Contrapositive of quantified statement

[Mr Davis 97]

I have the following statement: Let $a,b \in \mathbb{R}$. If $a \le b_1$, for every $b_1 > b$, then $a \le b$. I have put it into logical notation in the following way: $\forall a,b, b_1 \in \mathbb{R} ((b_1 > b \rightarrow a \le b_1) \rightarrow a \le b)$. My question is, if I want to write this implication as its contrapositive, how would the quantifiers change and why?

 

[andrewkirk]

To see how to write the contrapositive, first write the statement formally, in prenex normal form, as follows:

$$\forall a \forall b \exists b1\left( (a\in\mathbb R\wedge b\in\mathbb R\wedge b1\in\mathbb R\wedge a\leq b1 \wedge b1 > b)
\to a\leq b \right) $$

Then, to make the contrapositive, swap the antecedent with the consequent, and negate them both. Do that inside the outer parenthesis, leaving the quantifiers unchanged (there are logical axiom justifying that, but it's probably too much detail to go into those).

Having done that, you can convert from prenex normal form back into a more prosy form that may seem more intuitive.

 

[Mr Davis 97]

To see how to write the contrapositive, first write the statement formally, in prenex normal form, as follows:

$$\forall a,b,b1\left( (a\in\mathbb R\wedge b\in\mathbb R\wedge b1\in\mathbb R\wedge a\leq b1 \wedge b1 > b)
\to a\leq b \right) $$

Then, to make the contrapositive, swap the antecedent with the consequent, and negate them both. Do that inside the outer parenthesis, leaving the quantifiers unchanged (there are logical axiom justifying that, but it's probably too much detail to go into those).

Having done that, you can convert from prenex normal form back into a more prosy form that may seem more intuitive.

$\forall a,b, b_1 \in \mathbb{R} (a>b \rightarrow (b_1>b \wedge a > b_1))$. Would this be the logically equivalent statement? What is it really saying?

 

[andrewkirk]

$\forall a,b, b_1 \in \mathbb{R} (a>b \rightarrow (b_1>b \wedge a > b_1))$. Would this be the logically equivalent statement? What is it really saying?

No. You have negated the consequent correctly, but not the antecedent. You need to use de Morgan's laws when negating conjunctions.

 

[Stephen Tashi]

I have the following statement: Let $a,b \in \mathbb{R}$. If $a \le b_1$, for every $b_1 > b$, then $a \le b$. I have put it into logical notation in the following way: $\forall a,b, b_1 \in \mathbb{R} ((b_1 > b \rightarrow a \le b_1) \rightarrow a \le b)$.

Are you actually studying formal logic? Or are you only using concepts and notation from formal logic to express mathematical reasoning precisely in "natural language"?

I don't what syntax rules for formal logic you would use, but using correct syntax in a formal logic would produce something excruciating.

Let's take it for granted that all variables represent real numbers. The we have (using informal syntax):

$(\forall a) (\forall b) [ (\forall b_1)( (b_1 \gt b) \implies (a \le b_1)] \implies [a \le b]$

I want to write this implication as its contrapositive, how would the quantifiers change and why?

Which implication are you asking about?

 

[Mr Davis 97]

Are you actually studying formal logic? Or are you only using concepts and notation from formal logic to express mathematical reasoning precisely in "natural language"?

I don't what syntax rules for formal logic you would use, but using correct syntax in a formal logic would produce something excruciating.

Let's take it for granted that all variables represent real numbers. The we have (using informal syntax):

$(\forall a) (\forall b) [ (\forall b_1)( (b_1 \gt b) \implies (a \le b_1)] \implies [a \le b]$


Which implication are you asking about?

I am talking about the outer implication. But given your post, I have an additional question. You wrote the statement as $(\forall a) (\forall b) [ (\forall b_1)( (b_1 \gt b) \implies (a \le b_1)] \implies [a \le b]$, while I wrote it as $\forall a,b, b_1 \in \mathbb{R} ((b_1 > b \rightarrow a \le b_1) \rightarrow a \le b)$. Could you explain why you chose to place the $b_1$ in a different place? Is mine wrong and yours right, or are they both technically right?

 

[andrewkirk]

I am talking about the outer implication. But given your post, I have an additional question. You wrote the statement as $(\forall a) (\forall b) [ (\forall b_1)( (b_1 \gt b) \implies (a \le b_1)] \implies [a \le b]$, while I wrote it as $\forall a,b, b_1 \in \mathbb{R} ((b_1 > b \rightarrow a \le b_1) \rightarrow a \le b)$. Could you explain why you chose to place the $b_1$ in a different place? Is mine wrong and yours right, or are they both technically right?

Stephen's is correct. It says that if every number that is $>b$ is also $\geq a$ then $a\leq b$.

