    p V r implication of line 2 line 7
    q V s constructive dilemma using line 5,6,7 line 8
    ~q ^ v premise line 9
    ~q simplification of line 9 line 10
    s disjunctive syllogism using line 8,10
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  2. #2
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    That works. Here is what I would do.
    From 1, \neg q \Rightarrow \neg p.
    From 2 & 3and the above \neg q \Rightarrow s
    That with \neg q from 4 gives s
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  3. #3
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    thanks, I have another that is confusing me...

    (p)If it is cold this weekend or (q)the car is not fixed, then (r)Mike will not go
    shopping and (s)he will stay at home. If Mike does not go shopping, (t)he will
    order pizza. (u)The forecast for this weekend is cold. (v)The car will be fixed by
    Friday. Therefore, Mike will order pizza.

    I came up with this:

    (p V ~q) -> (~r ^ s)
    ~r -> t
    u
    v
    --------------------
    t

    I don't think this is provable. 'The forecast for this weekend is cold' does not affirm that it will be cold this weekend, therefore, another variable must be introduced. 'The car will be fixed by Friday' does not equate to 'the car is fixed', therefore, another variable must be introduced here, too. u and v do nothing to help prove or disprove the statement, and the first two lines of the proof cannot prove or disprove t, therefore, I believe that one cannot prove or disprove the statement t. Any thoughts?
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  4. #4
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    IF the car will be fixed by Friday then surely it will be fixed the weekend,so do not use another variable.

    Also if the forecast is for cold weather then it will be cold in the weekend ,so do not use another variable .

    Now rewrite you argument and try again
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  5. #5
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    okay

    (p V ~q) -> (~r ^ s) line 1
    ~r -> t line 2
    p line 3
    q line 4
    --------------------
    t

    proof:

    p premise of line 1
    p V ~q addition law of line 3 (line 6)
    ~r ^ s modus ponens of line 6 and line 1 (line 7)
    ~r simplification of line 7 (line 8)
    t modus ponens of line 8 and line 1.

    yes, it's provable.
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  6. #6
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    CORRECT!!!!!!
    But you must number each line of your proof ,so that you will not cause confusion to the reader.
    Now try a proof by contradiction .i.e assume not t (As an exercise).
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