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Proof Technique: Element Arguments

In the previous section, we provided proofs for two theorems (Theorem 3.2.1 and Theorem 3.2.2). Both proofs required we select an arbitrary element from some set A, and show that because that arbitrarily chosen element satisfies some property, then all elements of set A satisfy said property. This is simply a rehash of the concept of Universal Specification and Universal Generalization as discussed in Chapter 2.

In this section, we adapt the methods of Universal Specification and Universal Generalization to a new powerful proof technique we can use for sets.

The Underlying Argument

Here, we show off the basic structure of the argument that underlies an element argument.

Let’s suppose p(x) and q(x) are open statements defined on some set A (here the set A is acting like our universe set 𝒰.)

StepPropositionReason
(1)∀x ∈ A [p(x)]Premise
(2)c ∈ Ac is an arbitrarily chosen element from A
(3)p(c)Universal Specification on (1) and (2)
(4)p(c) → q(c)Axiom, Definition, Premise, or Theorem
(5)q(c)Modus Ponens on (3) and (4)
(6)∴ ∀x ∈ A [q(x)]Universal Generalization on (2) and (5)

Even though all element arguments follow this basic structure, pay close attention to step (4). Notice that the reason given was either an axiom, definition, premise, or theorem. This is the part of the argument that changes with each and every single property we’re trying to show is true for every element of some given set A. Depending on what we’re trying to show, we’ll need to bring in multiple axioms, definitions, premises, or theorems in order to reach the desired conclusion.

Typically, step (4) will be expanded into multiple steps, depending on how many axioms, definitions, and theorems are needed to get from proposition p(c) to q(c).

Some Straight-Forward Theorems

Here is one basic theorem whose proof relies on a pretty typical element argument.

Theorem 3.3.1

If A ⊆ B and B ⊆ C, then A ⊆ C.

Proof

Let x be an arbitrarily chosen element from A.

Because x ∈ A, and because A ⊆ B, we know that x ∈ B as well.

Furthermore, because x ∈ B and B ⊆ C, we must have that x ∈ C.

Thus, we have determined that for an arbitrarily chosen element x from A, x must also be an element of C, and so we must have that A ⊆ C as desired.

Let’s rewrite this proof using the structure presented above.

StepPropositionReason
(1)∀x ∈ A [x ∈ B]Premise A ⊆ B
(2)x ∈ Ax is an arbitrarily picked element from A
(3)x ∈ BUniversal Specification on (1) and (2)
(4)∀x ∈ B [x ∈ C]Premise B ⊆ C
(5)x ∈ CModus Ponens on (3) and (4)
(6)∴ ∀x ∈ A [x ∈ C]Universal Generalization on (2) and (5)

This particular proof was a one-to-one translation of the basic structure of an element argument. Let’s look at a theorem that takes a little more work to prove.

Theorem 3.3.2

If A ⊂ B and B ⊂ C, then A ⊂ C.

Proof

General Strategy: Looking at the definition of proper subset, there are essentially two things we need to do. First, we show that every element of A is also an element of C, meaning A ⊆ C. Next, we show that there is an element in C that is not in A. Thus, the definition of proper subset will be satisfied.

Step 1: Show that A ⊆ C

Let x ∈ A be an arbitrarily picked element of A. Because A ⊂ B, Theorem 3.2.3 also tells us that A ⊆ B. As such, we also have that x ∈ B.

Furthermore, because B ⊂ C, Theorem 3.2.3 tells us that B ⊆ C as well. As such, since x ∈ B and B ⊆ C, we also have that x ∈ C.

Thus, since any arbitrarily picked element of A is also an element of C, we know that A ⊆ C.

Step 2: Show that there is an element in C that is not in A.

Because we know that B ⊂ C, there exists an element y that is in C that is not in B, in other words,

∃y ∈ C [y ∉ B].

Furthermore, since we know that A ⊆ B and y ∉ B, we also know that y ∉ A as well.

Conclusion

At this point, we have shown that A ⊆ C, and that ∃y ∈ C [y ∉ A]. Thus by definition, we have that

A ⊂ C

as desired.

This proof required that we give a name to an element we knew existed, namely an element that was contained within C, but was not contained within B, and as such not not contained in A. There is a special name we give to this situation, which we describe below.

