How should I prove this direct proof of natural deduction?

Question

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!

Answer 1

In "my" (i.e. the version I learnt) version of natural deduction this could go :

1. $(B \land C)$ (assumption)
2. $C$ ($\land$ -elimination)
3. $(B \land C) \to C$ ($\to$-introduction,drop 1.)
4. $A \to C$ (axiom)
5. $(A \lor (B \land C))$ (axiom)
6. $C$ 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.

Answer 2

$$A \lor (B \land C)$$ $$\implies (A \lor B) \land (A \lor C)$$ Introducing $A\implies C$ means that: $$\implies (C \lor B) \land (C \lor C)$$ $$\implies (C \lor B) \land C$$ $$\implies C$$

Answer 3

$1\quad(A \lor(B\ \land C))\quad$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.