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

  • Context: Graduate 
  • Thread starter Thread starter poutsos.A
  • Start date Start date
  • Tags Tags
    Calculus Proof
Click For Summary
SUMMARY

This discussion provides a detailed proof in predicate calculus demonstrating the existence of a unique element \( x \) from the set \( A = \{2, 4, 6\} \) such that for all \( y \) in the set \( B = \{0, 1, 2\} \), the condition \( x^2y < 10 \) holds. The proof employs universal and existential quantifiers, along with logical deductions, to establish both the existence and uniqueness of \( x \). The final conclusion is expressed in quantifier form as \( \exists !x[ x \in A \land \forall y(y \in B \rightarrow x^2y < 10] \).

PREREQUISITES
  • Understanding of predicate calculus and quantifiers
  • Familiarity with propositional calculus
  • Basic knowledge of set theory and elements
  • Ability to manipulate algebraic expressions involving inequalities
NEXT STEPS
  • Study the laws of universal and existential quantifiers in predicate calculus
  • Learn about the uniqueness quantifier and its applications in mathematical proofs
  • Explore set theory concepts, particularly the properties of finite sets
  • Investigate algebraic inequalities and their implications in calculus
USEFUL FOR

Mathematicians, computer scientists, and students of logic who are interested in formal proofs and the application of predicate calculus in set theory.

poutsos.A
Messages
102
Reaction score
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}
 
Physics news on Phys.org
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}
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 4 ·
Replies
4
Views
6K
  • · Replies 23 ·
Replies
23
Views
5K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K
Replies
3
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 9 ·
Replies
9
Views
6K
  • · Replies 1 ·
Replies
1
Views
2K