$⊢p \land q \to (p\to q)$ - Natural deduction proof confusion I have the following:
$$⊢p \land q \to (p\to q)$$

I'm having a difficult time trying to figure out where to begin. I believe that I am supposed to assume p and q and then somehow use the copy rule to construct the equation, however I am not quite sure.

Can someone help me out?

Attempt taken from a comment: "...this is my attempt... first assume p, assume q, copy p, copy q, introduce →, introduce ∧, then introduce → between both... in that order"
To prove $A \to B$ in a natural deduction system you assume $A$ and derive $B$, and then apply Conditional Proof. OK? That's the standard way of proving a conditional.

To prove $(p \land q) \to (p \to q)$ in a natural deduction system, you need to assume $(p \land q)$ as a temporary assumption and aim for $(p \to q)$. 

So assume  $p \land q$. That gives you immediately $p$ and $q$. 

Can you now get to $p \to q$?

Well there's a standard trick for getting from $q$ to $A \to q$ for any $A$ at all. Do you know it? The fine details of how you set it out will depend, however, on your exact system. (Basically, in an application of Conditional Proof, we are usually allowed "vacuous discharge" of premisses that were never in fact used. You may need to check out how this is handled in the system you are working with! It will be, so to speak, in the fine print!)
Assuming the conjunction operator has higher precedence than the conditional operator, what needs to be proved is the same as this:

$$⊢(p∧q)→(p→q)$$

Here is a proof:



The OP made the following attempt:


  this is my attempt... first assume p, assume q, copy p, copy q, introduce →, introduce ∧, then introduce → between both... in that order


The above proof uses these steps:


Assume $P∧Q$ as the antecedent of the desired conditional.
Assume $P$ as the antecedent of the conditional in the conclusion.
Derive $Q$ from the first assumption using conjunction elimination.
Discharge the assumption on line 2 by rewriting the subproof in lines 2-3 as $P→Q$ on line 5 with conditional introduction as the justification.
Discharge the assumption on line 1 by rewriting the subproof in lines 1-4 as $(P∧Q)→(P→Q)$ on line 5 with conditional introduction as the justification.


That completes the proof.



Kevin Klement's JavaScript/PHP Fitch-style natural deduction proof editor and checker http://proofs.openlogicproject.org/

"Operator Precedence" Introduction to Logic http://intrologic.stanford.edu/glossary/operator_precedence.html