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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Question
Prove this argument with rules of inference
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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Answers
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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by: ozoPosted on 2009-04-25 at 20:23:32ID: 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