I Implication, Boolean Expression and Venn Diagrams

[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​

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

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

Thanks

 

[fresh_42]

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?

 

[DanielMB]

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"?

 

[jedishrfu]

To bolster your understanding Khan Academy comes to the rescue:

https://www.khanacademy.org/partner...pistemology/wiphi-language/v/conditionals-pt1
After seeing this video, you may well decide NOT to become a philosopher and so you will have give up the notion of idle deep thoughts on topics with no clear answers while imbibing some wine and grapes in a well pressed toga but I digress.

 

[fresh_42]

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.

 

[DanielMB]

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)

I understand your example

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?

 

[fresh_42]

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, $\left( A \Longrightarrow B \right) \Longleftrightarrow \left( \lnot B \Longrightarrow \lnot A \right)$

I understand your example

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 $A=B$.

 

[pbuk]

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?

 

[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?

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

 

[pbuk]

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

 

[DanielMB]

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

 

[pbuk]

In the diagram 4 (A implies B) each line in the truth table has its own representation

Including line 3 $A \wedge B$ ?

 

[DanielMB]

@pbuk

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