~(~pΛs) would be the same as (pv~s), correct?
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.
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
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
.