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Proof in predicate calculus

  1. Oct 22, 2008 #1
    [tex]\forall a\forall b[/tex][( a>0 & b>0)------> (a[tex]\leq b[/tex] <------>[tex]a^{2}[/tex][tex]\leq b^{2}[/tex])].

    or in words: for all a and for all b , if a>0 and b>0 then .a[tex]\leq b[/tex] iff [tex]a^{2}[/tex][tex]\leq b^{2}[/tex]

    is there a possibility for a proof within the predicate calculus??
    Last edited: Oct 23, 2008
  2. jcsd
  3. Oct 23, 2008 #2


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    Surely that's not what you meant? Surely you mean [itex]a^2\le b^2[/itex].
  4. Oct 23, 2008 #3
    yes thank you it is [tex]a^{2}[/tex][tex]\leq b^{2}[/tex],instead of [tex]a^{2}[/tex][tex]\leq a^{2}[/tex]
  5. Oct 23, 2008 #4


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    I'm not at all clear on what your question is.

    How do you define "<" in the "predicate calculus"?
  6. Oct 23, 2008 #5
    < is a two place predicate symbol x<y or y<x .

    The axioms and theorems concerning < are exactly those of the real Nos,i.e trichotomy law ,transitivity e.t.c.

    Now writing a proof within predicate calculus means writing a proof where one uses quantified formulas in which the introduction and elimination of quantifiers is done according to the axioms and theorems of predicate calculus.

    Note predicate calculus includes propositional logic as well
  7. Oct 23, 2008 #6
    [tex]\forall x\alpha\rightarrow\alpha^{x}_{t}[/tex], where t is substituable for x in [tex]\alpha[/tex].

    [tex]\alpha^{x}_{t}[/tex] is the expression obtained from the fomula [tex]\alpha[/tex] by replacing the variable x, whenever it occurs free in [tex]\alpha[/tex], by the term t (see "substitution section" of First-order logic (wiki).

    [tex]\forall a\forall b\phi[/tex], where [tex]\phi[/tex] is the wff [( a>0 & b>0)------> (a<= b <-> a^2 <= b^2)], a and b occurs free in [tex]\phi[/tex]. Thus we can replace "a" by a term( variable) "x" and "b" by a term(variable) "y", respectively.

    Now, it reduces to prove the below statement.
    ( x>0 & y>0)------> (x <=y <-> x^2 <= y^2).

    To prove ->
    Let y = x + k (k>=0) . Square y and compare x^2.
    To prove <-
    Use a contrapositive method.
  8. Oct 24, 2008 #7
    I don't see what there is to prove here, it's like p&~p contradiction, and pv~p a tautology. (-:

    anyway, a<=b means there exists a real c>=0 s.t b=a+c, after squaring you get:
    b^2=a^2+c^2+2ac, a is positive and so is c, denote d=c^2+2ac>0.
    for the other way around you just reverse what youv'e done so far cleverly.
    b^2=a^2+d, b=+-sqrt(a^2+d) we know that b>0 so we choose the plus sign, b=sqrt(a^2+d), now we need to show there exists e>0 such that b=a+e choose a+e=sqrt(a^2+d)
    then raise the square, and I let you fill the minor details... (-:
  9. Oct 24, 2008 #8
    Correction, e=sqrt(a^2+d)-a

  10. Oct 25, 2008 #9
    definitely the above proofs are not proofs within predicate calculus,perhaps an example from my notes will give you an idea.

  11. Dec 16, 2008 #10
    So how did you prove it eventually?
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