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Proof of Infinity

Logicism is the proposition that the foundations of mathematics are logical. Logicism has been viewed as unlikely by modern mathematicians for its inability to prove the axiom of infinity. An axiom is something assumed to be true or taken on faith because it seems valid. In standard set theory, one of the axioms is the axiom of infinity. Which states that there exist sets with infinite members. In this article, I will be deriving the axiom of infinity for the universe. To prove the axiom of infinity requires a non-standard ontology. One need not assume that the boolean concept of 1 and 0, (truth value) are equivalent to the integer concept of 1 and 0 (numeric value). It is for this reason that the axiom of infinity is proven for the boolean case, whilst the proof for the integer case remains to be seen.

Definition of a set: A is a set is equivalent to A is the set of all of its members.

(∀A)(Set(A) ≡ (A = {x | x ∈ A}))

The axiom of infinity is defined as:

(∃X)((1∈X)&(∀x∈X)(S(x)∈X))

There exists a set X, so that 1 is a member of X and for all members x of X, the successor of x is also a member of X.

The axiom of infinity is a definition for an infinite set which is also an inductive set meaning that it is defined via mathematical induction. Mathematical induction is used to prove properties (which may be set membership) for all natural numbers, and is defined via the following formula:

(∀P)((P(1) & (∀x)(P(x) → P(x+1))) → (∀n∈N) P(n))

For all predicates P, if 1 has property P and for all x, x has property P implies that x plus 1 has property P, then all natural numbers have property P.

The natural numbers are positive integers or whole numbers meaning that they start from 1, 2, 3 … etc. or 0, 1, 2, 3 … and so on.

The successor of any number is that number plus one. In Boolean algebra, plus is a disjunction, and 1 is true. So the successor of x in logic is equivalent to x and or true.

S(x) = (x + 1)

S(x) = (x V T)

The standard Von Neumann interpretation of successorship is that the successor of x equals x union a set containing only x. y is a member of the successor of x is equivalent to y is a member of x or y is x.

∀x (S(x) = x ∪ {x} = {y | y ∈ x V y = x})

∀(x, y)(y ∈ S(x) ≡ (y ∈ x V y = x))

This makes sense considering how the natural numbers are encoded. I will prove this axiom using this definition of successorship as the proof for successorship in the logical case is trivial. However, I will not be proving that the infinite set contains all natural numbers. Rather I will be proving that infinity is a property of truth. Originally the encoding for natural numbers in the infinite set would be:

0 = Ø = {}

1 = {0} = {Ø}

2 = {0, 1} = {Ø, {Ø}}

3 = {0, 1, 2} = {Ø, {Ø}, {Ø, {Ø}}}

4 = {0, 1, 2, 3} = {Ø, {Ø}, {Ø, {Ø}}, {Ø, {Ø}, {Ø, {Ø}}}}

… etc

Now it’s time to prove the following using mathematical induction:

(∃X)(1∈X & (∀x)(x∈X→S(x)∈X))

Starting with the basis step we must prove that one is a member of a set.

1 ∈ X

The truth value of a tautology is 1.

v(T) = 1

Propositions are equivalent to their truth values, so tautologies are equivalent to 1.

T ≡ 1

Tautologies are true.

T(T)

Using substitution on the previous two premises:

T(1)

Given the definition of the universe:

(∀x)(Tx ≡ (x ∈ U))

The universe is the set of all truths. It follows that 1 is a member of the universe.

1 ∈ U

So we’ve proven that there is a set that 1 is a member of:

1 ∈ X

Moving on to the second part of the definition:

(∃X)(∀x)(x∈X→S(x)∈X)

Now let’s start the inductive step by assuming that x is a member of the universe, and proving that the successor of x is also a member of the universe.

Given the theorem in deduction six from this article:

2. (∀y)((y ∈ U) ≡ (y = U))

It follows that:

3. x = U

By using substitution on premise 2 with premise 3:

4. (∀y)((y ∈ x) ≡ (y = x))

Given the definition of a set it follows that:

5. x = {y | y ∈ x}

Given the tautology of redundancy:

6. P ≡ (P V P)

And by substituting (y ∈ x) for P in premises 5, 6:

7. x = {y | y ∈ x V y ∈ x}

Using substitution on premise 7, and premise 4:

8. x = {y | y ∈ x V y = x}

Given the definition of successorship, we use substitution.

S(x) = {y | y ∈ x V y = x} ∴

9. S(x) = x

Via substitution of identity on premises 1 and 9:

10. S(x) ∈ U

Therefore, infinite sets exist.

(∃X)((1∈X)&(∀y)((x∈X)→(S(x)∈X))

Thus, truth is infinite.

One can also prove that the contraverse is infinite via a similar process, but for the base case we use zero instead of one.

(∃X)(0∈X & (∀x)(x∈X→S(x)∈X))

Starting with the basis step we must prove that zero is a member of a set.

0 ∈ X

The truth value of a falsehood is 0.

v(F) = 0

Propositions are equivalent to their truth values, so falsehoods are equivalent to 0.

F ≡ 0

Falsehoods are false.

F(F)

Using substitution on the previous two premises:

F(0)

Given the definition of the contraverse:

(∀x)(Fx ≡ (x ∈ C))

The contraverse is the set of all falsehoods. It follows that the boolean zero is a member of the contraverse.

0 ∈ C

So we’ve proven that there is a set that the zero is a member of:

0 ∈ X

Moving on to the second part of the definition:

(∃X)(∀x)(x∈X→S(x)∈X)

Now let’s start the inductive step by assuming that x is a member of the contraverse, and proving that the successor of x is also a member of the contraverse.

Given the theorem in deduction six from this article:

2. (∀y)((y ∈ C) ≡ (y = C))

It follows that:

3. x = C

By using substitution on premise 2 with premise 3:

4. (∀y)((y ∈ x) ≡ (y = x))

Given the definition of a set it follows that:

5. x = {y | y ∈ x}

Given the tautology of redundancy:

6. P ≡ (P V P)

And by substituting (y ∈ x) for P in premises 5, 6:

7. x = {y | y ∈ x V y ∈ x}

Using substitution on premise 7, and premise 4:

8. x = {y | y ∈ x V y = x}

Given the definition of successorship, we use substitution.

S(x) = {y | y ∈ x V y = x} ∴

9. S(x) = x

Via substitution of identity on premises 1 and 9:

10. S(x) ∈ C

(∃C)((0∈C)&(∀y)((x∈C)→(S(x)∈C)))

Thus, falsehood is infinite.

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