Natural deduction proof: $C, (C \land D)↔F \vdash (D \land E) \to F$

Question

I'm having trouble with proving

C,  (C Λ D) ↔ F  |-  (D Λ E) → F  

If it were $\lor$ instead of $\land$, then I would be able to do it. If I can prove that $(C \lor D) \leftrightarrow F$, I can prove that $(D \lor E) → F$ is true as well, because all that matters in this case is $D$ and I can ignore whether $C$ and $E$ are true or false. But it's actually $\land$ that is used in this problem.

I could prove some parts of the left side of the formula, but I'm not sure how this leads to the conclusion that $(D \land E) \to F$ holds true.

1      C                                premise
2      (C Λ D) ↔ F                      premise
3      F → (C Λ D)                      2 (↔E)
4      | F                              Hypothesis
5      | (C Λ D)                        3, 4 (→E)
6      |  D                             5 (ΛE)
-------------------------------------------------------
7      |  ????                          Hypothesis 
??????????

(UPDATED VERSION 1 BELOW, ONE MORE ATTEMPT)

1      C                             premise
2      (C Λ D) ↔ F                   premise
3      (C Λ D) → F                   2 (↔E)
4      | (D Λ E)                     Hypothesis
5      || (C Λ D)                    Hypothesis                
6      || F                          3, 5 (→E)  
---------------------------------------------------
7      (D Λ E)  → F                  4, 6  (→I)    

(UPDATED VERSION 2 BELOW)

1      C                             premiss
2      (C Λ D) ↔ F                   premiss
3      (C Λ D) → F                   2 (↔E)
4      | (D Λ E)                     Hypothesis
5      |  D                          4 (ΛE)
6      | C Λ D                       1, 5 (ΛI)              
7      | F                           3, 6 (→E)  
---------------------------------------------------
8      (D Λ E)  → F                  4, 6  (→I)   

Answer 1

HINT

Since you need to prove a statement whose main operator is a conditional ($\to$), you need to set this up as a conditional proof ($\to$ Intro).

That is, assume $D \land E$ as a Hypothesis, and then see if you can get to $F$:

\begin{array}{lll} 1 & | D \land E & Hypothesis\\ ... & | .... & \\ n & | F & ...\\ n+1 & (D \land E) \to F & 1-n \ (\to I) \end{array}

Answer 2

Your error is at the beginning of your attempt. Since you want to prove that $F$ follows from $D \land E$, you have to use the other elimination rule for $\leftrightarrow$, i.e. you have to infer $(C \land D) \to F$ from $(C \land D) \leftrightarrow F$. Inferring $F \to (C \land D)$ from $(C \land D) \leftrightarrow F$ is useless.

Below, you can find a correct derivation in natural deduction of $C, (C \land D) \leftrightarrow F \vdash (D \land E) \to F$.

\begin{align} \dfrac{\dfrac{(C \land D) \leftrightarrow F}{(C \land D) \to F}\leftrightarrow_e \qquad \dfrac{C \qquad \dfrac{[D \land E]^*}{D}\land_e}{C \land D}\land_i}{\dfrac{F}{(D \land E) \to F}\to_i^*}\to_e \end{align}

Answer 3

In the following proof I try to analyse the premises ( using ---> definition and DeMorgan's law) in order to do, in the end, the simplest/shortest as possible conditional proof.

The proof shows that the reason why " (D&E)--> F" is true under the given premises is that, under the premises, D by itself implies F. So, a fortiori, (D&E) imply F.