Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Proof in predicate calculus

  1. Sep 24, 2008 #1
    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}
     
  2. jcsd
  3. Oct 4, 2008 #2
    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}
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Similar Discussions: Proof in predicate calculus
Loading...