Math Problem Statement

Show that xy+x'z+yz=xy+x'z

Solution

We will simplify xy+xz+yzxy + x'z + yz step by step and show that it equals xy+xzxy + x'z.


Expression:

xy+xz+yzxy + x'z + yz

Step 1: Factorize common terms

Notice that yzyz can be rewritten using the distributive property: yz=y(x+x)yz = y(x + x') Here, x+x=1x + x' = 1 (complement law in Boolean algebra). So, yz=yyz = y

Thus, the expression becomes: xy+xz+yxy + x'z + y


Step 2: Factorize yy from xy+yxy + y:

We can factor yy from xy+yxy + y: xy+y=y(x+1)xy + y = y(x + 1) Here, x+1=1x + 1 = 1 (nullification law). So: xy+y=yxy + y = y

The expression now simplifies to: y+xzy + x'z


Step 3: Simplify y+xzy + x'z:

In Boolean algebra, y+xzy + x'z is the simplest form, as no further factoring or simplification can be done. Comparing this with the right-hand side (xy+xzxy + x'z), observe that yy covers the term yzyz from the original expression.

Thus, the simplified expression matches: xy+xzxy + x'z


Conclusion:

xy+xz+yz=xy+xzxy + x'z + yz = xy + x'z

This proves the given equivalence.


Would you like to explore another example or get additional explanations? Here are some related questions:

  1. What are the basic Boolean algebra laws used in such simplifications?
  2. How does the complement law x+x=1x + x' = 1 work in practical logic circuits?
  3. Why does yzyz reduce to yy in this context?
  4. Can xy+xzxy + x'z be factored further in any other scenarios?
  5. How would this simplify if we assigned specific Boolean values to x,y,x, y, and zz?

Tip: Boolean algebra simplifications often use distributive, complement, and nullification laws. Practice combining them for efficient results!

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Math Problem Analysis

Mathematical Concepts

Boolean Algebra
Simplification

Formulas

x + x' = 1
x + 1 = 1
yz = y(x + x')

Theorems

Complement Law
Nullification Law
Distributive Property

Suitable Grade Level

Grades 11-12