Proof verification for natural deduction in propositional logic I wanted to know if the following natural deduction proof is correct.
    (p ⇒ r) ∨ (q ⇒ r) ├ (p ∧ q) ⇒ r
1       (p ⇒ r) ∨ (q ⇒ r)      Assump o
2       (p ∧ q) ├ r        
2.1     p ∧ q                   Assump 2
2.2     r                       V E 1,2
3       (p ∧ q) ⇒ r            ⇒I 2       


The OP seeks a verification of a natural deduction proof.  One way to get confidence that one's proof is correct is to use a proof checker.  Here is a correct proof of the result that one can use for comparison. 

The goal is to show a conditional so line 2 starts with the antecedent $P∧Q$. The proof succeeds if one can derive the conclusion, $R$, which is done on line 9. 
The premise is a disjunction. One takes advantage of that by considering each of the two cases in the disjunction. If from both cases one can derive $R$, then one can use disjunction elimination ("vE") to derive $R$ as desired.
Further details on the deduction rules can be found in forallx. A link to the proof checker is provided below.

Kevin Klement's JavaScript/PHP Fitch-style natural deduction proof editor and checker http://proofs.openlogicproject.org/
P. D. Magnus, Tim Button with additions by J. Robert Loftis remixed and revised by Aaron Thomas-Bolduc, Richard Zach, forallx Calgary Remix: An Introduction to Formal Logic, Winter 2018. http://forallx.openlogicproject.org/
