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Prove this argument with rules of inference

Posted on 2009-04-25
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How do I use the rules of inference to show that the following is logically valid?

[(~p V q) -> r] /\ [s V ~q] /\ [~t] /\ [p -> t] /\ [(~p /\ r) -> ~s] => ~q

I've used Matlab to check that it is actually valid, but I need to show a proof using the rules of inference.
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Question by:rhianwen
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ozo earned 2000 total points
ID: 24234482
suppose
[(~p V q) -> r] /\ [s V ~q] /\ [~t] /\ [p -> t] /\ [(~p /\ r) -> ~s]  /\ q

then
[(~p V true) -> r] /\ [s V false] /\ [~t] /\ [p -> t] /\ [(~p /\ r) -> ~s]
[ r] /\ [s] /\ ~p /\ [(~p /\ r) -> ~s]
[s] /\ [~s]
is a contradiction
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by:rhianwen
ID: 24234524
Why can you put the q on the end there? Also if its a contradiction doesn't that prove the argument invalid? I know the argument is valid because I used Matlab to construct a truth table to check it.
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by:ozo
ID: 24234531
[(~p V q) -> r] /\ [s V ~q] /\ [~t] /\ [p -> t] /\ [(~p /\ r) -> ~s]  /\ q
is invalid, so
[(~p V q) -> r] /\ [s V ~q] /\ [~t] /\ [p -> t] /\ [(~p /\ r) -> ~s] => ~q
is valid
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Author Comment

by:rhianwen
ID: 24234724
So using the rules of inference I can say

H1     (~p V q) -> r
H2     s V ~q
H3     ~t
H4     p -> t
H5     (~p /\ r) -> ~s
----------------
            ~q

so

H3 & H4
             ~t /\ p -> t
              => ~p            (6)  (modus tollens)

H5 & (6)
           ~p /\ (~p /\ r) -> ~s
            => ~p /\ ~p -> (r -> ~s)               (exportation)
            =>  ~p -> (r -> ~s)
             => r -> ~s                (7)

this is where I get stuck. In my head I can do the proof in like 4 steps but I'm having trouble laying it out in terms of the rules.


 
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LVL 85

Expert Comment

by:ozo
ID: 24234841
6 & 1
=> r  (8)

8 & 7
=> ~s  (9)

9 & 2
=> ~q






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Author Comment

by:rhianwen
ID: 24234859
Ahh awesome thankyou!

For some reason I was thinking I could only use each statement once so I was getting stuck with the 6 & 1 bit.
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