I'm trying to prove that $\vdash p\land (q\lor r)\to(p\land q)\lor (p\land r)$ in natural deduction.
0. (no premises) 1.p and (q or r) (assumption) 2.p (and elimination from 1) 3.q or r (and elimination from 1) Case 1: 4. q (assumption) 5. q and p (and introduction) Case 2: 6. r 7. ??? . . . . (p and q) or (p and r)
I tried to obtain $p\land q$ in the subproof and then use $\lor$-introduction to get $(p\land q)\lor (p\land r)$. I considered cases, but in the case $r$ I don't see any way to get $p\land q$. Probably this is the wrong strategy?