What is BD+B(D+E)+D^'(D+F) ?

The solution to BD+B(D+E)+D^'(D+F) is 2BD+BE+D^{1+'}+D^'F

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simplify BD+B(D+E)+D^'(D+F)

simplify

B D + B (D + E) + D^{^{'}} (D + F)

2 B D + BE + D^{1 +'} + D^{'} F

Solution steps

B D + B (D + E) + D^{^{'}} (D + F)

Expand

B (D + E) : B D + BE

= B D + B D + BE + D^{^{'}} (D + F)

Expand

D^{^{'}} (D + F) : D^{1 +^{'}} + D^{^{'}} F

= B D + B D + BE + D^{1 +^{'}} + D^{^{'}} F

Add similar elements:

B D + B D = 2 B D

= 2 B D + BE + D^{^{^{'}} + 1} + D^{^{^{'}}} F

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What is BD+B(D+E)+D^'(D+F) ?
The solution to BD+B(D+E)+D^'(D+F) is 2BD+BE+D^{1+'}+D^'F