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MATH203-1403A-02Applications of Discrete

MathematicsPhase 1 / Individual

ProjectMark L. Simon II

Instructor: Doris SchantzJuly 08, 2014

InstructionsDemonstrate DeMorgan’s Laws using a Venn diagram. 1. Draw a Venn diagram showing the elements of sets A, B, and the universe for all 4 regions.2. Draw a second diagram showing only the elements of the complement of set A.3. Draw a third diagram showing only the elements of the complement of set B.4. Draw a fourth diagram showing the union of steps 2 and 3 for the first DeMorgan's law or the intersection of steps 2 and 3 for the second DeMorgan's law.5. Draw a fifth diagram showing the elements of the intersection of sets A and B for the first DeMorgan's law or the union of sets A and B for the second DeMorgan's law.6. Finally, draw the last diagram showing the complement of step 5. Compare the results from step 4 against those in step 6 to prove both DeMorgan's laws. In the absence of data elements, you can use 4 different colors to clearly indicate the regions of the universe and follow all of the steps.

InstructionsDemonstrate DeMorgan’s Laws using a Venn diagram. 1. Draw a Venn diagram showing the elements of

sets A, B, and the universe for all 4 regions.

SET – A = Birds with feathers that can fly {eagle, sparrow, robin, hawk, owl , chicken, road runner}

SET – B = Birds with feathers that cannot fly {kiwi, chicken, black footed penguin, emu, ostrich, chicken, road runner}

A u B = {eagle, sparrow, robin, hawk, owl , starling , kiwi, chicken, black footed penguin, emu, ostrich, chicken, road runner}

Eagle, Sparrow, Robin, Hawk, Owl

Kiwi, Black Footed Penguin , Emu, Ostrich

U

A B

Chicken, Road runner

(Region 1)

(Region 2) (Region 3)

(Region 4)

Instructions

2. Draw a second diagram showing only the elements of the complement of set A.

Compliment of A = Ac

Yellow - is designated as the compliment area.

NOTE: The Black Footed Penguin – The African Black Footed Penguin has neither the ability to fly and is not considered to have real feather-like qualities. The feathers have adapted to a feather-like dense fir for cooling and heating for the area and has mass quantities of oil within the pelt. Therefore, it will be considered a “Composite for certain slides.


Kiwi, Penguin, Emu, Ostrich

U

A B


Black Footed Penguin /composite

Compliment of A = Ac

Instructions3. Draw a third diagram showing only the elements of the complement of set B.

Compliment of B = Bc

Yellow - is designated as the compliment area.



U

A B


Black Footed Penguin /composite

Compliment of B = Bc

Instructions

4. Draw a fourth diagram showing the union of steps 2 and 3 for the first De Morgan's law or the intersection of steps 2 and 3 for the second De Morgan's law.

Complement of A U B = Birds with feathers that can fly and can’t fly In Demorgan’s Law of union compliment of the diagram on the next image is designated as light blue.

Intersection of Ac and Bc = Ac Bc

In Demorgan’s Law of Intersect, which is the second slide from this slide, depicts all areas where intersection takes place with complement. The complement is shown as the featherless / flightless Black Footed Penguin. However, it is noted that the Road Runner and Chicken do have feathers, but choose not to fly even though they are capable of flight for short periods.



U

A B


Black Footed Penguin /Composite

Complement of A U B = (A U B)c



UA B




Eagle, Sparrow, Robin,

Hawk, Owl Kiwi, Penguin, Emu,

OstrichChicken, Road runner

Intersection of Ac and Bc = Ac Bc






A B A B

Instructions

5. Draw a fifth diagram showing the elements of the intersection of sets A and B for the first De Morgan's law or the union of sets A and B for the second De Morgan's law. I will show both.

Complement of Ac Bc = (A U B)c

Complement of Ac Bc = (A B)c

. Those items not in the set are designated in light blue. SET – A = Birds with feathers that can fly {eagle, sparrow, robin, hawk, owl , chicken, road runner}

SET – B = Birds with feathers that cannot fly {kiwi, chicken, black footed penguin, emu, ostrich, chicken, road runner}

A U B = {eagle, sparrow, robin, hawk, owl , starling , kiwi, chicken, black footed penguin, emu, ostrich, chicken, road runner}

Eagle, Sparrow, Robin, Hawk, Owl Kiwi, Emu, Ostrich

U A B


Union of Ac Bc = (A U B)c

Chicken, Road Runner


Hawk, Owl



A B


Hawk, Owl





AB

The (A B)c shows the connection between the two ovals. However, the final outcome still encompasses the two ovals in the De Morgan Law leaving only the black footed penguin as the complement. As suggested on next slide.



