All math statements of the if A then B form?

[evagelos]

In a thread in philosophy it was pointed out by HalsofIvy that and i quote:

( Notice, by the way, that all statements in mathematics are of the form "If A then B").

There in my last post i used axioms from mathematical logic to oppose the above.

Since i got no answer there and since i used axioms from logic i present it here for a final decision

 

[CompuChip]

Can you please link to the relevant post or repeat the argument here? I don't feel like searching through the philosophy board just to find that one post.

 

[CRGreathouse]

When I learned formal logic (Enterton's text) the only connective we used was \Rightarrow; the others were abbreviations of some combination of this with negation. Of course, philosophically, every statement \phi is implicitly \mathcal{A}\Rightarrow\phi where A is the collection of axioms in the system...

 

[evagelos]

Originally Posted by HallsofIvy
Notice, by the way, that all statements in mathematics are of the form "If A then B". .

Originally Posted by evagelos
Here are the axioms of propositional calculus in mathematical logic:


......A----->( B------A)......... ...1

......( A----->( B-----C))-------->(( A----->B)------>(A---->C))...2


where A , B , C are statements.


Do you still insist that all statements in mathematics are of the form " If A then B"??

 

[HallsofIvy]

Since you give a collection of "if then" statements right there, what is your point?
(You have (B------A) and (B------C). I don't recognize that. Did you mean B----->A and B----->C?)

My point was that every statement in mathematics is based on axioms! Every statement, whether stated or not, starts "if the axioms are true then".

For example, the statement from Euclidean geometry, "The sum of the angle measures in any triangle is 180 degrees", which doesn't appear to be of the form "if A then B" really is "If all the postulates of Euclidean geometry are true and if ABC is a triangle with angles A, B, and C, then the measure of angle A + measure of angle B+ measure of angle C= 180 degrees.

 

[Focus]

Since you give a collection of "if then" statements right there, what is your point?
(You have (B------A) and (B------C). I don't recognize that. Did you mean B----->A and B----->C?)

My point was that every statement in mathematics is based on axioms! Every statement, whether stated or not, starts "if the axioms are true then".

For example, the statement from Euclidean geometry, "The sum of the angle measures in any triangle is 180 degrees", which doesn't appear to be of the form "if A then B" really is "If all the postulates of Euclidean geometry are true and if ABC is a triangle with angles A, B, and C, then the measure of angle A + measure of angle B+ measure of angle C= 180 degrees.

If you use not ¬ as well you can express any propositional logic statements.

 

[evagelos]

The 1st axiom on propositional calculus;

A----->( B------>A) is of the form:

if A then ( if B then A).

So where is the "if A then B" form ??

 

[CRGreathouse]

The 1st axiom on propositional calculus;

A----->( B------>A) is of the form:

if A then ( if B then A).

So where is the "if A then B" form ??

a\Rightarrow(b\Rightarrow c) is obviously of form A\Rightarrow B with A := a and B := (b\Rightarrow c).

 

[evagelos]

a\Rightarrow(b\Rightarrow c) is obviously of form A\Rightarrow B with A := a and B := (b\Rightarrow c).

1st of all there is a great difference between :

.......A----->( B------>A)..........1

and

........A------> ( B------->C)..........2

1 is a tautology whilst 2 is not
And as you may very well know we use only tautologies as axioms of predicate calculus.

if however in 1 we substitute B---->A By C the new formula : A----->C is just a conditional and not a tautology

So HallsofIvy your statement that:

All statements in mathematics are of the form " If A then B" is wrong

 

[HallsofIvy]

I see. What you are really saying is that you don't know what is meant by a "if then".

 

[evagelos]

I see. What you are really saying is that you don't know what is meant by a "if then".

yes sir i don't. But i can prove formally couple of theorems apart from those in symbolic logic.....can you?.For example i asked you in another thread to give a formal proof that:...nothing contains everything, in set theory and you did not respond.

Of course if you wish

 

[Focus]

yes sir i don't. But i can prove formally couple of theorems apart from those in symbolic logic.....can you?.For example i asked you in another thread to give a formal proof that:...nothing contains everything, in set theory and you did not respond.

