Question

with the premise (p&q)->r, and premise q is there any way you can simplify these two premises using some rule to get p->r?

Answer 1

Like I showed the last time you asked a version of this question
(p&q)->r
is equivalent to
q->(p->r)

So now, what's new is assuming q:

(p&q)->r and q
is equivalent to
q->(p->r) and q

and this implies

p->r