Yours is not correct, which can be observed by noting that it is false for the counter-example $a=3,b=2,B_1=4$, or indeed any case where $b<a\leq b_1$.

Yours is in pre-nex normal form, meaning all quantifiers are at the front. To convert Stephen's into that form so we can make a like-with-like comparison, we need to follow the rules set out here.

That gives us (using the first rule in the 'Implication' section):
$$
\forall a\forall b \exists b_1 \in \mathbb{R} ((b_1 > b \rightarrow a \le b_1) \rightarrow a \le b)
$$
which we see is different from your version:
$$
\forall a \forall b\forall b_1 \in \mathbb{R} ((b_1 > b \rightarrow a \le b_1) \rightarrow a \le b)
$$

The linked article about pre-nex form tells you all that's needed about how quantifiers change as we move them from one place to another.

 

[Mr Davis 97]

Stephen's is correct. It says that if every number that is $>b$ is also $\geq a$ then $a\leq b$.

Yours is not correct, which can be observed by noting that it is false for the counter-example $a=3,b=2,B_1=4$, or indeed any case where $b<a\leq b_1$.

Yours is in pre-nex normal form, meaning all quantifiers are at the front. To convert Stephen's into that form so we can make a like-with-like comparison, we need to follow the rules set out here.

That gives us (using the first rule in the 'Implication' section):
$$
\forall a\forall b \exists b_1 \in \mathbb{R} ((b_1 > b \rightarrow a \le b_1) \rightarrow a \le b)
$$
which we see is different from your version:
$$
\forall a \forall b\forall b_1 \in \mathbb{R} ((b_1 > b \rightarrow a \le b_1) \rightarrow a \le b)
$$

The linked article about pre-nex form tells you all that's needed about how quantifiers change as we move them from one place to another.

Okay, I see. I guess my most pressing question now is how could I have interpreted the original statement correctly and written $\forall a\forall b \exists b_1 \in \mathbb{R} ((b_1 > b \rightarrow a \le b_1) \rightarrow a \le b)$. From the original statement I don't really see where the existence quantifier comes from.

 

[andrewkirk]

Okay, I see. I guess my most pressing question now is how could I have interpreted the original statement correctly and written $\forall a\forall b \exists b_1 \in \mathbb{R} ((b_1 > b \rightarrow a \le b_1) \rightarrow a \le b)$. From the original statement I don't really see where the existence quantifier comes from.

Start with Stephen's version, which is a correct, literal formalisation of the natural language statement in the OP. Then convert it to pre-nex form using the method set out in the linked wiki section on how to convert to pre-nex (see my previous post). The $\exists$ arises from the rule I mentioned.

 

[Stephen Tashi]

Could you explain why you chose to place the $b_1$ in a different place?

@andrewkirk has explained details of the particular problem. As to how I, personally , approach such problems, I demand that each quantified variable have its own scope.

As I recall, in formal logic things go this way:

Each quantified variable has its own "scope". So the usual notation goes like:
$\forall x ( ...)$
$\exists x (...)$
Where the scope is put inside parenthesis or brackets of some sort.

If that notation is obeyed strictly then English phrases like "For each pair of real numbers x and y..." are denoted
$\forall x ( \ \forall y ( x \in \mathbb{R} \land y \in \mathbb{R} \implies ...)\ )$

It is common to see the above written in condensed notation like:
$\forall x,y ( x \in \mathbb{R} \land y \in \mathbb{R} \implies ...)$
Or even as:
$\forall x \in \mathbb{R}, y \in \mathbb{R} (...)$

But you must keep in mind that if condensed notation makes sense, then it must be possible to express the idea using the strict notation, where we precisely define the scope of each quantified variable. In the strict notation, scopes are nested inside other scopes rather than giving a set of several variables the same scope.

When you truly study formal logic, you study rules for manipulating strings of symbols. You have to use various theorems and conventions that justify using condensed notation and transforming statements with quantifiers in certain places to equivalent statement with quantifiers in different places. I don't remember the theorems, but I do retain a suspicious attitude about condensed notation such as $\forall x,y,z(...)$. When I translate an English language statement into symbolic form, I visualize it as a statement where each quantified variable has its own scope - even if I don't use the strict notation to indicate the scope.

Mathematicians who use the notations of formal logic just as abbreviations for common speech, become adept at manipulating quantified variables just by "commonsense" - meaning a trained commonsense. For example, they can see there is little danger in transforming a statement of the form $\forall x ( \forall y( \forall z(...)))$ to the form $\forall y,z,x (...)$ They know to be careful with situations like $\forall x ( \exists y ( \forall z (...) ) )$ and $\forall x ( \forall y ( \ (\forall z ( A(z,y)) \implies B(x,y))\ ))$.