Main derivation
S01 · definition
Before
\text{a set}\ S\ \text{in the universe}\ \mathbb{Z}After
A=B\iff\forall x\in\mathbb{Z}\,(x\in A\iff x\in B)Conditions: none
Rule: mko_set · Uses: G1
Leaf basis: mko_set + set_extensionality
By the MKO, within a specified universe a set is characterized by which objects are its elements, and sets with exactly the same elements are equal. The universe is ℤ (G1), so proving two sets equal means proving x∈A ⇔ x∈B for an arbitrary integer x, and proving them unequal means finding one integer in one set but not the other.
S02 · equivalent
Before
x\in A
After
x^2=4
Conditions: x\in\mathbb{Z}
Rule: set_builder · Uses: G2
Leaf basis: set_builder + and_intro_elim
By set-builder notation, x∈A ⇔ x∈ℤ and x²=4. In the universe ℤ, x∈ℤ already holds, so x∈A ⇔ x²=4. A is carved out of the existing set ℤ by a condition, which is a restricted form of set formation.
S03 · equivalent
Before
x^2=4
After
x^2-4=0
Conditions: none
Rule: eq_congruence · Uses: S02
Leaf basis: eq_congruence + add_inverse + add_assoc + add_zero
Add −4 to both sides; backwards, add 4 to both sides.
S04 · equivalent
Before
x^2-4=0
After
(x-2)(x+2)=0
Conditions: none
Rule: distributive · Uses: S03
Leaf basis: distributive + neg_mul + add_inverse + eq_replace
By distributivity (x−2)(x+2)=x²+2x−2x−4=x²−4 for every integer; replacing equals does not change whether the equation holds.
S05 · equivalent
Before
(x-2)(x+2)=0
After
x-2=0\ \lor\ x+2=0
Conditions: none
Rule: zero_product · Uses: S04
Leaf basis: zero_product
ℤ is an integral domain, so by the zero-product property a product of two integers is 0 exactly when at least one factor is 0.
S06 · equivalent
Before
x-2=0\ \lor\ x+2=0
After
x=2\ \lor\ x=-2
Conditions: none
Rule: eq_congruence · Uses: S05
Leaf basis: eq_congruence + add_inverse + numeric_evaluation
Add 2 to both sides of the first disjunct and −2 to both sides of the second; each is a two-way equivalence, so the disjunction before and after is true or false together.
S07 · equivalent
Before
x=2\ \lor\ x=-2
After
x\in B
Conditions: none
Rule: list_set · Uses: S06, G1, G2
Leaf basis: or_laws + list_set
By commutativity of disjunction rewrite as x=−2 ∨ x=2; by list notation (G1) this is exactly x∈B.
S08 · implies
Before
\text{for arbitrary}\ x\in\mathbb{Z}:\ x\in A\iff x\in BAfter
A=B
Conditions: none
Rule: set_extensionality · Uses: S02, S03, S04, S05, S06, S07
Leaf basis: logic_chain + forall_intro + set_extensionality
S02 to S07 hold for an arbitrary integer x; chaining them gives x∈A ⇔ x∈B, and by generalization and the criterion for set equality, A=B. As a contrasting non-element: 3²=9≠4, so 3∉A; also 3≠−2 and 3≠2, so 3∉B, and the two sides agree.
S09 · implies
Before
\forall x\in\mathbb{Z}:\ x\in C\iff x\in BAfter
C=B
Conditions: none
Rule: set_extensionality · Uses: L_C, G2
Leaf basis: set_extensionality + list_set + or_laws
By Lemma L_C (proved in S09.1 and S09.2), every integer x has x∈C ⇔ x∈B; by the criterion for set equality, C=B. C is written with three symbols in a different order, but a set depends only on its elements, so order and repetition do not create a new set.
S10 · implies
Before
-2\in B,\ -2\notin D
After
D\neq B
Conditions: none
Rule: contradiction · Uses: G1, G2, list_set
Leaf basis: contradiction + set_extensionality + list_set + numeric_evaluation
By list notation −2∈B (the first disjunct −2=−2 holds), while −2∈D ⇔ −2=2, and −2≠2, so −2∉D. If D=B the two would have the same elements and −2∈D would follow, a contradiction. So D={2}≠B.
S11 · implies
Before
2\ \text{versus}\ \{2\}After
\{2\}\subseteq B\iff 2\in B;\quad \{2\}\in B\ \text{is not a true statement}Conditions: none
Rule: subset_def · Uses: G1, G2, list_set, typed_membership
Leaf basis: subset_def + list_set + typed_membership
2 is an integer and {2}=D is a set of integers; they are not the same object. Their correct relation: by the definition of subset, {2}⊆B ⇔ for every y, y∈{2} ⇒ y∈B; since y∈{2} ⇔ y=2, this is equivalent to 2∈B. As 2∈B holds, {2}⊆B. On the other hand, by typed_membership the elements of B are integers and {2} is not an integer, so '{2}∈B' is not a true statement in this context (in the typed reading it is not even well formed). Confusing 2 with {2} leads to confusing ⊆ with ∈.
S12 · implies
Before
\exists R\in U\ \forall y\in U:\ (y\in R\iff y\notin y)
After
\text{fix such an}\ R\in UConditions: none
Rule: exists_elim · Uses: G3
Leaf basis: exists_elim
Turn to the second mistake identified in the MKO, that any condition defines a set without restriction. If the condition 'y∉y' defined a set in the domain U, there would be some R∈U whose elements are exactly the objects that are not elements of themselves. For a proof by contradiction, temporarily assume such an R exists and give it the fresh name R.
