is the inference pattern given below valid ?? (p^q)->~t w v r w->p r->q
w->p r->q ... w v r ............from the principle of constructive dilemma..... they can be written in single..as . p v q....... ........ but how to solve... (p^q)->~t p v q.......
the final conclusion given is.........(w v r)->~t
i wish i could help on it, but quite frankly, i wouldnt know where to begin with it :/
how much of it are you sure about?
i'm sure about...... the last three..... and by simplifying them........ i got.. p v q.... so..we have to solve (p^q)->~t and p v q..... and check whether if it is equal to......(w v r)->~t
would truth tables be helpful? and is this one big logical expression: ((p^q)->~t) ^ (p v q) to solve?
but i'm not allowed to use truth tables here... :(
do they stab you in the eye as well so that you have to work with impaired vision too??
if using truth tables gets you to the results, then you have something to work with to guide you
with any luck im remembering these correctly p q t p^q -> -t 1 1 1 1 0 =0 1 1 0 1 1 =1 1 0 1 0 0 =1 1 0 0 0 1 =1 0 1 1 0 0 =1 0 1 0 0 1 =1 0 0 1 0 0 =1 0 0 0 0 1 =1 w r t wvr ->-t 1 1 1 1 0 =0 1 1 0 1 1 =1 1 0 1 1 0 =0 1 0 0 1 1 =1 0 1 1 1 0 =0 0 1 0 1 1 =1 0 0 1 0 0 =1 0 0 0 0 1 =1
i think you should begin with truth table.
i see i didnt include the ^ (pvq) tho
so........ is it an invalid inference pattern??
w->p (one case false) r->q (one case false) w^r (one case true) w or r (one case false)
pvq has the same design as wvr soo ^ compared to 0 1 =0 0 1 1 =1 1 1 1 =1 0 1 1 =1 1 1 1 =1 0 1 1 =1 1 1 0 =0 1 1 0 =0 1 without knowing anything else, my gut tells me that since the patterns are different, then we cannot infer it
experiment tho seems to have a better grasp on what its asking for tho :)
plz.......... try to solve without using truth tables..
it is an ivalid inference .....it seems... let's leave it.... @experimentX @amistre64 thanks a lot.......
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