# Implication, Boolean Expression and Venn Diagrams

• I
• DanielMB
In summary: In the case of "A implies B" Boolean expression "-A + B" includes Venn diagrams 1, 3 and 5, Why those diagrams do not accomplish the truth table? Is that the Boolean expression is not equivalent to "A implies B"?We can't say anything about ##\lnot A \wedge B## if ##A \Longrightarrow B## is given. It makes a statement under the condition that ##A## is true. If ##A## is false, then we have no statement and anything can be, ##B## as well as ##\lnot B##.
DanielMB
TL;DR Summary
Relationship among implication, Boolean expression and Venn diagrams
Hello, I’m having difficulties understanding logical relations “A implies B” and “A if and only if” using Boolean expressions and Venn diagrams, there is something where I’m wrong, but I could not find it out. Please, be benevolent and tell me where I’m wrong. Thanks

Note : Obviously I’m supposing that A,B are not null sets

A implies B

Representing “A implies B” using Boolean expression as “–A+B” and the correlated Venn diagrams, knowing that the truth table for “A implies B” is (see attached image)

“A implies B” can be represented by the following Venn diagrams, all of them acomplishing “-A+B” expression (see attached image)

But I’ve found that those diagrams (Vn) are not full compatible with the corresponding truth table

V1: a, b, -c, -d -> Not according to truth table​
V2: a, b, -c, d -> According to truth table​
V3: a, -b, -c, d -> Not according to truth table (but according to “A if and only if B” truth table)​
V4: a, b, -c, d -> According to truth table​
V5: a, -b, -c, d -> Not according to truth table (but according to “A if and only if B” truth table)​

Should I discard (V1, V3, V5)? Are there additional hypotheses?, like the following

H1) A, B with not null intersection : to reject V1​
H2) B is not an A subset : to reject V3​
H3) A not equal to B : to reject V5​

A if and only if B

Representing “A if and only if B” using Boolean expression as “(A.B)+(-A.-B)” and the correlated Venn diagrams, knowing that the truth table for “A if and only if B” is (see attached image)

“A if and only if B” can be represented by the following Venn diagrams, all of them acomplishing “(A.B)+(-A.-B)” expression (see attached image)

Note : V3 and V4 are equivalent, because “A if and only if” is conmutative

But I’ve found that those diagrams (Vn) are not full compatible with the corresponding truth table

V1: a, -b, -c, -d -> Not according to truth table​
V2: a, -b, -c, d -> According to truth table​
V3: a, -b, -c, d -> According to truth table​
V4: a, -b, -c, d -> According to truth table​
V5: a, -b, -c, d -> According to truth table​

H1) A, B with not null intersection : to reject V1​

Thanks

I have difficulties to understand your diagrams. ##A \Longrightarrow B## means, that if ##x\in A## then ##x\in B##, i.e. ##A\subseteq B## (#4).

And for the other one we have ##A \subseteq B ## and ##B\subseteq A## (#5).

What is it that you don't understand?

fresh_42 said:
I have difficulties to understand your diagrams. ##A \Longrightarrow B## means, that if ##x\in A## then ##x\in B##, i.e. ##A\subseteq B## (#4).

And for the other one we have ##A \subseteq B ## and ##B\subseteq A## (#5).

What is it that you don't understand?
Thanks fresh_42,

In the case of "A implies B" Boolean expression "-A + B" includes Venn diagrams 1, 3 and 5, Why those diagrams do not accomplish the truth table? Is that the Boolean expression is not equivalent to "A implies B"?

DanielMB said:
Thanks fresh_42,

In the case of "A implies B" Boolean expression "-A + B" includes Venn diagrams 1, 3 and 5, Why those diagrams do not accomplish the truth table? Is that the Boolean expression is not equivalent to "A implies B"?
We can't say anything about ##\lnot A \wedge B## if ##A \Longrightarrow B## is given. It makes a statement under the condition that ##A## is true. If ##A## is false, then we have no statement and anything can be, ##B## as well as ##\lnot B##.

E.g.:
If it rains, the road will be wet. (##A## = it rains; ##B## = wet road; ##A \Longrightarrow B##)
Now if it does not rain, then the road can be dry, or wet because a cleaning machine ran over it, or someone has washed his car or whatever. Everything is possible, except that it rains.

Michael Price
Thanks Justin, I'll watch the videos

@fresh_42 Well, you can say that: if A(1) ⇒B(1) is also certain that B(0) ⇒ A(0)

Please, look at the "if and only if" image, specifically the diagram #2 (green parts are according to boolean expression), I can't assert that x ∈ A ⇒ x ∈ B and viz., the only valid graph for the biconditional is #5, representing the equivalence relationship between A and B. I know that Wikipedia is not a trustable source of information, but why uses diagram #2.? Am I wrong?

DanielMB said:
Thanks Justin, I'll watch the videos

@fresh_42 Well, you can say that: if A(1) ⇒B(1) is also certain that B(0) ⇒ A(0)
Yes, ##\left( A \Longrightarrow B \right) \Longleftrightarrow \left( \lnot B \Longrightarrow \lnot A \right)##

Please, look at the "if and only if" image, specifically the diagram #2 (green parts are according to boolean expression), I can't assert that x ∈ A ⇒ x ∈ B and viz.,...
Why not? ##\left( A \Longleftrightarrow B \right) \Longleftrightarrow \left( (A \Longrightarrow B ) \wedge (B \Longrightarrow A)\right)## or in set speak: ##A \subseteq B \wedge B\subseteq A
\Longleftrightarrow A=B##
... the only valid graph for the biconditional is #5, representing the equivalence relationship between A and B.
Right, two equal sets.
I know that Wikipedia is not a trustable source of information,...
The mathematical section isn't so bad. Wiki cannot be trusted for what you read about celebrities or politicians, because they want to look good. Venn diagrams don't care what people think. More professional is nLab but not as easy.
... but why uses diagram #2.? Am I wrong?
Please give us a precise link and reference. Diagram 2 in what you wrote about ##A\Longleftrightarrow B## shows ##A\cap B## which is ##A \wedge B##, or the exclusive OR, depending on whether you mean the intersection or the white sets. Neither is an equivalence.

