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.

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