We assume the negative of the conclusion viz E & ~a & ~b & d and show this leads to a contradiction:
Original Premisses:
01. ~s ⇒ r
02. ~m & h ⇒ ~d
03. E ⇒ a & ~k
04. l & n ⇒ v
05. z & m ⇒ ~c
06. t ⇒ A
07. ~s & ~n & c ⇒ ~r
08. z & v ⇒ m
09. E ⇒ w
10. v & k & m ⇒ t
11. E ⇒ s & c & ~n
12. A ⇒ ~B
13. k & ~h ⇒ n
14. r ⇒ c
15. ~v & ~a & ~h ⇒ l
16. w & ~t & s ⇒ n
17. d ⇒ e & k
18. s & n & ~b ⇒ B
19. w & e ⇒ z
20. ~t & ~A & ~r & ~c ⇒ ~n
H
E & ~a & ~b & d
Hypothesis
21
E ⇒ w
By 9
22
w ⇒ z ∨ ~e
By 19
23
d ⇒ e
By 17
24
e
By H, 23
25
w ⇒ z
By 22, 24
26
z ⇒ ~c ∨ ~m
By 5
27
E ⇒ c
By 11
28
c
By H, 27
29
z ⇒ ~m
By 26, 28
30
E ⇒ ~m
By 21, 25, 29
31
~m
By H, 30
32
E ⇒ z
By 21, 25
33
z
By H, 32
34
~m ⇒ ~z ∨ ~v
By 8
35
~v
By 31, 33, 34
36
~v ⇒ ~l ∨ ~n
By 4
37
~l ∨ ~n
By 35, 36
38
~l ⇒ v ∨ a ∨ h
By 15
39
~a
By H
40
~l ⇒ h
So by 35,38,39
41
h ⇒ ~d ∨ m
By 2
42
d
By H
43
~h
By 31, 42, 41
44
l
By 40
45
~v ⇒ ~n
By 36,44
46
~n
By 35, 45
47
~n ⇒ ~k ∨ h
By 13
48
k
By 42, 17
49
~n ⇒ h
By 47, 48
50
~n ⇒ ~d ∨ m
By 49, 41
51
d&~m
By 31, 42
52
n
By 50, 51
53
Contradiction
46, 52
54
~(E & ~a & ~b & d)
By H, 53
Note that the attribute b is not used in this derivation and therefore the broader conclusion ~(E & ~a & d) holds, as explained in the notes on the solution by method of trees