Equivalence means the same sets; at least here. Equivalence isn't equality, so it does not mean the same sets - only if you use those Venn diagrams. E.g. one could say that ##\frac{2}{4}=\frac{1}{2}##, or that they are equivalent, because it makes a difference whether you come home with ##\frac{2}{4}## pieces of cake or with half a cake. But if we only draw sets ##A=\{\,\frac{2}{4}\,\}## and ##B=\{\,\frac{1}{2}\,\}## and ask whether they are equivalent, then we have to say ##A=B##.

DanielMB
I think there is a simpler way to look at this: the truth tables have 4 rows; only 1 of the Venn diagrams in each set has 4 areas so only that diagram can represent the full truth table.

To put it another way, all of the diagrams except Diagram 2 illustrate a special case - for instance Diagram 1 illustrates the special case ## A \cap B = \emptyset ##, so the fourth line in the truth table is not represented by any part of the diagram. Can you identify the other special cases and the related rows in the truth tables?

pbuk said:
I think there is a simpler way to look at this: the truth tables have 4 rows; only 1 of the Venn diagrams in each set has 4 areas so only that diagram can represent the full truth table.

To put it another way, all of the diagrams except Diagram 2 illustrate a special case - for instance Diagram 1 illustrates the special case ## A \cap B = \emptyset ##, so the fourth line in the truth table is not represented by any part of the diagram. Can you identify the other special cases and the related rows in the truth tables?

Thanks pbuk,

I tried to identify for both cases the accomplishing or not for the related rows in the truth table

"A ⇒ B"
Diagram #1 : (1, 2 , nothing, nothing) ⇒ OK​
Diagram #2 : (1, 2, 3, 4) ⇒ OK​
Diagram #3 : (1, not 2, 3, 4) ⇒ Not according with the truth table !?​
Diagram #4 : (1, 2, 3, 4) ⇒ OK​
Diagram #5 : (1, nothing, nothing, 4) ⇒ OK​

"A ⇔ B"
Diagram #1 : (1, nothing , nothing, nothing) ⇒ OK​
Diagram #2 : (1, 2, 3, 4) ⇒ OK​
Diagram #3 : (1, 2, 3, 4) ⇒ OK​
Diagram #4 : (1, 2, 3, 4) ⇒ OK​
Diagram #5 : (1, nothing, nothing, 4) ⇒ OK​

Please, tell me if I'm wrong

Which area of Diagram 3 shows ## B \wedge A ##, the second line of the truth table? What about Diagram 4?

DanielMB
pbuk said:
Which area of Diagram 3 shows ## B \wedge A ##, the second line of the truth table? What about Diagram 4?

Yes, you are right, it doesn't appear represented in the diagram 3 (A implies B)
In the diagram 4 (A implies B) each line in the truth table has its own representation

DanielMB said:
In the diagram 4 (A implies B) each line in the truth table has its own representation
Including line 3 ## A \wedge B ## ?

@pbuk

The whole diagram is allowed (green) but it is not posible, since A is a subset of B, so A determines B value

## 1. What is an implication in Boolean expressions?

An implication in Boolean expressions is a logical statement that connects two propositions, known as the antecedent and consequent. It is represented by the symbol "→" and can be read as "if-then". The implication is true when the antecedent is false or when both the antecedent and consequent are true.

## 2. How do you simplify a Boolean expression?

To simplify a Boolean expression, you can use the laws of Boolean algebra, such as the commutative, associative, and distributive laws. You can also use De Morgan's laws to simplify expressions with negations. Additionally, you can use Karnaugh maps to visually simplify more complex Boolean expressions.

## 3. What is a Venn diagram and how is it used?

A Venn diagram is a graphical representation of the relationships between sets. It consists of overlapping circles, with each circle representing a set and the overlapping area representing the elements that are common to both sets. Venn diagrams are used to visually represent logical relationships and to solve problems involving set operations.

## 4. How do you determine the validity of a Venn diagram?

To determine the validity of a Venn diagram, you can use the rules of set theory. The diagram must accurately represent the elements of the sets and their relationships, such as intersections and unions. Additionally, the diagram should follow the laws of logic, such as the law of non-contradiction, which states that a statement and its negation cannot both be true.

## 5. What is the difference between a conditional and a biconditional statement?

A conditional statement, also known as an implication, is a logical statement that connects two propositions with the "if-then" relationship. A biconditional statement is a logical statement that connects two propositions with the "if and only if" relationship. In other words, a biconditional statement is true when both propositions have the same truth value, while a conditional statement can be true even if the two propositions have different truth values.

• Set Theory, Logic, Probability, Statistics
Replies
1
Views
2K
• Set Theory, Logic, Probability, Statistics
Replies
11
Views
3K
• Engineering and Comp Sci Homework Help
Replies
24
Views
3K
• General Math
Replies
2
Views
1K
• Engineering and Comp Sci Homework Help
Replies
14
Views
4K
• Precalculus Mathematics Homework Help
Replies
7
Views
2K
• Special and General Relativity
Replies
6
Views
1K
• Special and General Relativity
Replies
6
Views
1K
• Engineering and Comp Sci Homework Help
Replies
1
Views
1K
• Electrical Engineering
Replies
1
Views
2K