Show that the following argument is valid in SD.
(I will use "⊃" to show conditional.)

Code:
(E v (L v M)) & (E ≡ F)
L ⊃ D
D ⊃ ~ L
----------------------
E v M
I'm only allowed to use the basic derivation rules of SD:
Code:
Reiteration (R)
Conjunction Intro. (&I) and Conjunction Elim. (&E)
Conditional Intro. (⊃I) and Conditional Elim. (⊃E)
Negation Intro. (~I) and Negation Elim. (~E)
Disjunction Intro. (vI) and Disjunction Elim. (vE)
Biconditional Intro. (≡I) and Biconditional Elim. (≡E)
My attempts at a solution have not been very successful, but this is what I have come up with so far:

Code:
1. (E v (L v M)) & (E ≡ F) [assumption]
2. L ⊃ D [assumption]
3. D ⊃ ~ L [assumption]
------
4. ~(E v M) [subproof1open: assumption]
---
5. L [subproof2open: assumption]
---
6. L [subproof2close: 5R]

7. M [subproof2open: assumption]
---
8. ~L [subproof3open: assumption]
---
9. L [subproof3: ????]
10. ~ L [subproof3close: 8R]
11. L [subproof2close: 8-10~E]
12. D [subproof1: 3-13⊃E]
13. L [subproof1: 5-6,7-11vE]
14. ~ L [subproof1close: 3-12⊃E]
15. E v M [4-14~E]
Any help at all would be great, thanks in advance.