How should I prove this direct proof of natural deduction? The question is:
(A|(B&C)), (A->C)

and the goal is to get,
C

I made (A|(B&C)) into ((A|C)&(A|B)) by using distribution. Then by commutation as well as simplification, I could get (A|C). If I can get ~A, I can do Modus Tollens with (A|C) and directly get the goal, but I spent like 3 hours but could not get the answer... Also I cannot have more premises or assumptions...
How should I resolve this question?
Any help would be very thankful!
 A: $1\quad(A \lor(B\ \land C))\quad$Premise
$2\quad A \to C\quad$ Premise
$3\quad(A \lor B) \land (A \lor C)$ Distribution Law (1)
$4\quad(A \lor C)\quad$ Simplification (3)
$5\quad(\neg \neg C \lor A)\quad$ Double negation and Commutative Law (4)
$6\quad \neg C \to A\quad$ Equivalence for Implication and Disjunction (5)
$7\quad \neg C \to C\quad$ Hypothetical Syllogism (2, 6)
$8\quad C \lor  C\quad$ Equivalence for Implication and Disjunction (7)
$9\quad C\quad$ (8) 
It's easy to see that $C \lor C \equiv C $ is a tautological equivalence. Numbers to the right correspond to the premises on which that line depends. Notation is based on Patrick Suppes  book Introduction to Logic.
A: In "my" (i.e. the version I learnt) version of natural deduction this could go :


*

*$(B \land C)$ (assumption)

*$C$ ($\land$ -elimination)

*$(B \land C) \to C$ ($\to$-introduction,drop 1.)

*$A \to C$ (axiom)

*$(A \lor (B \land C))$ (axiom)

*$C$ from $(\lor$-elimination and 3,4,5)


Done. 
This corresponds nicely to how you would give the argument to a person: 
either $A$ holds, and then $C$ does, by $A \to C$, or otherwise even $B \land C$ holds, so certainly $C$ holds. So always $C$ holds.
A: $$A \lor (B \land C)$$
$$\implies (A \lor B) \land (A \lor C)$$
Introducing $A\implies C$ means that:
$$\implies (C \lor B) \land (C \lor C)$$
$$\implies (C \lor B) \land C$$
$$\implies C$$
A: Here is a direct natural deduction proof in a Fitch-style proof checker.

The two premises are on the first two lines. The first premise is a disjunction so I attempt to use disjunction elimination (vE) by considering both cases. The first case $A$ and conditional elimination or modus ponens (→E) allow me to derive the goal $C$. The second case $B \land C$ using conjunction elimination (∧E) allow me to also derive the goal $C$.
Since I can derive $C$ in both cases, I use disjunction elimination on line 7 to derive the goal.

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