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 poutsos.A Sep24-08 11:30 PM

proof in predicate calculus

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}

 evagelos Oct4-08 05:40 PM

Re: proof in predicate calculus

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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