Use the rules of inference to show that the following argument is valid.
(pvq)→(~qΛr)
(~qΛr)→~(~pΛs)
pvq
s
q→~p
------------------
∴ ~q
OK, so in order to show that it is valid, do I need to make a truth table? This just seems so vague to me.
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Use the rules of inference to show that the following argument is valid.
(pvq)→(~qΛr)
(~qΛr)→~(~pΛs)
pvq
s
q→~p
------------------
∴ ~q
OK, so in order to show that it is valid, do I need to make a truth table? This just seems so vague to me.
~(~pΛs) would be the same as (pv~s), correct?
Quote:
Originally Posted by ezwind72
1) (pvq)---->(~q^r)........................................ass umption
2) (~qΛr)→~(~pΛs).................................... ......assumption
3)pvq............................................. .................assumption
4) s................................................. ...............assumption
5)q→~p............................................ ................assumption
6) (~qΛr)....................................(3),(1) , M.Ponens
7) ~(~pΛs)...................................(2),(6), M.Ponens
8) pv~s..........................................(7), De Morgan
9) s---->p..........................................(8),ma terial implication ,or relation between connectives
10) p................................................. .......(4),(9), M.Ponens
11)p---->~q..........................................(5),a nd contrapositive law
12) ~q................................................ .....(10),(11),M.Ponens
So these are the rules you gave me? So how does that prove it is valid? Also, what are the numbers in the () before the rule you have written? Thanks for your help!
Quote:
Originally Posted by ezwind72
Take for example line 6 where i have written...
.......................... (3),(1) ,M.Ponens. It means i used the assumption
(pvq)---->(~q^r). and the assumption,
pvq and the rule of M.Ponens to get line 6 i.e ~q^r.
The general form of M.Ponens rule is:
A---->B and A THEN this implies B ,AND if we let A=(pvq) , B=(~q^r),using the rule we get (~qvr).
You also asked for the validity of the proof .Proofs in propositional calculus are pure formal proofs where every line of the proof can be easily checked and double checked and if no mistake is found then the proof is correct.
Unlikely with ordinary mathematical proofs where if one is asked for its validity or correctness one must rely on the superiority of the instructor .
Also in propositional logic you can check the validity of the argument by writing the truth table of the argument and if you get a tautology then the argument is valid hence provable.
In our case the truth table of the conditional:
([(pvg)→(~qΛr)] & [(~qΛr)→~(~pΛs)] & [pvq] & s & [ q→~p])------> ~q
should be a tautology
.
poutsos.B, I think I understand what you are saying. The only step I don't understand how you got is step 9. Could you please explain that? Thanks again!!
pv~s is equivalent to ~svp (by commutativity of propositions AVB <===>BvA) ,~svp is equivalent to s---->p (by material implication law or as is otherwised called ,relation between connectives).Quote:
Originally Posted by ezwind72
The general form of material implication is ~AvB <====> A---->B
OK, I fully understand what you did. Thanks very much for all your help!!