Existential Specification and Existential Generalization

When we discussed quantifiers, we talked about two different kinds: those of the universal variety, and those of the existential variety. The Universal Quantifier has its own specification and generalization schemes, and the Existential Quantifier naturally has its own counterparts.

The concept of Existential Specification comes into play when we know some element exists, and in order to manipulate that element, we simply give it a name. In other words, if we know that

∃x ∈ 𝒰 [p(x)] = 1

then we can simply use a symbol to refer to that particular element where p(x) is a true statement. The exact symbol chosen does not matter: it could be the letter c, a Greek letter like β, or it could be some geometrical symbol like ☆, ☺, or even ☒ just as long as we are consistent with what symbol is used. For example, knowing that ∃x ∈ 𝒰 [p(x)] is true, we could use the letter c to denote the particular element within 𝒰 such that

p(c) = 1.

The concept of Universal Generalization is essentially the reverse of Existential Specification. If we know that the particular element c ∈ 𝒰 makes the open statement p(x) true, meaning p(c) = 1, then we know that some element exists that makes p(x) a true statement (because we identified such an element), meaning we know that

∃x ∈ 𝒰 [p(x)] = 1.

We implicitly used Existential Specification and Existential Generalization in the proof for Theorem 3.3.2. Since we are discussing a general proof strategy, let’s write out a formal argument in tabular format as we’ve previously done for Theorem 3.3.2:

StepPropositionReason
(1)A ⊂ BPremise
(2)A ⊆ BTheorem 3.2.3 on (1)
(3)B ⊂ CPremise
(4)B ⊆ CTheorem 3.2.3 on (3)
(5)A ⊆ CTheorem 3.3.1 on (2) and (4)
(6)∀b ∈ B [b ∈ C] ∧ ∃c ∈ C [c ∉ B]Definition of B ⊂ C on (3)
(7)∃c ∈ C [c ∉ B]Conjunctive Simplification on (6)
(8)y ∈ C ∧ y ∉ BExistential Specification on (7)
(9)y ∉ BConjunctive Simplification on (8)
(10)y ∉ AModus Tollens on (2) and (9)
(11)∃c ∈ C [c ∉ A]Existential Generalization on (10)
(12)A ⊆ C ∧ ∃c ∈ C [c ∉ A]Conjunction on (5) and (11)
(13)∀a ∈ A [a ∈ C] ∧ ∃c ∈ C [c ∉ A]Definition of A ⊆ C on (12)
(14)∴ A ⊂ CDefinition of A ⊂ C on (13)

As can be seen, even though we can formally lay out every step of our proof in a tabular format, perhaps the paragraph format presented above is more readable and less cumbersome to write.

Two Additional Theorems

We’ve already examined the theorems

A ⊂ B ∧ B ⊂ C ⟹ A ⊂ C

and

A ⊆ B ∧ B ⊆ C ⟹ A ⊆ C.

We could naturally examine situations involving both regular subsets and proper subsets.

Theorem 3.3.3

If A ⊆ B and B ⊂ C, then A ⊂ C.

Proof

Because B ⊂ C, invoking Theorem 3.2.3 yields the fact that B ⊆ C. Thus, because we know that A ⊆ B and B ⊆ C, Theorem 3.3.1 tells us that A ⊆ C.

Now, because B ⊂ C, we know there exists some c ∈ C such that c ∉ B. But because we also know that A ⊆ B, we also know that c ∉ A as well.

Thus, because we know that A ⊆ C and there is an element c ∈ C such that c ∉ A, we know that A ⊂ C by definition as desired.

Theorem 3.3.4

If A ⊂ B and B ⊆ C, then A ⊂ C.

Proof

Just as in the proof for Theorem 3.3.3, we know that A ⊆ C.

Because A ⊂ B, we know there is some element b ∈ B such that b ∉ A. But because B ⊆ C, we know that b ∈ C as well.

Thus, because we know that A ⊆ C and there is an element b ∈ C such that b ∉ A, we know that A ⊂ C by definition as desired.

Element Arguments represent a powerful proof technique because they give us a way to account for what elements are in what sets, which will be an important skill in the upcoming sections when we start to talk about how we can combine and operate on sets.

Remember that when we talk about a set, what we care about is whether some given object is a member of that set or not, so by picking an arbitrary element, we can use the subset relationships discussed so far to determine if that element is a member of any other set.