# Proof in predicate calculus

1. Sep 24, 2008

### poutsos.A

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}

2. Oct 4, 2008

### evagelos

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}