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poutsos.A Sep24-08 10:30 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[tex]^{2}[/tex]y<10.
or in quantifier form:


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

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

evagelos Oct4-08 04:40 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.

[tex]\forall z\exists !x[/tex](x=z).................................................................. ..............................................1

and for z=2 we have

[tex]\exists x[/tex](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[tex]^{2}[/tex]y=0<10................................................................. ........................................................12

and hence y=0 =====> x[tex]^{2}[/tex]y<10................................................................... ......................................................13

in a similar way we prove .

y=1 ====>x[tex]^{2}[/tex]y<10................................................................... .................14

y=2 =====>x[tex]^{2}[/tex]y<10................................................................... ...................15

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

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

hence : yεB======>x[tex]^{2}[/tex]y<10................................................................... ........18

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

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

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

NOW for the uniqueness part you have to prove that.


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


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