Question

If you have (p&q) -> r, can I use the rule of conjunctive simplification to reduce the p&q to just p?

Answer 1

this is discrete mathematics btw

Answer 2

I think so....Yes?

Answer 3

If you're asking whether
(p&q) -> r
implies
p -> r

then the answer is most definitely no.

E.g., p = (x=1), q = (y=x), r = (y=1)

Answer 4

oh darn you are right.. any way to get (p&q) -> r using q to become p->r?

Answer 5

Well
((p^q) -> r) = ~((p^q)^-r)
= ~(p^q)v r
= (~p v ~q) v r
= ~p v (~q v r)
= ~p v ~(q ^ ~r)
= ~p v (q -> r)
= ~ (p^~(q -> r))
= p -> (q -> r)

So yes

I.e., p^q -> r is equivalent to p -> (q ->r)
and by symmetry of the argument also equivalent to
q -> (p -> r)

Answer 6

you misread my question I was trying to get it to p->r but thanks for trying

Answer 7

In short, if the only thing you have is ( p^q -> r), then you cannot conclude (p -> r)