How Can Predicate Calculus Prove a Unique Solution in Set Theory?

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How do we prove in predicate calculus using the laws of universal end existential quantifiers,propositional calculus,and those of algebra the following??

There exists a unique x, xε{ 2,4,6} such that if yε{ 0,1,2} then x^{2}y<10.
or in quantifier form:


\exists !x[ xεA & \forall y(yεB------> x^{2}y<10)]

where A={ 2,4,6} and B={ 0,1,2}
 
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poutsos.A

I will try to give a proof of the above problem without mentioning the laws of logic, theorems or axioms and definitions used,you will have to do that.

By a theorem in predicate calculus (with equality) we have.

\forall z\exists !x(x=z)....................1

and for z=2 we have

\exists x(x=2)......................2

drop the existential quantifier and

x=2.....................3

but x=2 ====> x=2 v x=4 v x=6......................4

and from 3 and 4 we have : x=2 v x=4 v x=6............5

but xεA <====> x=2 v x=4 v x=6......................6

and from 5 and 6 we get: xεA......................7

now let yεB............................8

But yεB <====> y=0 v y=1 v y=2.......................9

and from 8 and 9 we get: y=0 v y=1 v y=2.....................10

Now let y=0.....................11

but y=0===> y^2=0====>.x^{2}y=0<10.....................12

and hence y=0 =====> x^{2}y<10.....................13

in a similar way we prove .

y=1 ====>x^{2}y<10................14

y=2 =====>x^{2}y<10................15

hence: y=0 v y=1 v y=2=======>x^{2}y<10............16

and from 10 and 16 we get: x^{2}y<10..............17

hence : yεB======>x^{2}y<10..............18

And introducing universal quantification: \forall y( yεB====>x^{2}y<10)..............19

And thus: xεA & \forall y( yεB====>x^{2}y<10).........20

And introducing existential quantification we get; \exists x[ xεA & \forall y( yεB====>x^{2}y<10)]...............21

NOW for the uniqueness part you have to prove that.


\forall x\forall w{[ xεA & \forall y(yεΒ=====>x^{2}y<10)] & [ wεA & \forall y(yεΒ=====>w^{2}y<10)] =====> x=w}
 
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