U

A B





Intersection of Ac U Bc = (A B)c OR Union of Ac Bc = (A U B)c OR Both






AB A B


The outcome is the same for both De Morgan Laws with the end result being the black footed Penguin being the complement.

Instructions.6. Finally, draw the last diagram showing the complement of step 5. Compare the results from step 4 against those in step 6 to prove both De Morgan's laws. In the absence of data elements, you can use 4 different colors to clearly indicate the regions of the universe and follow all of the steps.

Within these slides I have shown that both De Morgan LawsComplement of Ac Bc = (A U B)c


Are true. One law works from the outside in and the other law works form the inside out with the same result in the end.

Eagle, Sparrow, Robin, Hawk, Owl Kiwi, Emu, Ostrich

U



Hawk, Owl



A B


Hawk, Owl





AB

U U


A

Ac Bc

Intersect A and B compliment should be this.

Chicken, Road Runner

B

Instructions / Part II

Define two propositions (simple statements that can be either true or false). Give a real-world example of 2 propositions that r and s can represent. Call them r and s. Create a truth table that shows all values of the following:

Propositions – These are basically described as a declarative statements added into a function

Proposition Definition Your Answer Explained

r Set - r Is a basic compilation of values within the r-set. In this case “ I am very cold”

s Set -s Is a basic compilation of values within the s - set. In this case “ I need to start a fire”.

¬ r Negation of set - r This refers to “It is not the case that”

It “is not” the case that set – r “I am very cold” is true or false. When applied to a table this set can be either true or false.

¬ s Negation of set – s This refers to “It is not the case that”

It “is not” the case that set – r “I need to start a fire” is true or false. When applied to a table this set can be either true or false.

r ^ s (^) stands for the conjunction of “r” AND “s”

This is considered a truth value and determines weather the values are true or false. It also helps decide the outcome or conclusion

CONTINUED – NEXT SLIDE



r Set - r Is a basic proposition of value within the r-set. In this case “ I am very cold”

s Set -s Is a basic proposition of value within the s - set. In this case “ I need to start a fire”.

Set R = I am really cold

Set S = I need to start a fire

These are considered the propositions we will use when inputting the functions of ^ =and, v = or, and the codes ¬ or ~ = “it is not the case that” are placed within the function. They define the hypothesis of the outcome.

RT

F

ST

F


Proposition Definition Your Answer Explained¬ r Negation of set - r This refers to “It is

not the case that”It “is not” the case that set – r “I am very cold” is true or false. When applied to a table this set can be either true or false using the “and” and “or”.

¬ s Negation of set – s This refers to “It is not the case that”

It “is not” the case that set – s “I need to start a fire” is true or false. When applied to a table this set can be either true or false using the “and” and “or”.

R ¬R

T F

F T

S ¬S

T F

F T

Each table shows the negation of Set ~R and Set ~ S. The negation of this is Set – S and Set – R. The opposite truth or fallacy exists way function the proposition against itself.



r ^ s (^) stands for the conjunction of “r” AND “s”

This table runs simple input/output to come to a comclusion as follows:1. It is true that I am cold and it’s true I need a fire to get warm (True)2. It is true that I am cold and I don’t need to start a fire as I am inside a

tent. (False)3. It’s false that I am cold and I do need to start a fire. (True)4. It’s false that I am cold and it’s false that I need to start a fire. Don’t

know how to start a fire. (false)

R S R ^ ST T T

T F F

F T F

F F F



¬ r v s “It is not the case” that “I am cold OR I need to start a fire.Negation is stated.

This can go one of four ways1. It is true that I am not cold and it’s true I need a fire to cook with.2. It is true that I am not cold and I don’t need a fire as I am inside a tent.3. It’s false that I am not cold and I do need to start a fire.4. It’s false that I am not cold and it’s false that I can start a fire as I don’t

know how to start a fire.

¬R S ¬R v ST T T

T F T

F T T

F F F



r ^ ¬ s I am cold and “it is not the case that I can start a fire

This can go one of four ways1. It is true that I am cold or it’s not the case that I need a fire, This is false as I

am cold. 2. It is true that I am cold or it is not the case that I need to start a fire as I am

inside a tent. This is true.3. It’s false that I am cold or it is not the case that I need to start a fire. This is

true. I’m not cold.4. It’s true that I am cold or it’s true that I can’t start a fire as I don’t know how to

start a fire. This is true.

R ¬S R ^ ¬S

T F F

T T T

F F F

F T F

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