S13 · implies
Before
\forall y\in U:\ (y\in R\iff y\notin y)
After
R\in R\iff R\notin R
Conditions: R\in U
Rule: forall_elim · Uses: S12
Leaf basis: forall_elim
The property in S12 holds for every y∈U, and R itself is in U, so we may take y=R and get R∈R ⇔ R∉R.
S14 · case_split
Before
R\in R\iff R\notin R
After
R\in R\ \lor\ R\notin R
Conditions: none
Rule: case_analysis · Uses: S13, excluded_middle
Leaf basis: case_analysis + excluded_middle
By excluded middle, R∈R or R∉R. In S14.1 and S14.2 each case yields a contradiction from S13, and the two branches cover every possibility.
S15 · implies
Before
\text{the assumption of S12 leads to a contradiction}After
\neg\exists R\in U\ \forall y\in U:\ (y\in R\iff y\notin y)
Conditions: none
Rule: contradiction · Uses: S12, S14, S14.1, S14.2
Leaf basis: contradiction + exists_elim
The temporary assumption of S12 leads to a contradiction in both cases, so by proof by contradiction no such R exists. This holds for every domain U and every membership relation and uses no axiom of set theory. Thus no object R∈U represents the condition y∉y for all y∈U: one cannot require every condition to be represented by a set within U. This does not rule out forming the relative subset R={y∈U | y∉y}⊆U by separation in ordinary set theory; that R is not an element of U. By contrast, S02 used the restricted form that carves elements out of the existing set ℤ.
S16 · implies
Before
\text{conclusions of S08, S09, S10, S11, S15}After
A=B=C,\ D\neq B,\ \{2\}\subseteq B,\ \neg\exists R\in U\ \forall y\in U:\ (y\in R\iff y\notin y)Conditions: none
Rule: and_intro_elim · Uses: S08, S09, S10, S11, S15
Leaf basis: and_intro_elim
Together: A written by a condition, B written as a list and C written with a repetition are one and the same set; D={2} differs from B; the element 2 and the set {2} relate to B by ∈ and by ⊆ respectively; no R∈U satisfies y∈R iff y∉y for every y∈U.
Substeps and lemmas
S09.1 · equivalent
Before
x\in C
After
x=2\lor x=-2\lor x=2
Conditions: none
Rule: list_set · Uses: G2, list_set
Leaf basis: list_set
Unfold C={2,−2,2} by list notation: x∈C ⇔ x=2 ∨ x=−2 ∨ x=2.
S09.2 · equivalent
Before
x=2\lor x=-2\lor x=2
After
x=-2\lor x=2,\ \text{i.e.}\ x\in BConditions: none
Rule: or_laws · Uses: S09.1, or_laws
Leaf basis: or_laws + list_set
By commutativity and associativity of disjunction rewrite as (x=2∨x=2)∨x=−2; by idempotence x=2∨x=2 ⇔ x=2, giving x=2∨x=−2; commuting gives x=−2∨x=2, which by list notation is x∈B. S09.1 and S09.2 together prove Lemma L_C.
S14.1 · implies
Before
R\in R
After
R\notin R,\ \text{contradiction}Conditions: none
Rule: modus_ponens · Uses: S13, S14
Leaf basis: modus_ponens + iff_intro
Case one: R∈R. The '⇒' direction of S13 gives R∉R, which together with R∈R is a contradiction.
S14.2 · implies
Before
R\notin R
After
R\in R,\ \text{contradiction}Conditions: none
Rule: modus_ponens · Uses: S13, S14
Leaf basis: modus_ponens + iff_intro
Case two: R∉R. The '⇐' direction of S13 gives R∈R, which together with R∉R is a contradiction.
Error controls and negative controls
An element and the set containing only it
The MKO names the common mistake of confusing an element x with the set containing only x. 2 is an integer and {2} is a set of integers: 2∈B and {2}⊆B both hold and are equivalent (S11), but '{2}∈B' is not true and {2} is not equal to B (S10). On meeting notation such as 2 or {2}, first identify whether it denotes an integer element or a set of integers in this context, then decide between ∈ and ⊆.
Not every condition forms a set
Another mistake identified in the MKO is believing that any condition defines a set without restriction. S12 to S15 prove that for any domain U, no R∈U satisfies y∈R iff y∉y for every y∈U. This does not exclude forming the relative subset R={y∈U | y∉y}⊆U by separation in ordinary set theory; that R has R∉U. So set formation needs principles, for example carving elements out of a known set (as with {x∈ℤ ∣ x²=4} in S02); this guide does not choose a particular axiom system.
Order and repetition do not create new sets
C={2,−2,2} is written with three symbols, but by extensionality C=B={−2,2} (S09). Distinguishing order or multiplicity needs a different kind of object, such as the ordered triple (2,−2,2) or a multiset; those are not sets in the sense of this MKO.
- {2,−2,2} has three elements, so it differs from {−2,2} →
reject — x∈{2,−2,2} ⇔ x=−2∨x=2, the same elements as B, so they are equal by extensionality (S09). - 2 and {2} are the same thing →
reject — 2 is an integer and {2} is a set of integers; 2∈B holds while '{2}∈B' is not true (S11). - For every condition P(y), {y ∣ P(y)} is a set →
reject — Take P(y) to be y∉y; S12 to S15 derive a contradiction. - A is written with a condition and B as a list, so A≠B →
reject — A set is determined by its elements, not by how it is written; S08 proves A=B.