Math Problem Statement

Simplify the following boolean expressions:

  1. F= ((A+BC') + D(E+F))
  2. F = ((A+B+C)D)'
  3. F = AB'+ A(B+C)' + B(B+C)'
  4. F = [AB'(C+BD)A'B'|C
  5. F= A'BC + AB'C' + A'B'C' + AB'C + АВС

Solution

Let's simplify the given Boolean expressions step by step:


1. F=(A+BC)+D(E+F)F = (A + BC') + D(E + F)

Solution:

  • A+BCA + BC' is already in its simplest form.
  • D(E+F)D(E + F) expands to DE+DFDE + DF.
  • So, F=(A+BC)+DE+DFF = (A + BC') + DE + DF.

Thus, the simplified form is:

F=A+BC+DE+DFF = A + BC' + DE + DF


2. F=((A+B+C)D)F = ((A + B + C)D)'

Solution:

  • Apply De Morgan's Law:

    F=(A+B+C)+DF = (A + B + C)' + D'

  • Apply De Morgan’s Law on (A+B+C)(A + B + C)':

    F=ABC+DF = A'B'C' + D'

Thus, the simplified form is:

F=ABC+DF = A'B'C' + D'


3. F=AB+A(B+C)+B(B+C)F = AB' + A(B + C)' + B(B + C)'

Solution:

  • A(B+C)A(B + C)' can be expanded by De Morgan's Law as A(BC)A(B'C').
  • B(B+C)B(B + C)' can be expanded similarly as B(BC)=0B(B'C') = 0 (since BB=0BB' = 0).

So, the expression simplifies to:

F=AB+ABCF = AB' + AB'C'

Finally, factoring out ABAB', we get:

F=AB(1+C)=ABF = AB'(1 + C') = AB'


4. F=AB(C+BD)+AB+CF = AB'(C + BD) + A'B' + C

Solution:

  • First expand AB(C+BD)AB'(C + BD):

    ABC+ABBD=ABC+ABBD=ABC+ABDAB'C + AB'BD = AB'C + AB'BD = AB'C + AB'D

  • Now the expression becomes:

    F=ABC+ABD+AB+CF = AB'C + AB'D + A'B' + C

  • Combine like terms:

    F=C+ABD+ABF = C + AB'D + A'B'

Thus, the simplified form is:

F=C+ABD+ABF = C + AB'D + A'B'


5. F=ABC+ABC+ABC+ABC+ABCF = A'BC + AB'C' + A'B'C' + AB'C + ABC

Solution:

Group terms and look for common factors:

  • Group ABCA'BC and ABCA'B'C' together, and ABCAB'C', ABCAB'C, and ABCABC together:

    F=A(BC+BC)+AB(C+C)+ABCF = A'(BC + B'C') + AB'(C + C') + ABC

  • Simplify:

    • BC+BC=(BC)BC + B'C' = (B \oplus C)'
    • C+C=1C + C' = 1

So, the expression simplifies to:

F=A(BC)+AB+ABCF = A'(B \oplus C)' + AB' + ABC

Thus, the final simplified form is:

F=A(BC)+AB+ABCF = A'(B \oplus C)' + AB' + ABC


Summary of Simplified Expressions:

  1. F=A+BC+DE+DFF = A + BC' + DE + DF
  2. F=ABC+DF = A'B'C' + D'
  3. F=ABF = AB'
  4. F=C+ABD+ABF = C + AB'D + A'B'
  5. F=A(BC)+AB+ABCF = A'(B \oplus C)' + AB' + ABC

Would you like more details on any of these steps or further clarification? Here are 5 related questions:

  1. How does De Morgan's law help in simplifying Boolean expressions?
  2. Can Boolean expressions be simplified further after applying De Morgan’s law?
  3. What are the key properties of Boolean algebra used for simplification?
  4. How is factoring useful in simplifying Boolean expressions?
  5. How do we approach simplifying expressions with multiple terms like ABC+ABC+ABCA'BC + AB'C' + A'B'C'?

Tip: Using Karnaugh maps (K-maps) is another helpful tool to simplify Boolean expressions visually.

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Math Problem Analysis

Mathematical Concepts

Boolean Algebra
Logic Simplification
De Morgan's Law

Formulas

De Morgan's Law
Distribution and Factoring in Boolean Algebra

Theorems

De Morgan's Theorems
Distributive and Complement Laws of Boolean Algebra

Suitable Grade Level

Grades 11-12, University level