Of course if you wish

I don't like where this is going. If you don't know what an "if" statement is, then don't argue about it . Its not really been a very constructive thread, more like a pissing contest.

 

[CRGreathouse]

1st of all there is a great difference between :

.......A----->( B------>A)..........1

and

........A------> ( B------->C)..........2

1 is a tautology whilst 2 is not
And as you may very well know we use only tautologies as axioms of predicate calculus.

if however in 1 we substitute B---->A By C the new formula : A----->C is just a conditional and not a tautology

So HallsofIvy your statement that:

All statements in mathematics are of the form " If A then B" is wrong

a\Rightarrow(b\Rightarrow a) is obviously a special case of a\Rightarrow(b\Rightarrow c). Do I need to do the translation for you?

a\Rightarrow(b\Rightarrow a) is of the form A\Rightarrow B with A := a and B := (b\Rightarrow a).

And don't taunt Halls, who clearly knows how to work from axioms on up: Number systems

 

[evagelos]

So a--->(b---->a) is a special case of a---->(b----->c)?

so the very 1st axiom of propositional calculus is special case of a---->(b---->c)

show me any book of logic that says so.

also it seems to me that you do not know what an identity is,because a----->(b---->a) is an identity while a---->(b----->c) is not.

Do i have to repeat myself??

Now try to prove one of the very 1sts theorems in propositional calculus: p----->p,by using as an axiom ...A------>B instead .....a------>( b----->a) by doing any substitution you like along the proof.

in this case M.Ponens is the only rule of inference.

 

[HallsofIvy]

You are repeating yourself: you keep saying over and over again that you don't understand what I am saying. a---->(b---->a) is itself of the form "If ... then"!

 

[evagelos]

GRGreathouse , in the axiom ...a----->( b----->a) you substitute :

...a with A,...b----->a , with B and you got :...A----->B.


Why did you not curry on the substitution ,and substitute ...A------->B ,with C ??

What would you get then?? Only ...C?

And where is the ' If A then B" statement now??

 

[evagelos]

You are repeating yourself: you keep saying over and over again that you don't understand what I am saying. a---->(b---->a) is itself of the form "If ... then"!

The a----->(b----->a) axiom is not of the "If...then" form if you reduce it to that form you cannot prove any theorems in propositional logic,and you convert it from a
tautology to a simple conditional statement.

Besides as i showed in my post #16 if you substitute a with A ...(b---->a)...with B,you get ...A----->B.And if you substitute now...A------>B...With C you get C Where isthe "If A then B' form now??

 

[Moo Of Doom]

You don't seem to understand that A and B are simply NOTATION. When you define some notation, you can always refer back to it. So when you say

"Let C be the statement 'A => B'"

and then ask where is the "If A then B" form now, looking at C, I can simply reply: refer to the definition of C. C is the statement 'A => B', and therefore clearly of that form.

Just because the symbol "C" is not written in that form doesn't mean that the statement C is not. C represents a statement, and that statement is of the form.

We also don't care about how we denote the hypothesis and the conclusion by symbols.

k => g
R => J
SOMETHING => SOMETHING ELSE

These are all of the form A => B, even though there is no "A" and no "B" there. This is what we mean by "form." Any collection of symbols you choose to consider a statement is of the form A => B precisely if there is the symbol "=>" somewhere in the middle of the collection. This holds also for

a => (b => a)

and

a => (b => c)

---

It's true that if you're simply given

"A => B"

as an axiom, you cannot conclude the same things as with the axiom

"a => (b => a)".

But the KEY here is that you're NOT given just "A => B" as an axiom but also the definitions of A and B, thus we are given:

"A => B, where A is the statement a, and B is the statement (b => a)".

This is sufficient to prove all the same theorems, since these are the same statement, written differently.

 

[HallsofIvy]

GRGreathouse , in the axiom ...a----->( b----->a) you substitute :

...a with A,...b----->a , with B and you got :...A----->B.


Why did you not curry on the substitution ,and substitute ...A------->B ,with C ??

What would you get then?? Only ...C?

And where is the ' If A then B" statement now??

