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Math Help - Fitch :(

  1. #1
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    Fitch :(

    Hi.

    Please help
    How to prove:
    1.p=>q - permission goal ~q=>~p
    2. ~p=>q - permission and q=>r goal (~p=>~r)=>p
    3. ~pI~q goal ~(p^q)

    Thanks
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  2. #2
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    Re: Fitch :(

    Since you did not edit your post to add a note that it has been solved, was the [SOLVED] tag added intentionally?
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  3. #3
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    Re: Fitch :(

    Thanks. I found a solution
    Only the last item leaves me wrong, for example: ~ pI goal ~ q ~ (p ^ q) - I can do up to 17, then later I got wrongI do not know why? Maybe I do not understand something. From the pictures that I have noted: 1, 2-9, 10 -17 and or eliminations. - Only 18 leaves me wrong: (
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  4. #4
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    Re: Fitch :(

    Quote Originally Posted by anastazya View Post
    3. ~pI~q goal ~(p^q)
    Does "I" mean disjunction, i.e., "or"? Usually it is denoted by ∨. In ASCII one can also write \/ or the letter "v".

    Quote Originally Posted by anastazya View Post
    Only the last item leaves me wrong
    By "last", do you mean number 3 in the quote above?

    Quote Originally Posted by anastazya View Post
    for example: ~ pI goal ~ q ~ (p ^ q)
    This is hard to understand. I am not sure what "pI" is and why there is no connective between "~q" and "~(p ^ q)".

    Quote Originally Posted by anastazya View Post
    I can do up to 17, then later I got wrongI do not know why? Maybe I do not understand something. From the pictures that I have noted: 1, 2-9, 10 -17 and or eliminations. - Only 18 leaves me wrong: (
    This is also hard to understand without seeing the first 17 steps.
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  5. #5
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    Re: Fitch :(

    do u mean
    premise ~p|~q
    to prove ~(p&q)???

    I'm also looking for the proof of this.
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  6. #6
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    Re: Fitch :(

    1.~p | ~q Premise
    2.~p Assumption
    3.p & q Assumption
    4.p And Elimination: 3
    5.p & q => p Implication Introduction: 4
    6.~p Reiteration: 2
    7.p & q Assumption
    8.~p Reiteration: 6
    9.p & q => ~p Implication Introduction: 8
    10.~(p & q) Negation Introduction: 5, 9
    11.~p => ~(p & q) Implication Introduction: 10
    12.~q Assumption
    13.p & q Assumption
    14.q And Elimination: 13
    15.p & q => q Implication Introduction: 14
    16.p & q Assumption
    17.~q Reiteration: 12
    18.p & q => ~q Implication Introduction: 17
    19.~(p & q) Negation Introduction: 15, 18
    20.~q => ~(p & q) Implication Introduction: 19
    21.~(p & q) Or Elimination: 1, 11, 20

    This is the solution I got. But it's too long.
    Still it works
    Thanks from emakarov
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  7. #7
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    Re: Fitch :(

    Quote Originally Posted by Thejanie View Post
    This is the solution I got. But it's too long.
    I believe this is the correct derivation. I don't see how it can be shortened except by removing step 6 and making step 8 a reiteration of 2.

    A more standard variant of natural deduction (of which Fitch calculus is a particular notation) has a symbol ⊥ for contradiction. Then negation introduction is shorter: you assume ~p and p & q, derive p, derive ⊥ from ~p and p, then close the assumption p & q and derive ~(p & q) in one step. There is no need to derive p & q => p and p & q => ~p. Similarly, or elimination does not require implications ~p => ~(p & q) and ~q => ~(p & q): you just derive ~(p & q) two times from assumptions ~p and ~q, respectively, and or elimination derives ~(p & q) and closes the ~p and ~q in one step.
    Thanks from Thejanie and MarkFL
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  8. #8
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    Re: Fitch :(

    The program I used doesn,t have contradiction. BTW can you help me to prove this the other way around? I mean starting from ~(p & q) as the premise to prove ~p | ~q.
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  9. #9
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    Re: Fitch :(

    Quote Originally Posted by Thejanie View Post
    BTW can you help me to prove this the other way around? I mean starting from ~(p & q) as the premise to prove ~p | ~q.
    Ah, this is more complicated. This requires the rule of double-negation elimination or the law of excluded middle. Using the latter is easier. Which one do you have and what does it look like?
    Thanks from MarkFL
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  10. #10
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    Re: Fitch :(

    Fitch :(-fitch.png
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  11. #11
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    Re: Fitch :(

    I see that you don't have the law of excluded middle, but I meant to ask what the negation elimination rule looks like. What formulas does it take and produce? I would guess it takes ~~A and produce A. Is this correct?
    Thanks from MarkFL
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  12. #12
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    Re: Fitch :(

    Yeah...
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  13. #13
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    Re: Fitch :(

    We have a premise ~(p & q). The proof of ~p | ~q is by contradiction, i.e., we assume ~(~p | ~q) and prove p & q (described below), which contradicts the premise. This gives ~~(~p | ~q) by negation introduction and then ~p | ~q by negation elimination.

    So, assume ~(~p | ~q). We need to prove p and q. These two subderivations are similar. To prove p, we assume ~p and derive ~p | ~q, which contradicts the assumption. Therefore, ~~p and we conclude p by negation elimination. The second part, q, is proved similarly.
    Thanks from Thejanie
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  14. #14
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    Re: Fitch :(

    Thank you so much!!!
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  15. #15
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    Re: Fitch :(

    hi emakarov
    I'm struggling with this one still ...
    this is what I have so far:
    1. ~(p & q) Premise
    2. ~(~p | ~q) Assumption
    3. ~p Assumption
    4. ~p | ~q Or Introduction: 3
    5. ~p => ~p | ~q Implication Introduction: 4
    6. ~p Assumption

    Now I'm stuck as to how to get to the conclusion of ~p|~q.

    Can you please help?

    Also ... please bear in mind I don't have the necessary background in discrete mathematics .. so is there a specific textbook or site you can recommend also so I can understand better and learn?

    Thank you so much in advance
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