Reduce the boolean expression to two literals

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The discussion revolves around simplifying the boolean expression ABC'D + A'BD + ABCD. The initial attempts at simplification led to an intermediate expression of BD(AC' + A' + AC), but errors were identified in the transformation process. After correcting these mistakes, the final simplification resulted in just BD, aligning with the solution provided by Wolfram Alpha. The user also referenced a boolean algebra identity, A + BC = (A + B)(A + C), but ultimately decided it wasn't helpful for their solution. The thread concludes with the user successfully solving the problem.
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Homework Statement



ABC'D + A'BD + ABCD

The Attempt at a Solution



Here is as far as I have gotten:

BD (AC' + A' + AC)
BD (AC' + A')

According to Wolfram Alpha, the answer should just be BD. Can someone help me get there?
 
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Given the problem statement is correct you have two errors in the first line (a missing negation and a case of swapped symbol). I recommend you start from the beginning and take each step slowly, making sure all the symbols are transformed correctly along the way.
 
I fixed the errors, sorry. Can someone tell me where to go next?
 
Never mind, just solved it!

Last steps:

BD (A(C+C') + A')
BD (A+A')
BD
 
Are you familiar with A+BC=(A+B)(A+C)?

EDIT: Nvm, this isn't going to help.
 

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