It's hard to believe you are serious. If you CALL the statement "If A then B" "C", it is still an "if ... then" statement.

 

[evagelos]

It's hard to believe you are serious. If you CALL the statement "If A then B" "C", it is still an "if ... then" statement.

In the same way that you do a substitution of ..b------>a by B in a---->( b---->a) in exactly the same way i can do a substitution of...A------>B by C.

And let the "If...then" form disappear

 

[atyy]

The 1st axiom on propositional calculus;

A----->( B------>A) is of the form:

if A then ( if B then A).

So where is the "if A then B" form ??

It's true that if you're simply given

"A => B"

as an axiom, you cannot conclude the same things as with the axiom

"a => (b => a)".

Are "statements" and "axioms" different?

 

[evagelos]

You don't seem to understand that A and B are simply NOTATION.


But the KEY here is that you're NOT given just "A => B" as an axiom but also the definitions of A and B, thus we are given:

"A => B, where A is the statement a, and B is the statement (b => a)".

This is sufficient to prove all the same theorems, since these are the same statement, written differently.

Let me then define by C : a------>(b----->a) and use C as an axiom together with the said definition.

isn't that o.k

 

[CRGreathouse]

So 2*200 is even because it's in the form 2n, but when we write it 400 it's not anymore?

 

[CompuChip]

you convert it from a
tautology to a simple conditional statement.

Also note that you are the one who introduced the word tautology into the thread.
The original suggestion was, that even tautologies are just conditional statements. It had nothing to do with being able or unable to prove certain things from them. Just with writing them in that form.

And, to get on-topic again, I think it is hard to argue against any statement being formulated under certain axioms only, even if they are tacitly assumed instead of explicitly mentioned most of the time. But if anyone has some serious objection, I hope they will come forward with it and maybe get a real discussion (as in: two-way communication) going.

 

[atyy]

And, to get on-topic again, I think it is hard to argue against any statement being formulated under certain axioms only

How about the axioms themselves? Are they counted as statements?

 

[evagelos]

.......Yes..........

 

[Tac-Tics]

If you are a philosophy major, then all math statements are of the form "if A then B".

If you are not a philosophy major, then you that this statement and the one previous are poorly specified.

The axioms of mathematics can be thought of implicit. They can be thought of explicit. It depends on how you're laying out your problem. If you consider whether "laying out the problem" constitutes an conditional statement, then the answer degenerates with the flavor of "yes, it does" hanging in the air. But like all good philosophy questions, the answer is meaningless.

 

[HallsofIvy]

Let me then define by C : a------>(b----->a) and use C as an axiom together with the said definition.

isn't that o.k

I am reminded of Abraham Lincoln's question: "If you call a Lamb's tail a leg, how many legs does it have?" and answer: "Four. Calling a tail a leg doesn't make it one".

You can "hide" the conditional but it is still there.

Is "A----> B" and "if, then" statement? Of course it is.

If I defind "C= A---->B", is "C" and "if, then" statement? Of course it is.

 

[HallsofIvy]

Let me then define by C : a------>(b----->a) and use C as an axiom together with the said definition.

isn't that o.k

I am reminded of Abraham Lincoln's question: "If you call a Lamb's tail a leg, how many legs does it have?" and answer: "Four. Calling a tail a leg doesn't make it one".

You can "hide" the conditional but it is still there.

Is "A----> B" an "if, then" statement? Of course it is.

If I define "C= A---->B", is "C" and "if, then" statement? Of course it is.

 

[evagelos]

I am reminded of Abraham Lincoln's question: "If you call a Lamb's tail a leg, how many legs does it have?" and answer: "Four. Calling a tail a leg doesn't make it one".

You can "hide" the conditional but it is still there.

Is "A----> B" an "if, then" statement? Of course it is.

If I define "C= A---->B", is "C" and "if, then" statement? Of course it is.

Why when you substitute ...b---->a ,by B in ...a---->(b---->a) don't you hide the conditional ...a----->b?While is still there??

Hence...a----->(b----->a) it is not of the (if A then B ) form.

Your example is not very appropriate try to find another one