Boolean Simplification

Question

I'm trying to simplify the following booleans:

Y = ¬A¬B¬C¬D + ¬A¬B¬CD + ¬A¬BC¬D + ¬A¬BCD + A¬B¬C¬D + A¬BC¬D + ABC¬D

The solution is:

Y = ¬A¬B + ¬B¬D + AC¬D

My solution is:

Y = ¬A¬B (¬C¬D + ¬CD + C¬D + CD) + A¬B¬C¬D + A¬BC¬D + ABC¬D
  = ¬A¬B (¬C (¬D + D) + C (¬D + D)) + A¬B¬C¬D + A¬BC¬D + ABC¬D
  = ¬A¬B (¬C + C) + A¬B¬C¬D + A¬BC¬D + ABC¬D
  = ¬A¬B + A¬D (¬B¬C + ¬BC + BC) 
  = ¬A¬B + A¬D (¬B¬C + (¬BC + ¬BC) + BC) // law of idempotency
  = ¬A¬B + A¬D (¬B (¬C + C) + C (¬B + B))
  = ¬A¬B + A¬D (¬B + C)
  = ¬A¬B + A¬B¬D + AC¬D

As you can see.. my answer varies with the solution by the 2nd term. Any ideas? Thanks.

Answer 1

All your work is right. The first two terms of your answer simplify to:

Y = ¬A¬B + A¬B¬D
  = ¬B(¬A + A¬D)
  = ¬B(¬AD + ¬A¬D + A¬D) // split the first term
  = ¬B(¬AD + ¬A¬D + ¬A¬D + A¬D) // idempotence
  = ¬B(¬A + ¬D) // combine the first two and last two terms
  = ¬A¬B + ¬B¬D.

I wouldn't have believed it myself at first, but idempotence lets the ¬A "spit out" a ¬A¬D, which is, after all, just a subset.

Answer 2

I agree with you. The last three terms all have A required, so anything that comes out of them (your last two terms) need an A. Are you sure the given solution is right?