Can This Predicate Calculus Proof Be Established?

[poutsos.A]

\forall a\forall b[( a>0 & b>0)------> (a\leq b <------>a^{2}\leq b^{2})].

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

is there a possibility for a proof within the predicate calculus??

 

[HallsofIvy]

Surely that's not what you meant? Surely you mean a^2\le b^2.

 

[poutsos.A]

\forall a\forall b[( a>0 & b>0)------> (a\leq b <------>a^{2}\leq b^{2})].

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

is there a possibility for a proof within the predicate calculus??

yes thank you it is a^{2}\leq b^{2},instead of a^{2}\leq a^{2}

 

[HallsofIvy]

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

How do you define "<" in the "predicate calculus"?

 

[poutsos.A]

< 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

 

[enigmahunter]

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

\forall a\forall b\phi, where \phi is the wff [( a>0 & b>0)------> (a<= b <-> a^2 <= b^2)], a and b occurs free in \phi. 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.

 

[MathematicalPhysicist]

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... (-:

 

[MathematicalPhysicist]

Correction, e=sqrt(a^2+d)-a

QED.
cheers...

 

[poutsos.A]

definitely the above proofs are not proofs within predicate calculus,perhaps an example from my notes will give you an idea.

thanks

 

[MathematicalPhysicist]

So how did you prove it eventually?