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All math statements of the if A then B form?

  1. Sep 9, 2008 #1
    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
  2. jcsd
  3. Sep 9, 2008 #2


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    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.
  4. Sep 9, 2008 #3


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    When I learned formal logic (Enterton's text) the only connective we used was [tex]\Rightarrow[/tex]; the others were abbreviations of some combination of this with negation. Of course, philosophically, every statement [tex]\phi[/tex] is implicitly [tex]\mathcal{A}\Rightarrow\phi[/tex] where A is the collection of axioms in the system...
  5. Sep 9, 2008 #4
    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"??
  6. Sep 9, 2008 #5


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    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.
  7. Sep 9, 2008 #6
    If you use not ¬ as well you can express any propositional logic statements.
  8. Sep 9, 2008 #7
    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 ??
  9. Sep 10, 2008 #8


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    [tex]a\Rightarrow(b\Rightarrow c)[/tex] is obviously of form [tex]A\Rightarrow B[/tex] with [tex]A := a[/tex] and [tex]B := (b\Rightarrow c).[/tex]
  10. Sep 10, 2008 #9
    1st of all there is a great difference between :

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


    ..................................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
  11. Sep 10, 2008 #10


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    I see. What you are really saying is that you don't know what is meant by a "if then".
  12. Sep 10, 2008 #11
    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
  13. Sep 10, 2008 #12
    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.
  14. Sep 10, 2008 #13


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

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

    And don't taunt Halls, who clearly knows how to work from axioms on up: Number systems
  15. Sep 10, 2008 #14
    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.
  16. Sep 10, 2008 #15


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    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"!
  17. Sep 10, 2008 #16
    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??
  18. Sep 10, 2008 #17
    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??
  19. Sep 11, 2008 #18
    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

    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)


    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.
  20. Sep 11, 2008 #19


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    It's hard to believe you are serious. If you CALL the statment "If A then B" "C", it is still an "if ... then" statement.
  21. Sep 11, 2008 #20
    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
  22. Sep 11, 2008 #21


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    Are "statements" and "axioms" different?
  23. Sep 11, 2008 #22
    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
  24. Sep 11, 2008 #23


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    So 2*200 is even because it's in the form 2n, but when we write it 400 it's not anymore?
  25. Sep 12, 2008 #24


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    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.
  26. Sep 13, 2008 #25


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