Simple proof of Boolean Algebra

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SUMMARY

The discussion focuses on proving the Boolean algebra equation (x+y)(y+z)(x'+z) = (x+y)(x'+z). The user successfully applies the distributive property to simplify the left-hand side to (y+xz)(x'+z) and further to x'y + yz + xz. The final step requires the application of the absorption law by adding the term xx', which does not alter the equation due to the property xx' = 0, allowing for the factorization to reach the desired result.

PREREQUISITES
  • Understanding of Boolean algebra principles
  • Familiarity with the distributive property in Boolean expressions
  • Knowledge of the absorption law in Boolean algebra
  • Ability to manipulate Boolean expressions using identities
NEXT STEPS
  • Study the absorption law in Boolean algebra
  • Learn about the distributive property in Boolean expressions
  • Explore Boolean algebra identities and their applications
  • Practice simplifying Boolean expressions with various examples
USEFUL FOR

Students studying digital logic design, computer science learners, and anyone interested in mastering Boolean algebra for applications in circuit design and optimization.

nahanksh
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Homework Statement


Prove (x+y)(y+z)(x'+z) = (x+y)(x'+z)


Homework Equations





The Attempt at a Solution



On LHS, i have done till

=(y+xz)(x'+z) 1st- distributive

=x'y+yz+xz 2nd- distributive(2ce)

=(x+y)(x'+z) 3rd-?


For third procedure, what properties should be used?
I think it could be done in one step but i can't figure it out...

Could someone please help me out here?
 
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Add xx' so you can factor it. It won't change anything since xx'=0.
 

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