How to solve this using boolean algebra?

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Hello! How to prove RHS from LHS?

A+B'C+A'BD'+AD+B'D = A + B’C + BD’ + B’D
 

Consider LHS
A+B'C+A'B'D'+AD+B'D
=>A+B'C+A'BD'+AD(B+B')+BD' (SINCE B+B'=1)
=>A+B'C+A'BD'+ABD'+ABD+B'D
=>A+B'C+BD'(A+A')+ABD+B'D
=>A+B'C+BD'+ABD+B'D
=>A(1+BD)+B'C+BD'+B'D
=>A+B'C+BD'+B'D=LHS
 
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    ADGAN

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Thank you very much!

- - - Updated - - -

When going from 1st line to 2nd line, how did AD(B+B') = ADB+ABD'? Isn't it ADB + ADB'?
 

Sorry i have done mistake Please check this sol.
CONSIDER RHS
A+B'C+BD'+B'D
=>A(D+D')+B'C+BD'(A+A')+B'D
=>AD+AD'+B'C+ABD'+A'BD'+B'D
=>AD+AD'(1+B)+B'C+A'BD'+B'D
=>AD+AD'+BC'+A'BD'+B'D
=>A(D+D')+BC'+A'BD'+B'D
=>A+BC'+A'BD'+B'D
=>A(1+D)+BC'+A'BD'+B'D
=>A+AD+BC'+A'BD'+B'D=LHS
 
Last edited:
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    ADGAN

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Thanks again ! Pls can you help me with this one also.

A'D'+ABC+BCD' = ABC+A'D'

The original one for the LHS was (A'+BC)((A'D)'). I simplified it to A'D'+ABC+BCD'.
 

whenever you find any boolean expression that cannot be solved directly, try to add variables like (1+b),(b+b').....
consider LHS
A'D'+ABC+BCD'
=>A'D'(B+B')+BCD'(A+A')+ABC
=>A'D'B+A'D'B'+ABCD'+A'BCD'+ABC
=>A'D'B(1+C)+A'D'B'+ABC(1+D')
=>A'D'B+A'B'D'+ABC
=>A'D'(B+B')+ABC
=>A'D'+ABC=RHS
 
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