Split on the law of excluded middle for , then use disjunction elimination:
 1  D ∧ E          premise
 2  (A ∧ D) → B    premise
 3  (¬A ∧ E) → B   premise
 4  D              ∧-elim-L 1
 5  E              ∧-elim-R 1
 6  A ∨ ¬A         excluded middle
 7  | A            assumption
 8  | A ∧ D        ∧-intro 7,4
 9  | B            →-elim 2,8
10  | ¬A           assumption
11  | ¬A ∧ E       ∧-intro 10,5
12  | B            →-elim 3,11
13  B              ∨-elim 6,7–9,10–12
The two occurrences of have identical type and depend only on the corresponding temporary branch assumption and the original premises.
Now derive the needed or inside its respective subproof:
 1  A → D          premise
 2  ¬A → E         premise
 3  (A ∧ D) → B    premise
 4  (¬A ∧ E) → B   premise
 5  A ∨ ¬A         excluded middle
 6  | A            assumption
 7  | D            →-elim 1,6
 8  | A ∧ D        ∧-intro 6,7
 9  | B            →-elim 3,8
10  | ¬A           assumption
11  | E            →-elim 2,10
12  | ¬A ∧ E       ∧-intro 10,11
13  | B            →-elim 4,12
14  B              ∨-elim 5,6–9,10–13
Implication elimination supplies the branch-specific premise before conjunction introduction packages it for the final implication.

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