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COS1501
Assignment 2 2024
(653506) - 14 June
2024
DETAILED ANSWERS
, 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.
COS1501
Assignment 2 2024
(653506) - 14 June
2024
DETAILED ANSWERS
, 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.
