COS1501 Assignment
2 2024 (653506) - 14
June 2024
COMPLETE ANSWERS
[School]
[Course title]
, COS1501 Assignment 2 2024 (653506) - 14 June 2024
Question 1: Venn Diagram for [(A ⋂ B)' - C] ⋂ [(A + B) - C]
To solve this, let's break down the expression step by step and draw the Venn
diagrams accordingly:
1. [(A ⋂ B)' - C]:
o First, find the complement of A∩BA \cap BA∩B: This is everything
outside A∩BA \cap BA∩B.
o Subtract CCC from this complement.
2. [(A + B) - C]:
o Find A+BA + BA+B, which is A∪BA \cup BA∪B.
o Subtract CCC from A∪BA \cup BA∪B.
Combining these two results using intersection gives us the desired set.
Since I don't have the Venn diagrams to choose from (options a, b, c, d), you would
need to refer to those to identify which diagram correctly represents the set.
Question 2: Counterexample for (A - B) U C' = (C' - B) + A
To find a counterexample:
• Substitute each given set of A,B,A, B,A,B, and CCC into both sides of the
equation.
• Check if both sides are equal for any given set. If they are not equal for one set,
that set is a valid counterexample.
Let's go through the options:
2 2024 (653506) - 14
June 2024
COMPLETE ANSWERS
[School]
[Course title]
, COS1501 Assignment 2 2024 (653506) - 14 June 2024
Question 1: Venn Diagram for [(A ⋂ B)' - C] ⋂ [(A + B) - C]
To solve this, let's break down the expression step by step and draw the Venn
diagrams accordingly:
1. [(A ⋂ B)' - C]:
o First, find the complement of A∩BA \cap BA∩B: This is everything
outside A∩BA \cap BA∩B.
o Subtract CCC from this complement.
2. [(A + B) - C]:
o Find A+BA + BA+B, which is A∪BA \cup BA∪B.
o Subtract CCC from A∪BA \cup BA∪B.
Combining these two results using intersection gives us the desired set.
Since I don't have the Venn diagrams to choose from (options a, b, c, d), you would
need to refer to those to identify which diagram correctly represents the set.
Question 2: Counterexample for (A - B) U C' = (C' - B) + A
To find a counterexample:
• Substitute each given set of A,B,A, B,A,B, and CCC into both sides of the
equation.
• Check if both sides are equal for any given set. If they are not equal for one set,
that set is a valid counterexample.
Let's go through the options: