Logic and the set theory Lecture 11,12: Quantifiers (The set theory) in How to Prove It.
S. Choi Department of Mathematical Science KAIST, Daejeon, South Korea
Fall semester, 2012
S. Choi (KAIST)
Logic and set theory
October 7, 2012
1 / 26
Introduction
About this lecture
Sets (HTP Sections 1.3, 1.4)
S. Choi (KAIST)
Logic and set theory
October 7, 2012
2 / 26
Introduction
About this lecture
Sets (HTP Sections 1.3, 1.4) Quantifiers and sets (HTP 2.1)
S. Choi (KAIST)
Logic and set theory
October 7, 2012
2 / 26
Introduction
About this lecture
Sets (HTP Sections 1.3, 1.4) Quantifiers and sets (HTP 2.1) Equivalences involving quantifiers (HTP 2.2)
S. Choi (KAIST)
Logic and set theory
October 7, 2012
2 / 26
Introduction
About this lecture
Sets (HTP Sections 1.3, 1.4) Quantifiers and sets (HTP 2.1) Equivalences involving quantifiers (HTP 2.2) More operations on sets (HTP 2.3)
S. Choi (KAIST)
Logic and set theory
October 7, 2012
2 / 26
Introduction
About this lecture
Sets (HTP Sections 1.3, 1.4) Quantifiers and sets (HTP 2.1) Equivalences involving quantifiers (HTP 2.2) More operations on sets (HTP 2.3) Course homepages: http://mathsci.kaist.ac.kr/~schoi/logic.html and the moodle page http://moodle.kaist.ac.kr
S. Choi (KAIST)
Logic and set theory
October 7, 2012
2 / 26
Introduction
About this lecture
Sets (HTP Sections 1.3, 1.4) Quantifiers and sets (HTP 2.1) Equivalences involving quantifiers (HTP 2.2) More operations on sets (HTP 2.3) Course homepages: http://mathsci.kaist.ac.kr/~schoi/logic.html and the moodle page http://moodle.kaist.ac.kr Grading and so on in the moodle. Ask questions in moodle.
S. Choi (KAIST)
Logic and set theory
October 7, 2012
2 / 26
Introduction
Some helpful references
Sets, Logic and Categories, Peter J. Cameron, Springer. Read Chapters 3,4,5.
S. Choi (KAIST)
Logic and set theory
October 7, 2012
3 / 26
Introduction
Some helpful references
Sets, Logic and Categories, Peter J. Cameron, Springer. Read Chapters 3,4,5. A mathematical introduction to logic, H. Enderton, Academic Press.
S. Choi (KAIST)
Logic and set theory
October 7, 2012
3 / 26
Introduction
Some helpful references
Sets, Logic and Categories, Peter J. Cameron, Springer. Read Chapters 3,4,5. A mathematical introduction to logic, H. Enderton, Academic Press. http://plato.stanford.edu/contents.html has much resource.
S. Choi (KAIST)
Logic and set theory
October 7, 2012
3 / 26
Introduction
Some helpful references
Sets, Logic and Categories, Peter J. Cameron, Springer. Read Chapters 3,4,5. A mathematical introduction to logic, H. Enderton, Academic Press. http://plato.stanford.edu/contents.html has much resource. Introduction to set theory, Hrbacek and Jech, CRC Press.
S. Choi (KAIST)
Logic and set theory
October 7, 2012
3 / 26
Introduction
Some helpful references
http://en.wikipedia.org/wiki/Truth_table,
S. Choi (KAIST)
Logic and set theory
October 7, 2012
4 / 26
Introduction
Some helpful references
http://en.wikipedia.org/wiki/Truth_table, http://logik.phl.univie.ac.at/~chris/gateway/ formular-uk-zentral.html, complete (i.e. has all the steps)
S. Choi (KAIST)
Logic and set theory
October 7, 2012
4 / 26
Introduction
Some helpful references
http://en.wikipedia.org/wiki/Truth_table, http://logik.phl.univie.ac.at/~chris/gateway/ formular-uk-zentral.html, complete (i.e. has all the steps) http://svn.oriontransfer.org/TruthTable/index.rhtml, has xor, complete.
S. Choi (KAIST)
Logic and set theory
October 7, 2012
4 / 26
Sets
Sets A set is a collection....This naive notion is fairly good.
S. Choi (KAIST)
Logic and set theory
October 7, 2012
5 / 26
Sets
Sets A set is a collection....This naive notion is fairly good. The set theory is compatible with logic.
S. Choi (KAIST)
Logic and set theory
October 7, 2012
5 / 26
Sets
Sets A set is a collection....This naive notion is fairly good. The set theory is compatible with logic. Symbols ∈, {}. (belong, included)
S. Choi (KAIST)
Logic and set theory
October 7, 2012
5 / 26
Sets
Sets A set is a collection....This naive notion is fairly good. The set theory is compatible with logic. Symbols ∈, {}. (belong, included) {{}, {{}}, {{{}}}}
S. Choi (KAIST)
Logic and set theory
October 7, 2012
5 / 26
Sets
Sets A set is a collection....This naive notion is fairly good. The set theory is compatible with logic. Symbols ∈, {}. (belong, included) {{}, {{}}, {{{}}}} {a}. We hold that a ∈ {a, b, c, ...}.
S. Choi (KAIST)
Logic and set theory
October 7, 2012
5 / 26
Sets
Sets A set is a collection....This naive notion is fairly good. The set theory is compatible with logic. Symbols ∈, {}. (belong, included) {{}, {{}}, {{{}}}} {a}. We hold that a ∈ {a, b, c, ...}. The main thrust of the set theory is the theory of description by Russell.
S. Choi (KAIST)
Logic and set theory
October 7, 2012
5 / 26
Sets
Sets A set is a collection....This naive notion is fairly good. The set theory is compatible with logic. Symbols ∈, {}. (belong, included) {{}, {{}}, {{{}}}} {a}. We hold that a ∈ {a, b, c, ...}. The main thrust of the set theory is the theory of description by Russell. P(x): x is a variable. P(x) is the statement that x is a prime number
S. Choi (KAIST)
Logic and set theory
October 7, 2012
5 / 26
Sets
Sets A set is a collection....This naive notion is fairly good. The set theory is compatible with logic. Symbols ∈, {}. (belong, included) {{}, {{}}, {{{}}}} {a}. We hold that a ∈ {a, b, c, ...}. The main thrust of the set theory is the theory of description by Russell. P(x): x is a variable. P(x) is the statement that x is a prime number y ∈ {x|P(x)} is equivalent to P(y ). That is the truth set of P.
S. Choi (KAIST)
Logic and set theory
October 7, 2012
5 / 26
Sets
Sets A set is a collection....This naive notion is fairly good. The set theory is compatible with logic. Symbols ∈, {}. (belong, included) {{}, {{}}, {{{}}}} {a}. We hold that a ∈ {a, b, c, ...}. The main thrust of the set theory is the theory of description by Russell. P(x): x is a variable. P(x) is the statement that x is a prime number y ∈ {x|P(x)} is equivalent to P(y ). That is the truth set of P. Sets ↔ Properties
S. Choi (KAIST)
Logic and set theory
October 7, 2012
5 / 26
Sets
Sets A set is a collection....This naive notion is fairly good. The set theory is compatible with logic. Symbols ∈, {}. (belong, included) {{}, {{}}, {{{}}}} {a}. We hold that a ∈ {a, b, c, ...}. The main thrust of the set theory is the theory of description by Russell. P(x): x is a variable. P(x) is the statement that x is a prime number y ∈ {x|P(x)} is equivalent to P(y ). That is the truth set of P. Sets ↔ Properties D(p, q): p is divisible by q.
S. Choi (KAIST)
Logic and set theory
October 7, 2012
5 / 26
Sets
Sets A set is a collection....This naive notion is fairly good. The set theory is compatible with logic. Symbols ∈, {}. (belong, included) {{}, {{}}, {{{}}}} {a}. We hold that a ∈ {a, b, c, ...}. The main thrust of the set theory is the theory of description by Russell. P(x): x is a variable. P(x) is the statement that x is a prime number y ∈ {x|P(x)} is equivalent to P(y ). That is the truth set of P. Sets ↔ Properties D(p, q): p is divisible by q. A set B = {x|x is a prime number }.
S. Choi (KAIST)
Logic and set theory
October 7, 2012
5 / 26
Sets
Sets A set is a collection....This naive notion is fairly good. The set theory is compatible with logic. Symbols ∈, {}. (belong, included) {{}, {{}}, {{{}}}} {a}. We hold that a ∈ {a, b, c, ...}. The main thrust of the set theory is the theory of description by Russell. P(x): x is a variable. P(x) is the statement that x is a prime number y ∈ {x|P(x)} is equivalent to P(y ). That is the truth set of P. Sets ↔ Properties D(p, q): p is divisible by q. A set B = {x|x is a prime number }. x ∈ B. What does this mean?
S. Choi (KAIST)
Logic and set theory
October 7, 2012
5 / 26
Sets
Axioms of the set theory (Naive version)
There exists a set which has no elements. (Existence)
S. Choi (KAIST)
Logic and set theory
October 7, 2012
6 / 26
Sets
Axioms of the set theory (Naive version)
There exists a set which has no elements. (Existence) Two sets are equal if and only if they have the same elements. (Extensionality)
S. Choi (KAIST)
Logic and set theory
October 7, 2012
6 / 26
Sets
Axioms of the set theory (Naive version)
There exists a set which has no elements. (Existence) Two sets are equal if and only if they have the same elements. (Extensionality) There exists a set B = {x ∈ A|P(x)} if A is a set. (Comprehension)
S. Choi (KAIST)
Logic and set theory
October 7, 2012
6 / 26
Sets
Axioms of the set theory (Naive version)
There exists a set which has no elements. (Existence) Two sets are equal if and only if they have the same elements. (Extensionality) There exists a set B = {x ∈ A|P(x)} if A is a set. (Comprehension) For any two sets, there exists a set that they both belong to. That is, if A and B are sets, there is {A, B}. (Pairing)
S. Choi (KAIST)
Logic and set theory
October 7, 2012
6 / 26
Sets
Axioms of the set theory (Naive version)
There exists a set which has no elements. (Existence) Two sets are equal if and only if they have the same elements. (Extensionality) There exists a set B = {x ∈ A|P(x)} if A is a set. (Comprehension) For any two sets, there exists a set that they both belong to. That is, if A and B are sets, there is {A, B}. (Pairing) For any collection of sets, there exists a unique set that contains all the elements that belong to at least one set in the collection. (Union)
S. Choi (KAIST)
Logic and set theory
October 7, 2012
6 / 26
Sets
Axioms of the set theory (Naive version)
Given each set, there exists a collection of sets that contains among its elements all the subset of the given set. (Power set)
S. Choi (KAIST)
Logic and set theory
October 7, 2012
7 / 26
Sets
Axioms of the set theory (Naive version)
Given each set, there exists a collection of sets that contains among its elements all the subset of the given set. (Power set) An inductive set exists (Infinity)
S. Choi (KAIST)
Logic and set theory
October 7, 2012
7 / 26
Sets
Axioms of the set theory (Naive version)
Given each set, there exists a collection of sets that contains among its elements all the subset of the given set. (Power set) An inductive set exists (Infinity) Let P(x, y ) be a property that for every x, there exists unique y so that P(x, y ) holds. Then for every set A, there is a set B such that for every x ∈ A, there is y ∈ B so that P(x, y ) holds. (Substitution)
S. Choi (KAIST)
Logic and set theory
October 7, 2012
7 / 26
Sets
Axioms of the set theory (Naive version)
Given each set, there exists a collection of sets that contains among its elements all the subset of the given set. (Power set) An inductive set exists (Infinity) Let P(x, y ) be a property that for every x, there exists unique y so that P(x, y ) holds. Then for every set A, there is a set B such that for every x ∈ A, there is y ∈ B so that P(x, y ) holds. (Substitution) Zermelo-Fraenkel theory has more axioms...The axiom of foundation, the axiom of choice.(ZFC)
S. Choi (KAIST)
Logic and set theory
October 7, 2012
7 / 26
Sets
Example
N = {x|x is a natual number.}.
S. Choi (KAIST)
Logic and set theory
October 7, 2012
8 / 26
Sets
Example
N = {x|x is a natual number.}. Z = {x|x is an integer.}.
S. Choi (KAIST)
Logic and set theory
October 7, 2012
8 / 26
Sets
Example
N = {x|x is a natual number.}. Z = {x|x is an integer.}. Q = {x|x is a rational number.}
S. Choi (KAIST)
Logic and set theory
October 7, 2012
8 / 26
Sets
Example
N = {x|x is a natual number.}. Z = {x|x is an integer.}. Q = {x|x is a rational number.} R = {x|x is a real number. }.
S. Choi (KAIST)
Logic and set theory
October 7, 2012
8 / 26
Sets
Example
N = {x|x is a natual number.}. Z = {x|x is an integer.}. Q = {x|x is a rational number.} R = {x|x is a real number. }. {x|x 2 > 9, x ∈ R}.
S. Choi (KAIST)
Logic and set theory
October 7, 2012
8 / 26
Sets
Example
N = {x|x is a natual number.}. Z = {x|x is an integer.}. Q = {x|x is a rational number.} R = {x|x is a real number. }. {x|x 2 > 9, x ∈ R}. y ∈ {x ∈ A|P(x)} is equivalent to y ∈ A ∧ P(y ).
S. Choi (KAIST)
Logic and set theory
October 7, 2012
8 / 26
Sets
Example
N = {x|x is a natual number.}. Z = {x|x is an integer.}. Q = {x|x is a rational number.} R = {x|x is a real number. }. {x|x 2 > 9, x ∈ R}. y ∈ {x ∈ A|P(x)} is equivalent to y ∈ A ∧ P(y ). ∅ is the empty set.
S. Choi (KAIST)
Logic and set theory
October 7, 2012
8 / 26
Sets
Operations on sets
A ⊂ B if and only if ∀x(x ∈ A → x ∈ B).
S. Choi (KAIST)
Logic and set theory
October 7, 2012
9 / 26
Sets
Operations on sets
A ⊂ B if and only if ∀x(x ∈ A → x ∈ B). A ∩ B = {x|x ∈ A ∧ x ∈ B}.
S. Choi (KAIST)
Logic and set theory
October 7, 2012
9 / 26
Sets
Operations on sets
A ⊂ B if and only if ∀x(x ∈ A → x ∈ B). A ∩ B = {x|x ∈ A ∧ x ∈ B}. A ∪ B = {x|x ∈ A ∨ x ∈ B}.
S. Choi (KAIST)
Logic and set theory
October 7, 2012
9 / 26
Sets
Operations on sets
A ⊂ B if and only if ∀x(x ∈ A → x ∈ B). A ∩ B = {x|x ∈ A ∧ x ∈ B}. A ∪ B = {x|x ∈ A ∨ x ∈ B}. A ∩ B ⊂ A ∪ B.
S. Choi (KAIST)
Logic and set theory
October 7, 2012
9 / 26
Sets
Operations on sets
A ⊂ B if and only if ∀x(x ∈ A → x ∈ B). A ∩ B = {x|x ∈ A ∧ x ∈ B}. A ∪ B = {x|x ∈ A ∨ x ∈ B}. A ∩ B ⊂ A ∪ B. A − B = {x|x ∈ A ∧ x ∈ / B}.
S. Choi (KAIST)
Logic and set theory
October 7, 2012
9 / 26
Sets
Operations on sets
A ⊂ B if and only if ∀x(x ∈ A → x ∈ B). A ∩ B = {x|x ∈ A ∧ x ∈ B}. A ∪ B = {x|x ∈ A ∨ x ∈ B}. A ∩ B ⊂ A ∪ B. A − B = {x|x ∈ A ∧ x ∈ / B}. A = ∅ if and only if ¬∃x(x ∈ A).
S. Choi (KAIST)
Logic and set theory
October 7, 2012
9 / 26
Sets
Set theoretic problem When is the set empty?
S. Choi (KAIST)
Logic and set theory
October 7, 2012
10 / 26
Sets
Set theoretic problem When is the set empty? How can one verify two sets are disjoint, same, smaller, bigger, or none of the above?
S. Choi (KAIST)
Logic and set theory
October 7, 2012
10 / 26
Sets
Set theoretic problem When is the set empty? How can one verify two sets are disjoint, same, smaller, bigger, or none of the above? Answer: We use logic and the model theory.
S. Choi (KAIST)
Logic and set theory
October 7, 2012
10 / 26
Sets
Set theoretic problem When is the set empty? How can one verify two sets are disjoint, same, smaller, bigger, or none of the above? Answer: We use logic and the model theory. A ⊂ B means x ∈ A → x ∈ B.
S. Choi (KAIST)
Logic and set theory
October 7, 2012
10 / 26
Sets
Set theoretic problem When is the set empty? How can one verify two sets are disjoint, same, smaller, bigger, or none of the above? Answer: We use logic and the model theory. A ⊂ B means x ∈ A → x ∈ B. Equality of A and B means x ∈ A if and only if x ∈ B.
S. Choi (KAIST)
Logic and set theory
October 7, 2012
10 / 26
Sets
Set theoretic problem When is the set empty? How can one verify two sets are disjoint, same, smaller, bigger, or none of the above? Answer: We use logic and the model theory. A ⊂ B means x ∈ A → x ∈ B. Equality of A and B means x ∈ A if and only if x ∈ B. A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C)?
S. Choi (KAIST)
Logic and set theory
October 7, 2012
10 / 26
Sets
Set theoretic problem When is the set empty? How can one verify two sets are disjoint, same, smaller, bigger, or none of the above? Answer: We use logic and the model theory. A ⊂ B means x ∈ A → x ∈ B. Equality of A and B means x ∈ A if and only if x ∈ B. A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C)? x ∈ A ∪ (B ∩ C)
S. Choi (KAIST)
Logic and set theory
October 7, 2012
10 / 26
Sets
Set theoretic problem When is the set empty? How can one verify two sets are disjoint, same, smaller, bigger, or none of the above? Answer: We use logic and the model theory. A ⊂ B means x ∈ A → x ∈ B. Equality of A and B means x ∈ A if and only if x ∈ B. A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C)? x ∈ A ∪ (B ∩ C) x ∈ A ∨ (x ∈ B ∧ x ∈ C).
S. Choi (KAIST)
Logic and set theory
October 7, 2012
10 / 26
Sets
Set theoretic problem When is the set empty? How can one verify two sets are disjoint, same, smaller, bigger, or none of the above? Answer: We use logic and the model theory. A ⊂ B means x ∈ A → x ∈ B. Equality of A and B means x ∈ A if and only if x ∈ B. A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C)? x ∈ A ∪ (B ∩ C) x ∈ A ∨ (x ∈ B ∧ x ∈ C). (x ∈ A ∨ x ∈ B) ∧ (x ∈ A ∨ x ∈ C). DIST
S. Choi (KAIST)
Logic and set theory
October 7, 2012
10 / 26
Sets
Set theoretic problem When is the set empty? How can one verify two sets are disjoint, same, smaller, bigger, or none of the above? Answer: We use logic and the model theory. A ⊂ B means x ∈ A → x ∈ B. Equality of A and B means x ∈ A if and only if x ∈ B. A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C)? x ∈ A ∪ (B ∩ C) x ∈ A ∨ (x ∈ B ∧ x ∈ C). (x ∈ A ∨ x ∈ B) ∧ (x ∈ A ∨ x ∈ C). DIST Thus, x ∈ A ∪ (B ∩ C) ↔ (x ∈ A ∨ x ∈ B) ∧ (x ∈ A ∨ x ∈ C).
S. Choi (KAIST)
Logic and set theory
October 7, 2012
10 / 26
Sets
Set theoretic problem When is the set empty? How can one verify two sets are disjoint, same, smaller, bigger, or none of the above? Answer: We use logic and the model theory. A ⊂ B means x ∈ A → x ∈ B. Equality of A and B means x ∈ A if and only if x ∈ B. A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C)? x ∈ A ∪ (B ∩ C) x ∈ A ∨ (x ∈ B ∧ x ∈ C). (x ∈ A ∨ x ∈ B) ∧ (x ∈ A ∨ x ∈ C). DIST Thus, x ∈ A ∪ (B ∩ C) ↔ (x ∈ A ∨ x ∈ B) ∧ (x ∈ A ∨ x ∈ C). One can use Venn diagrams....
S. Choi (KAIST)
Logic and set theory
October 7, 2012
10 / 26
Sets
More set theoretic problem
Compare (A − B) − C, (A − B) ∩ (A − C), (A − B) ∪ (A − C).
S. Choi (KAIST)
Logic and set theory
October 7, 2012
11 / 26
Sets
More set theoretic problem
Compare (A − B) − C, (A − B) ∩ (A − C), (A − B) ∪ (A − C). x ∈ (A − B) ∧ x ∈ / C. (x ∈ A ∧ x ∈ / B) ∧ x ∈ / C.
S. Choi (KAIST)
Logic and set theory
October 7, 2012
11 / 26
Sets
More set theoretic problem
Compare (A − B) − C, (A − B) ∩ (A − C), (A − B) ∪ (A − C). x ∈ (A − B) ∧ x ∈ / C. (x ∈ A ∧ x ∈ / B) ∧ x ∈ / C. (x ∈ A ∧ x ∈ / B) ∨ (x ∈ A ∧ x ∈ / C).
S. Choi (KAIST)
Logic and set theory
October 7, 2012
11 / 26
Sets
More set theoretic problem
Compare (A − B) − C, (A − B) ∩ (A − C), (A − B) ∪ (A − C). x ∈ (A − B) ∧ x ∈ / C. (x ∈ A ∧ x ∈ / B) ∧ x ∈ / C. (x ∈ A ∧ x ∈ / B) ∨ (x ∈ A ∧ x ∈ / C). (A − B) ∩ (A − C).
S. Choi (KAIST)
Logic and set theory
October 7, 2012
11 / 26
Sets
More set theoretic problem
Compare (A − B) − C, (A − B) ∩ (A − C), (A − B) ∪ (A − C). x ∈ (A − B) ∧ x ∈ / C. (x ∈ A ∧ x ∈ / B) ∧ x ∈ / C. (x ∈ A ∧ x ∈ / B) ∨ (x ∈ A ∧ x ∈ / C). (A − B) ∩ (A − C). We can show (A − B) − C ⊂ (A − B) ∪ (A − C).
S. Choi (KAIST)
Logic and set theory
October 7, 2012
11 / 26
Sets
More set theoretic problem
Compare (A − B) − C, (A − B) ∩ (A − C), (A − B) ∪ (A − C). x ∈ (A − B) ∧ x ∈ / C. (x ∈ A ∧ x ∈ / B) ∧ x ∈ / C. (x ∈ A ∧ x ∈ / B) ∨ (x ∈ A ∧ x ∈ / C). (A − B) ∩ (A − C). We can show (A − B) − C ⊂ (A − B) ∪ (A − C). Is (A − B) ∪ (A − C) ⊂ (A − B) − C?
S. Choi (KAIST)
Logic and set theory
October 7, 2012
11 / 26
Sets
More set theoretic problem
Compare (A − B) − C, (A − B) ∩ (A − C), (A − B) ∪ (A − C). x ∈ (A − B) ∧ x ∈ / C. (x ∈ A ∧ x ∈ / B) ∧ x ∈ / C. (x ∈ A ∧ x ∈ / B) ∨ (x ∈ A ∧ x ∈ / C). (A − B) ∩ (A − C). We can show (A − B) − C ⊂ (A − B) ∪ (A − C). Is (A − B) ∪ (A − C) ⊂ (A − B) − C?
S. Choi (KAIST)
Logic and set theory
October 7, 2012
11 / 26
Sets
More set theoretic problem
Compare (A − B) − C, (A − B) ∩ (A − C), (A − B) ∪ (A − C). x ∈ (A − B) ∧ x ∈ / C. (x ∈ A ∧ x ∈ / B) ∧ x ∈ / C. (x ∈ A ∧ x ∈ / B) ∨ (x ∈ A ∧ x ∈ / C). (A − B) ∩ (A − C). We can show (A − B) − C ⊂ (A − B) ∪ (A − C). Is (A − B) ∪ (A − C) ⊂ (A − B) − C? Use logic to find examples.
S. Choi (KAIST)
Logic and set theory
October 7, 2012
11 / 26
Sets
More set theoretic problem
Comparing (A − B) − C and (A − B) ∪ (A − C).
S. Choi (KAIST)
Logic and set theory
October 7, 2012
12 / 26
Sets
More set theoretic problem
Comparing (A − B) − C and (A − B) ∪ (A − C). x ∈ (A − B) ∧ x ∈ / C and (x ∈ A ∧ x ∈ / B)∧ ∈ / C.
S. Choi (KAIST)
Logic and set theory
October 7, 2012
12 / 26
Sets
More set theoretic problem
Comparing (A − B) − C and (A − B) ∪ (A − C). x ∈ (A − B) ∧ x ∈ / C and (x ∈ A ∧ x ∈ / B)∧ ∈ / C. (x ∈ A ∧ x ∈ / B) ∨ (x ∈ A ∧ x ∈ / C).
S. Choi (KAIST)
Logic and set theory
October 7, 2012
12 / 26
Sets
More set theoretic problem
Comparing (A − B) − C and (A − B) ∪ (A − C). x ∈ (A − B) ∧ x ∈ / C and (x ∈ A ∧ x ∈ / B)∧ ∈ / C. (x ∈ A ∧ x ∈ / B) ∨ (x ∈ A ∧ x ∈ / C). ∀x((x ∈ A ∧ x ∈ / B) ∨ (x ∈ A ∧ x ∈ / C)) → (x ∈ A ∧ x ∈ / B) ∧ x ∈ / C) is invalid.
S. Choi (KAIST)
Logic and set theory
October 7, 2012
12 / 26
Sets
More set theoretic problem
Comparing (A − B) − C and (A − B) ∪ (A − C). x ∈ (A − B) ∧ x ∈ / C and (x ∈ A ∧ x ∈ / B)∧ ∈ / C. (x ∈ A ∧ x ∈ / B) ∨ (x ∈ A ∧ x ∈ / C). ∀x((x ∈ A ∧ x ∈ / B) ∨ (x ∈ A ∧ x ∈ / C)) → (x ∈ A ∧ x ∈ / B) ∧ x ∈ / C) is invalid. Find the counter-example...(Using what?)
S. Choi (KAIST)
Logic and set theory
October 7, 2012
12 / 26
Quantifiers and sets
Quantifiers and sets
A ∩ B ⊂ B − C. Translate this to logic
S. Choi (KAIST)
Logic and set theory
October 7, 2012
13 / 26
Quantifiers and sets
Quantifiers and sets
A ∩ B ⊂ B − C. Translate this to logic ∀x((x ∈ A ∧ x ∈ B) → (x ∈ B ∧ x ∈ / C)).
S. Choi (KAIST)
Logic and set theory
October 7, 2012
13 / 26
Quantifiers and sets
Quantifiers and sets
A ∩ B ⊂ B − C. Translate this to logic ∀x((x ∈ A ∧ x ∈ B) → (x ∈ B ∧ x ∈ / C)). If A ⊂ B, then A and C − B are disjoint.
S. Choi (KAIST)
Logic and set theory
October 7, 2012
13 / 26
Quantifiers and sets
Quantifiers and sets
A ∩ B ⊂ B − C. Translate this to logic ∀x((x ∈ A ∧ x ∈ B) → (x ∈ B ∧ x ∈ / C)). If A ⊂ B, then A and C − B are disjoint. ∀x(x ∈ A → x ∈ B) → ¬∃x(x ∈ A ∧ x ∈ (C − B)).
S. Choi (KAIST)
Logic and set theory
October 7, 2012
13 / 26
Quantifiers and sets
Quantifiers and sets
A ∩ B ⊂ B − C. Translate this to logic ∀x((x ∈ A ∧ x ∈ B) → (x ∈ B ∧ x ∈ / C)). If A ⊂ B, then A and C − B are disjoint. ∀x(x ∈ A → x ∈ B) → ¬∃x(x ∈ A ∧ x ∈ (C − B)). ∀x(x ∈ A → x ∈ B) → ¬∃x(x ∈ A ∧ x ∈ C ∧ x ∈ / B).
S. Choi (KAIST)
Logic and set theory
October 7, 2012
13 / 26
Quantifiers and sets
Examples
For every number a, the equation ax 2 + 4x − 2 = 0 has a solution if and only if a ≥ −2.
S. Choi (KAIST)
Logic and set theory
October 7, 2012
14 / 26
Quantifiers and sets
Examples
For every number a, the equation ax 2 + 4x − 2 = 0 has a solution if and only if a ≥ −2. Use R.
S. Choi (KAIST)
Logic and set theory
October 7, 2012
14 / 26
Quantifiers and sets
Examples
For every number a, the equation ax 2 + 4x − 2 = 0 has a solution if and only if a ≥ −2. Use R. ∀a(a ≥ −2 ↔ ∃x ∈ R(ax 2 + 4x − 2 = 0)).
S. Choi (KAIST)
Logic and set theory
October 7, 2012
14 / 26
Quantifiers and sets
Examples
For every number a, the equation ax 2 + 4x − 2 = 0 has a solution if and only if a ≥ −2. Use R. ∀a(a ≥ −2 ↔ ∃x ∈ R(ax 2 + 4x − 2 = 0)). Is this true? How does one verify this...
S. Choi (KAIST)
Logic and set theory
October 7, 2012
14 / 26
Equivalences involving quantifiers
Equivalences involving quantifiers
¬∀x
P(x) ↔ ∃x¬P(x).
S. Choi (KAIST)
Logic and set theory
October 7, 2012
15 / 26
Equivalences involving quantifiers
Equivalences involving quantifiers
¬∀x
P(x) ↔ ∃x¬P(x).
¬∃x
P(x) ↔ ∀x¬P(x).
S. Choi (KAIST)
Logic and set theory
October 7, 2012
15 / 26
Equivalences involving quantifiers
Equivalences involving quantifiers
¬∀x
P(x) ↔ ∃x¬P(x).
¬∃x
P(x) ↔ ∀x¬P(x).
Negation of A ⊂ B.
S. Choi (KAIST)
Logic and set theory
October 7, 2012
15 / 26
Equivalences involving quantifiers
Equivalences involving quantifiers
¬∀x
P(x) ↔ ∃x¬P(x).
¬∃x
P(x) ↔ ∀x¬P(x).
Negation of A ⊂ B. ¬∀x(x ∈ A → x ∈ B).
S. Choi (KAIST)
Logic and set theory
October 7, 2012
15 / 26
Equivalences involving quantifiers
Equivalences involving quantifiers
¬∀x
P(x) ↔ ∃x¬P(x).
¬∃x
P(x) ↔ ∀x¬P(x).
Negation of A ⊂ B. ¬∀x(x ∈ A → x ∈ B). ∃x¬(x ∈ A → x ∈ B).
S. Choi (KAIST)
Logic and set theory
October 7, 2012
15 / 26
Equivalences involving quantifiers
Equivalences involving quantifiers
¬∀x
P(x) ↔ ∃x¬P(x).
¬∃x
P(x) ↔ ∀x¬P(x).
Negation of A ⊂ B. ¬∀x(x ∈ A → x ∈ B). ∃x¬(x ∈ A → x ∈ B). ∃x¬(x ∈ / A ∨ x ∈ B). MI. (conditional law)
S. Choi (KAIST)
Logic and set theory
October 7, 2012
15 / 26
Equivalences involving quantifiers
Equivalences involving quantifiers
¬∀x
P(x) ↔ ∃x¬P(x).
¬∃x
P(x) ↔ ∀x¬P(x).
Negation of A ⊂ B. ¬∀x(x ∈ A → x ∈ B). ∃x¬(x ∈ A → x ∈ B). ∃x¬(x ∈ / A ∨ x ∈ B). MI. (conditional law) ∃x(x ∈ A ∧ x ∈ / B). DM.
S. Choi (KAIST)
Logic and set theory
October 7, 2012
15 / 26
Equivalences involving quantifiers
Equivalences involving quantifiers
¬∀x
P(x) ↔ ∃x¬P(x).
¬∃x
P(x) ↔ ∀x¬P(x).
Negation of A ⊂ B. ¬∀x(x ∈ A → x ∈ B). ∃x¬(x ∈ A → x ∈ B). ∃x¬(x ∈ / A ∨ x ∈ B). MI. (conditional law) ∃x(x ∈ A ∧ x ∈ / B). DM. There exists an element of A not in B.
S. Choi (KAIST)
Logic and set theory
October 7, 2012
15 / 26
Equivalences involving quantifiers
∃x ∈ A
P(x) is defined as ∃x(x ∈ A ∧ P(x)).
S. Choi (KAIST)
Logic and set theory
October 7, 2012
16 / 26
Equivalences involving quantifiers
∃x ∈ A
P(x) is defined as ∃x(x ∈ A ∧ P(x)).
∀x ∈ A
P(x) is defined as ∀x(x ∈ A → P(x)).
S. Choi (KAIST)
Logic and set theory
October 7, 2012
16 / 26
Equivalences involving quantifiers
∃x ∈ A
P(x) is defined as ∃x(x ∈ A ∧ P(x)).
∀x ∈ A P(x) is defined as ∀x(x ∈ A → P(x)). ¬∀x ∈ A P(x) ↔ ∃x ∈ A¬P(x).
S. Choi (KAIST)
Logic and set theory
October 7, 2012
16 / 26
Equivalences involving quantifiers
∃x ∈ A
P(x) is defined as ∃x(x ∈ A ∧ P(x)).
∀x ∈ A P(x) is defined as ∀x(x ∈ A → P(x)). ¬∀x ∈ A P(x) ↔ ∃x ∈ A¬P(x). proof: ¬∀x(x ∈ A → P(x)).
S. Choi (KAIST)
Logic and set theory
October 7, 2012
16 / 26
Equivalences involving quantifiers
∃x ∈ A
P(x) is defined as ∃x(x ∈ A ∧ P(x)).
∀x ∈ A P(x) is defined as ∀x(x ∈ A → P(x)). ¬∀x ∈ A P(x) ↔ ∃x ∈ A¬P(x). proof: ¬∀x(x ∈ A → P(x)). ∃x¬(x ∈ A → P(x)).
S. Choi (KAIST)
Logic and set theory
October 7, 2012
16 / 26
Equivalences involving quantifiers
∃x ∈ A
P(x) is defined as ∃x(x ∈ A ∧ P(x)).
∀x ∈ A P(x) is defined as ∀x(x ∈ A → P(x)). ¬∀x ∈ A P(x) ↔ ∃x ∈ A¬P(x). proof: ¬∀x(x ∈ A → P(x)). ∃x¬(x ∈ A → P(x)). ∃x¬(x ∈ / A ∨ P(x)).
S. Choi (KAIST)
Logic and set theory
October 7, 2012
16 / 26
Equivalences involving quantifiers
∃x ∈ A
P(x) is defined as ∃x(x ∈ A ∧ P(x)).
∀x ∈ A P(x) is defined as ∀x(x ∈ A → P(x)). ¬∀x ∈ A P(x) ↔ ∃x ∈ A¬P(x). proof: ¬∀x(x ∈ A → P(x)). ∃x¬(x ∈ A → P(x)). ∃x¬(x ∈ / A ∨ P(x)). ∃x(x ∈ A ∧ ¬P(x)).
S. Choi (KAIST)
Logic and set theory
October 7, 2012
16 / 26
Equivalences involving quantifiers
∃x ∈ A
P(x) is defined as ∃x(x ∈ A ∧ P(x)).
∀x ∈ A P(x) is defined as ∀x(x ∈ A → P(x)). ¬∀x ∈ A P(x) ↔ ∃x ∈ A¬P(x). proof: ¬∀x(x ∈ A → P(x)). ∃x¬(x ∈ A → P(x)). ∃x¬(x ∈ / A ∨ P(x)). ∃x(x ∈ A ∧ ¬P(x)). ∃x ∈ A¬P(x).
S. Choi (KAIST)
Logic and set theory
October 7, 2012
16 / 26
Equivalences involving quantifiers
∃x ∈ A
P(x) is defined as ∃x(x ∈ A ∧ P(x)).
∀x ∈ A P(x) is defined as ∀x(x ∈ A → P(x)). ¬∀x ∈ A P(x) ↔ ∃x ∈ A¬P(x). proof: ¬∀x(x ∈ A → P(x)). ∃x¬(x ∈ A → P(x)). ∃x¬(x ∈ / A ∨ P(x)). ∃x(x ∈ A ∧ ¬P(x)). ∃x ∈ A¬P(x). These are all equivalent statements
S. Choi (KAIST)
Logic and set theory
October 7, 2012
16 / 26
Equivalences involving quantifiers
¬∃x ∈ A
P(x) ↔ ∀x ∈ A¬P(x).
S. Choi (KAIST)
Logic and set theory
October 7, 2012
17 / 26
Equivalences involving quantifiers
¬∃x ∈ A
P(x) ↔ ∀x ∈ A¬P(x).
proof: ¬∃x(x ∈ A ∧ P(x)).
S. Choi (KAIST)
Logic and set theory
October 7, 2012
17 / 26
Equivalences involving quantifiers
¬∃x ∈ A
P(x) ↔ ∀x ∈ A¬P(x).
proof: ¬∃x(x ∈ A ∧ P(x)). ∀x¬(x ∈ A ∧ P(x)).
S. Choi (KAIST)
Logic and set theory
October 7, 2012
17 / 26
Equivalences involving quantifiers
¬∃x ∈ A
P(x) ↔ ∀x ∈ A¬P(x).
proof: ¬∃x(x ∈ A ∧ P(x)). ∀x¬(x ∈ A ∧ P(x)). ∀x(x ∈ / A ∨ ¬P(x)).
S. Choi (KAIST)
Logic and set theory
October 7, 2012
17 / 26
Equivalences involving quantifiers
¬∃x ∈ A
P(x) ↔ ∀x ∈ A¬P(x).
proof: ¬∃x(x ∈ A ∧ P(x)). ∀x¬(x ∈ A ∧ P(x)). ∀x(x ∈ / A ∨ ¬P(x)). ∀x(x ∈ A → ¬P(x).
S. Choi (KAIST)
Logic and set theory
October 7, 2012
17 / 26
Equivalences involving quantifiers
¬∃x ∈ A
P(x) ↔ ∀x ∈ A¬P(x).
proof: ¬∃x(x ∈ A ∧ P(x)). ∀x¬(x ∈ A ∧ P(x)). ∀x(x ∈ / A ∨ ¬P(x)). ∀x(x ∈ A → ¬P(x). ∀x ∈ A¬P(x).
S. Choi (KAIST)
Logic and set theory
October 7, 2012
17 / 26
Equivalences involving quantifiers
¬∃x ∈ A
P(x) ↔ ∀x ∈ A¬P(x).
proof: ¬∃x(x ∈ A ∧ P(x)). ∀x¬(x ∈ A ∧ P(x)). ∀x(x ∈ / A ∨ ¬P(x)). ∀x(x ∈ A → ¬P(x). ∀x ∈ A¬P(x). These are all equivalent statements
S. Choi (KAIST)
Logic and set theory
October 7, 2012
17 / 26
More operations on sets
Indexed sets
Let I be the set of indices i = 1, 2, 3, ...
S. Choi (KAIST)
Logic and set theory
October 7, 2012
18 / 26
More operations on sets
Indexed sets
Let I be the set of indices i = 1, 2, 3, ... p1 = 2, p2 = 3, p3 = 5,...
S. Choi (KAIST)
Logic and set theory
October 7, 2012
18 / 26
More operations on sets
Indexed sets
Let I be the set of indices i = 1, 2, 3, ... p1 = 2, p2 = 3, p3 = 5,... {p1 , p2 , ...} = {pi |i ∈ I} is another set, called, an indexed set. (Actually this is an axiom)
S. Choi (KAIST)
Logic and set theory
October 7, 2012
18 / 26
More operations on sets
Indexed sets
Let I be the set of indices i = 1, 2, 3, ... p1 = 2, p2 = 3, p3 = 5,... {p1 , p2 , ...} = {pi |i ∈ I} is another set, called, an indexed set. (Actually this is an axiom) In fact I could be any set.
S. Choi (KAIST)
Logic and set theory
October 7, 2012
18 / 26
More operations on sets
Indexed sets
Let I be the set of indices i = 1, 2, 3, ... p1 = 2, p2 = 3, p3 = 5,... {p1 , p2 , ...} = {pi |i ∈ I} is another set, called, an indexed set. (Actually this is an axiom) In fact I could be any set. {n2 |n ∈ N}, {n2 |n ∈ Z}.
S. Choi (KAIST)
Logic and set theory
October 7, 2012
18 / 26
More operations on sets
Indexed sets
Let I be the set of indices i = 1, 2, 3, ... p1 = 2, p2 = 3, p3 = 5,... {p1 , p2 , ...} = {pi |i ∈ I} is another set, called, an indexed set. (Actually this is an axiom) In fact I could be any set. {n2 |n ∈ N}, {n2 |n ∈ Z}. √ { x|x ∈ Q}
S. Choi (KAIST)
Logic and set theory
October 7, 2012
18 / 26
More operations on sets
Family of sets
A set whose elements are sets is said to be a family of sets.
S. Choi (KAIST)
Logic and set theory
October 7, 2012
19 / 26
More operations on sets
Family of sets
A set whose elements are sets is said to be a family of sets. We can also write {Ai |i ∈ I} for Ai a set and I an index set.
S. Choi (KAIST)
Logic and set theory
October 7, 2012
19 / 26
More operations on sets
Family of sets
A set whose elements are sets is said to be a family of sets. We can also write {Ai |i ∈ I} for Ai a set and I an index set. F = {{}, {{}}, {{{}}}}
S. Choi (KAIST)
Logic and set theory
October 7, 2012
19 / 26
More operations on sets
Family of sets
A set whose elements are sets is said to be a family of sets. We can also write {Ai |i ∈ I} for Ai a set and I an index set. F = {{}, {{}}, {{{}}}} Given a set A, the power set is defined: P(A) = {x|x ⊂ A}.
S. Choi (KAIST)
Logic and set theory
October 7, 2012
19 / 26
More operations on sets
Family of sets
A set whose elements are sets is said to be a family of sets. We can also write {Ai |i ∈ I} for Ai a set and I an index set. F = {{}, {{}}, {{{}}}} Given a set A, the power set is defined: P(A) = {x|x ⊂ A}. x ∈ P(A) is equivalent to x ⊂ A and to ∀y (y ∈ x → y ∈ A).
S. Choi (KAIST)
Logic and set theory
October 7, 2012
19 / 26
More operations on sets
The power set
P(A) ⊂ P(B). Analysis
S. Choi (KAIST)
Logic and set theory
October 7, 2012
20 / 26
More operations on sets
The power set
P(A) ⊂ P(B). Analysis ∀x(x ∈ P(A) → x ∈ P(B)).
S. Choi (KAIST)
Logic and set theory
October 7, 2012
20 / 26
More operations on sets
The power set
P(A) ⊂ P(B). Analysis ∀x(x ∈ P(A) → x ∈ P(B)). ∀x((∀y (y ∈ x → y ∈ A)) → (∀y (y ∈ x → y ∈ B))).
S. Choi (KAIST)
Logic and set theory
October 7, 2012
20 / 26
More operations on sets
The power set
P(A) ⊂ P(B). Analysis ∀x(x ∈ P(A) → x ∈ P(B)). ∀x((∀y (y ∈ x → y ∈ A)) → (∀y (y ∈ x → y ∈ B))). If A ⊂ B, then is P(A) ⊂ P(B)?
S. Choi (KAIST)
Logic and set theory
October 7, 2012
20 / 26
More operations on sets
The power set
P(A) ⊂ P(B). Analysis ∀x(x ∈ P(A) → x ∈ P(B)). ∀x((∀y (y ∈ x → y ∈ A)) → (∀y (y ∈ x → y ∈ B))). If A ⊂ B, then is P(A) ⊂ P(B)? To check this what should we do? Use our inference rules....
S. Choi (KAIST)
Logic and set theory
October 7, 2012
20 / 26
More operations on sets
A ⊂ B ` ∀x((∀y (y ∈ x → y ∈ A)) → (∀y (y ∈ x → y ∈ B))).
S. Choi (KAIST)
Logic and set theory
October 7, 2012
21 / 26
More operations on sets
A ⊂ B ` ∀x((∀y (y ∈ x → y ∈ A)) → (∀y (y ∈ x → y ∈ B))). 1. ∀x(x ∈ A → x ∈ B). A.
S. Choi (KAIST)
Logic and set theory
October 7, 2012
21 / 26
More operations on sets
A ⊂ B ` ∀x((∀y (y ∈ x → y ∈ A)) → (∀y (y ∈ x → y ∈ B))). 1. ∀x(x ∈ A → x ∈ B). A. 2: ∀y (y ∈ a → y ∈ A) H.
S. Choi (KAIST)
Logic and set theory
October 7, 2012
21 / 26
More operations on sets
A ⊂ B ` ∀x((∀y (y ∈ x → y ∈ A)) → (∀y (y ∈ x → y ∈ B))). 1. ∀x(x ∈ A → x ∈ B). A. 2: ∀y (y ∈ a → y ∈ A) H. 3: b ∈ a → b ∈ A.
S. Choi (KAIST)
Logic and set theory
October 7, 2012
21 / 26
More operations on sets
A ⊂ B ` ∀x((∀y (y ∈ x → y ∈ A)) → (∀y (y ∈ x → y ∈ B))). 1. ∀x(x ∈ A → x ∈ B). A. 2: ∀y (y ∈ a → y ∈ A) H. 3: b ∈ a → b ∈ A. 4: b ∈ A → b ∈ B.
S. Choi (KAIST)
Logic and set theory
October 7, 2012
21 / 26
More operations on sets
A ⊂ B ` ∀x((∀y (y ∈ x → y ∈ A)) → (∀y (y ∈ x → y ∈ B))). 1. ∀x(x ∈ A → x ∈ B). A. 2: ∀y (y ∈ a → y ∈ A) H. 3: b ∈ a → b ∈ A. 4: b ∈ A → b ∈ B. 5: b ∈ a → b ∈ B.
S. Choi (KAIST)
Logic and set theory
October 7, 2012
21 / 26
More operations on sets
A ⊂ B ` ∀x((∀y (y ∈ x → y ∈ A)) → (∀y (y ∈ x → y ∈ B))). 1. ∀x(x ∈ A → x ∈ B). A. 2: ∀y (y ∈ a → y ∈ A) H. 3: b ∈ a → b ∈ A. 4: b ∈ A → b ∈ B. 5: b ∈ a → b ∈ B. 6.: ∀y (y ∈ a → y ∈ B).
S. Choi (KAIST)
Logic and set theory
October 7, 2012
21 / 26
More operations on sets
A ⊂ B ` ∀x((∀y (y ∈ x → y ∈ A)) → (∀y (y ∈ x → y ∈ B))). 1. ∀x(x ∈ A → x ∈ B). A. 2: ∀y (y ∈ a → y ∈ A) H. 3: b ∈ a → b ∈ A. 4: b ∈ A → b ∈ B. 5: b ∈ a → b ∈ B. 6.: ∀y (y ∈ a → y ∈ B). 7. (∀y (y ∈ a → y ∈ A)) → ∀y (y ∈ a → y ∈ B). 2-6
S. Choi (KAIST)
Logic and set theory
October 7, 2012
21 / 26
More operations on sets
A ⊂ B ` ∀x((∀y (y ∈ x → y ∈ A)) → (∀y (y ∈ x → y ∈ B))). 1. ∀x(x ∈ A → x ∈ B). A. 2: ∀y (y ∈ a → y ∈ A) H. 3: b ∈ a → b ∈ A. 4: b ∈ A → b ∈ B. 5: b ∈ a → b ∈ B. 6.: ∀y (y ∈ a → y ∈ B). 7. (∀y (y ∈ a → y ∈ A)) → ∀y (y ∈ a → y ∈ B). 2-6 8. ∀x((∀y (y ∈ x → y ∈ A)) → ∀y (y ∈ a → y ∈ B)).
S. Choi (KAIST)
Logic and set theory
October 7, 2012
21 / 26
More operations on sets
∀x((∀y (y ∈ x → y ∈ A)) → (∀y (y ∈ x → y ∈ B))) ` A ⊂ B.
S. Choi (KAIST)
Logic and set theory
October 7, 2012
22 / 26
More operations on sets
∀x((∀y (y ∈ x → y ∈ A)) → (∀y (y ∈ x → y ∈ B))) ` A ⊂ B. 1. ∀x((∀y (y ∈ x → y ∈ A)) → (∀y (y ∈ x → y ∈ B))) A.
S. Choi (KAIST)
Logic and set theory
October 7, 2012
22 / 26
More operations on sets
∀x((∀y (y ∈ x → y ∈ A)) → (∀y (y ∈ x → y ∈ B))) ` A ⊂ B. 1. ∀x((∀y (y ∈ x → y ∈ A)) → (∀y (y ∈ x → y ∈ B))) A. 2.: a ∈ A H.
S. Choi (KAIST)
Logic and set theory
October 7, 2012
22 / 26
More operations on sets
∀x((∀y (y ∈ x → y ∈ A)) → (∀y (y ∈ x → y ∈ B))) ` A ⊂ B. 1. ∀x((∀y (y ∈ x → y ∈ A)) → (∀y (y ∈ x → y ∈ B))) A. 2.: a ∈ A H. 3.:: a ∈ {a}. H (used as a hypothesis)
S. Choi (KAIST)
Logic and set theory
October 7, 2012
22 / 26
More operations on sets
∀x((∀y (y ∈ x → y ∈ A)) → (∀y (y ∈ x → y ∈ B))) ` A ⊂ B. 1. ∀x((∀y (y ∈ x → y ∈ A)) → (∀y (y ∈ x → y ∈ B))) A. 2.: a ∈ A H. 3.:: a ∈ {a}. H (used as a hypothesis) 4.:: a ∈ A.
S. Choi (KAIST)
Logic and set theory
October 7, 2012
22 / 26
More operations on sets
∀x((∀y (y ∈ x → y ∈ A)) → (∀y (y ∈ x → y ∈ B))) ` A ⊂ B. 1. ∀x((∀y (y ∈ x → y ∈ A)) → (∀y (y ∈ x → y ∈ B))) A. 2.: a ∈ A H. 3.:: a ∈ {a}. H (used as a hypothesis) 4.:: a ∈ A. 5.: a ∈ {a} → a ∈ A. 3-4
S. Choi (KAIST)
Logic and set theory
October 7, 2012
22 / 26
More operations on sets
∀x((∀y (y ∈ x → y ∈ A)) → (∀y (y ∈ x → y ∈ B))) ` A ⊂ B. 1. ∀x((∀y (y ∈ x → y ∈ A)) → (∀y (y ∈ x → y ∈ B))) A. 2.: a ∈ A H. 3.:: a ∈ {a}. H (used as a hypothesis) 4.:: a ∈ A. 5.: a ∈ {a} → a ∈ A. 3-4 6.: (∀y (y ∈ {a} → y ∈ A)) → (∀y (y ∈ {a} → y ∈ B))
S. Choi (KAIST)
Logic and set theory
October 7, 2012
22 / 26
More operations on sets
∀x((∀y (y ∈ x → y ∈ A)) → (∀y (y ∈ x → y ∈ B))) ` A ⊂ B. 1. ∀x((∀y (y ∈ x → y ∈ A)) → (∀y (y ∈ x → y ∈ B))) A. 2.: a ∈ A H. 3.:: a ∈ {a}. H (used as a hypothesis) 4.:: a ∈ A. 5.: a ∈ {a} → a ∈ A. 3-4 6.: (∀y (y ∈ {a} → y ∈ A)) → (∀y (y ∈ {a} → y ∈ B)) 7.: (a ∈ {a} → a ∈ A) → (a ∈ {a} → a ∈ B).
S. Choi (KAIST)
Logic and set theory
October 7, 2012
22 / 26
More operations on sets
∀x((∀y (y ∈ x → y ∈ A)) → (∀y (y ∈ x → y ∈ B))) ` A ⊂ B. 1. ∀x((∀y (y ∈ x → y ∈ A)) → (∀y (y ∈ x → y ∈ B))) A. 2.: a ∈ A H. 3.:: a ∈ {a}. H (used as a hypothesis) 4.:: a ∈ A. 5.: a ∈ {a} → a ∈ A. 3-4 6.: (∀y (y ∈ {a} → y ∈ A)) → (∀y (y ∈ {a} → y ∈ B)) 7.: (a ∈ {a} → a ∈ A) → (a ∈ {a} → a ∈ B). 8.: a ∈ {a} → a ∈ B.
S. Choi (KAIST)
Logic and set theory
October 7, 2012
22 / 26
More operations on sets
∀x((∀y (y ∈ x → y ∈ A)) → (∀y (y ∈ x → y ∈ B))) ` A ⊂ B. 1. ∀x((∀y (y ∈ x → y ∈ A)) → (∀y (y ∈ x → y ∈ B))) A. 2.: a ∈ A H. 3.:: a ∈ {a}. H (used as a hypothesis) 4.:: a ∈ A. 5.: a ∈ {a} → a ∈ A. 3-4 6.: (∀y (y ∈ {a} → y ∈ A)) → (∀y (y ∈ {a} → y ∈ B)) 7.: (a ∈ {a} → a ∈ A) → (a ∈ {a} → a ∈ B). 8.: a ∈ {a} → a ∈ B. 9.: a ∈ {a} (True statement)
S. Choi (KAIST)
Logic and set theory
October 7, 2012
22 / 26
More operations on sets
∀x((∀y (y ∈ x → y ∈ A)) → (∀y (y ∈ x → y ∈ B))) ` A ⊂ B. 1. ∀x((∀y (y ∈ x → y ∈ A)) → (∀y (y ∈ x → y ∈ B))) A. 2.: a ∈ A H. 3.:: a ∈ {a}. H (used as a hypothesis) 4.:: a ∈ A. 5.: a ∈ {a} → a ∈ A. 3-4 6.: (∀y (y ∈ {a} → y ∈ A)) → (∀y (y ∈ {a} → y ∈ B)) 7.: (a ∈ {a} → a ∈ A) → (a ∈ {a} → a ∈ B). 8.: a ∈ {a} → a ∈ B. 9.: a ∈ {a} (True statement) 9.: a ∈ B.
S. Choi (KAIST)
Logic and set theory
October 7, 2012
22 / 26
More operations on sets
∀x((∀y (y ∈ x → y ∈ A)) → (∀y (y ∈ x → y ∈ B))) ` A ⊂ B. 1. ∀x((∀y (y ∈ x → y ∈ A)) → (∀y (y ∈ x → y ∈ B))) A. 2.: a ∈ A H. 3.:: a ∈ {a}. H (used as a hypothesis) 4.:: a ∈ A. 5.: a ∈ {a} → a ∈ A. 3-4 6.: (∀y (y ∈ {a} → y ∈ A)) → (∀y (y ∈ {a} → y ∈ B)) 7.: (a ∈ {a} → a ∈ A) → (a ∈ {a} → a ∈ B). 8.: a ∈ {a} → a ∈ B. 9.: a ∈ {a} (True statement) 9.: a ∈ B. 10. a ∈ A → a ∈ B.
S. Choi (KAIST)
Logic and set theory
October 7, 2012
22 / 26
More operations on sets
F = {Cs |s ∈ S} a family of sets.
S. Choi (KAIST)
Logic and set theory
October 7, 2012
23 / 26
More operations on sets
F = {Cs |s ∈ S} a family of sets. S Define F as the set of elements in at least one element of F.
S. Choi (KAIST)
Logic and set theory
October 7, 2012
23 / 26
More operations on sets
F = {Cs |s ∈ S} a family of sets. S Define F as the set of elements in at least one element of F. S F = {x|∃A(A ∈ F ∧ x ∈ A)} = {x|∃A ∈ F(x ∈ A)}.
S. Choi (KAIST)
Logic and set theory
October 7, 2012
23 / 26
More operations on sets
F = {Cs |s ∈ S} a family of sets. S Define F as the set of elements in at least one element of F. S F = {x|∃A(A ∈ F ∧ x ∈ A)} = {x|∃A ∈ F(x ∈ A)}. T Define F as the set of common elements of elements of F.
S. Choi (KAIST)
Logic and set theory
October 7, 2012
23 / 26
More operations on sets
F = {Cs |s ∈ S} a family of sets. S Define F as the set of elements in at least one element of F. S F = {x|∃A(A ∈ F ∧ x ∈ A)} = {x|∃A ∈ F(x ∈ A)}. T Define F as the set of common elements of elements of F. T F = {x|∀A(A ∈ F → x ∈ A)} = {x|∀A ∈ F(x ∈ A)}.
S. Choi (KAIST)
Logic and set theory
October 7, 2012
23 / 26
More operations on sets
F = {Cs |s ∈ S} a family of sets. S Define F as the set of elements in at least one element of F. S F = {x|∃A(A ∈ F ∧ x ∈ A)} = {x|∃A ∈ F(x ∈ A)}. T Define F as the set of common elements of elements of F. T F = {x|∀A(A ∈ F → x ∈ A)} = {x|∀A ∈ F(x ∈ A)}. Alternate notations: F = {Ai |i ∈ I}.
S. Choi (KAIST)
Logic and set theory
October 7, 2012
23 / 26
More operations on sets
F = {Cs |s ∈ S} a family of sets. S Define F as the set of elements in at least one element of F. S F = {x|∃A(A ∈ F ∧ x ∈ A)} = {x|∃A ∈ F(x ∈ A)}. T Define F as the set of common elements of elements of F. T F = {x|∀A(A ∈ F → x ∈ A)} = {x|∀A ∈ F(x ∈ A)}. Alternate notations: F = {Ai |i ∈ I}. T T F = i∈I Ai = {x|∀i ∈ I(x ∈ Ai )}.
S. Choi (KAIST)
Logic and set theory
October 7, 2012
23 / 26
More operations on sets
F = {Cs |s ∈ S} a family of sets. S Define F as the set of elements in at least one element of F. S F = {x|∃A(A ∈ F ∧ x ∈ A)} = {x|∃A ∈ F(x ∈ A)}. T Define F as the set of common elements of elements of F. T F = {x|∀A(A ∈ F → x ∈ A)} = {x|∀A ∈ F(x ∈ A)}. Alternate notations: F = {Ai |i ∈ I}. T T F = i∈I Ai = {x|∀i ∈ I(x ∈ Ai )}. S S F = i∈I Ai = {x|∃i ∈ I(x ∈ Ai )}.
S. Choi (KAIST)
Logic and set theory
October 7, 2012
23 / 26
More operations on sets
Example S x ∈ P( F). Analysis:
S. Choi (KAIST)
Logic and set theory
October 7, 2012
24 / 26
More operations on sets
Example S x ∈ P( F). Analysis: S x ⊂ F.
S. Choi (KAIST)
Logic and set theory
October 7, 2012
24 / 26
More operations on sets
Example S x ∈ P( F). Analysis: S x ⊂ F. S ∀y (y ∈ x → y ∈ F).
S. Choi (KAIST)
Logic and set theory
October 7, 2012
24 / 26
More operations on sets
Example S x ∈ P( F). Analysis: S x ⊂ F. S ∀y (y ∈ x → y ∈ F). ∀y (y ∈ x → ∃A ∈ F(y ∈ A)).
S. Choi (KAIST)
Logic and set theory
October 7, 2012
24 / 26
More operations on sets
Example S x ∈ P( F). Analysis: S x ⊂ F. S ∀y (y ∈ x → y ∈ F). ∀y (y ∈ x → ∃A ∈ F(y ∈ A)). S Prove that x ∈ F ` x ∈ P( F).
S. Choi (KAIST)
Logic and set theory
October 7, 2012
24 / 26
More operations on sets
Example S x ∈ P( F). Analysis: S x ⊂ F. S ∀y (y ∈ x → y ∈ F). ∀y (y ∈ x → ∃A ∈ F(y ∈ A)). S Prove that x ∈ F ` x ∈ P( F). x ∈ F ` ∀y (y ∈ x → ∃A ∈ F(y ∈ A)).
S. Choi (KAIST)
Logic and set theory
October 7, 2012
24 / 26
More operations on sets
Example S x ∈ P( F). Analysis: S x ⊂ F. S ∀y (y ∈ x → y ∈ F). ∀y (y ∈ x → ∃A ∈ F(y ∈ A)). S Prove that x ∈ F ` x ∈ P( F). x ∈ F ` ∀y (y ∈ x → ∃A ∈ F(y ∈ A)). 1. x ∈ F. A.
S. Choi (KAIST)
Logic and set theory
October 7, 2012
24 / 26
More operations on sets
Example S x ∈ P( F). Analysis: S x ⊂ F. S ∀y (y ∈ x → y ∈ F). ∀y (y ∈ x → ∃A ∈ F(y ∈ A)). S Prove that x ∈ F ` x ∈ P( F). x ∈ F ` ∀y (y ∈ x → ∃A ∈ F(y ∈ A)). 1. x ∈ F. A. 2.: a ∈ x H.
S. Choi (KAIST)
Logic and set theory
October 7, 2012
24 / 26
More operations on sets
Example S x ∈ P( F). Analysis: S x ⊂ F. S ∀y (y ∈ x → y ∈ F). ∀y (y ∈ x → ∃A ∈ F(y ∈ A)). S Prove that x ∈ F ` x ∈ P( F). x ∈ F ` ∀y (y ∈ x → ∃A ∈ F(y ∈ A)). 1. x ∈ F. A. 2.: a ∈ x H. 3.: ∃A ∈ F(a ∈ A).
S. Choi (KAIST)
Logic and set theory
October 7, 2012
24 / 26
More operations on sets
Example S x ∈ P( F). Analysis: S x ⊂ F. S ∀y (y ∈ x → y ∈ F). ∀y (y ∈ x → ∃A ∈ F(y ∈ A)). S Prove that x ∈ F ` x ∈ P( F). x ∈ F ` ∀y (y ∈ x → ∃A ∈ F(y ∈ A)). 1. x ∈ F. A. 2.: a ∈ x H. 3.: ∃A ∈ F(a ∈ A). 4. a ∈ x → (∃A ∈ F(a ∈ A)) 2-3.
S. Choi (KAIST)
Logic and set theory
October 7, 2012
24 / 26
More operations on sets
Example S x ∈ P( F). Analysis: S x ⊂ F. S ∀y (y ∈ x → y ∈ F). ∀y (y ∈ x → ∃A ∈ F(y ∈ A)). S Prove that x ∈ F ` x ∈ P( F). x ∈ F ` ∀y (y ∈ x → ∃A ∈ F(y ∈ A)). 1. x ∈ F. A. 2.: a ∈ x H. 3.: ∃A ∈ F(a ∈ A). 4. a ∈ x → (∃A ∈ F(a ∈ A)) 2-3. 5. ∀y (y ∈ x → (∃A ∈ F(y ∈ A)))
S. Choi (KAIST)
Logic and set theory
October 7, 2012
24 / 26
More operations on sets
Example S x ∈ P( F) ` x ∈ F. Is this valid? Try to use refutation tree test. S x ∈ P( F). x ∈ / F. 1 ∀y (y ∈ x → ∃A ∈ F(y ∈ A)). 2 x∈ / F. negation first.
S. Choi (KAIST)
Logic and set theory
October 7, 2012
25 / 26
More operations on sets
Example S x ∈ P( F) ` x ∈ F. Is this valid? Try to use refutation tree test. S x ∈ P( F). x ∈ / F. 1 ∀y (y ∈ x → ∃A ∈ F(y ∈ A)). 2 x∈ / F. negation first.
S. Choi (KAIST)
1 ∀y (y ∈ x → ∃A ∈ F(y ∈ A)). 2 x∈ / F. 3 a ∈ x → ∃A ∈ F(a ∈ A).
Logic and set theory
October 7, 2012
25 / 26
More operations on sets
Example S x ∈ P( F) ` x ∈ F. Is this valid? Try to use refutation tree test. S x ∈ P( F). x ∈ / F. 1 ∀y (y ∈ x → ∃A ∈ F(y ∈ A)).
1 ∀y (y ∈ x → ∃A ∈ F(y ∈ A)).
2 x∈ / F.
2 x∈ / F. negation first.
3 a ∈ x → ∃A ∈ F(a ∈ A).
1 ∀y (y ∈ x → ∃A ∈ F(y ∈ A)). 2 x∈ / F. 3 check a ∈ x → ∃A ∈ F(a ∈ A). 4 (i) a ∈ / x 4(ii) ∃A(a ∈ A ∧ A ∈ F).
S. Choi (KAIST)
Logic and set theory
October 7, 2012
25 / 26
More operations on sets
Example S x ∈ P( F) ` x ∈ F. Is this valid? Try to use refutation tree test. S x ∈ P( F). x ∈ / F. 1 ∀y (y ∈ x → ∃A ∈ F(y ∈ A)).
1 ∀y (y ∈ x → ∃A ∈ F(y ∈ A)).
2 x∈ / F.
2 x∈ / F. negation first.
3 a ∈ x → ∃A ∈ F(a ∈ A). 1 ∀y (y ∈ x → ∃A ∈ F(y ∈ A)).
1 ∀y (y ∈ x → ∃A ∈ F(y ∈ A)).
2 x∈ / F.
2 x∈ / F. 3 check a ∈ x → ∃A ∈ F(a ∈ A).
4 (i) a ∈ / x open 4(ii) check ∃A(a ∈ A ∧ A ∈ F)
4 (i) a ∈ / x 4(ii) ∃A(a ∈ A ∧ A ∈ F).
5 (ii) a ∈ A0 6 (ii) A0 ∈ F.
S. Choi (KAIST)
Logic and set theory
October 7, 2012
25 / 26
More operations on sets
How do one obtain a counter-example? x ∈ / F and a ∈ / x. F = {{1, 2}, {1, 3}}. x = {1, 2, 3}. a = 4. F = {{1, 2}, {1, 3}}. x = {1, 2, 3}. a = 3. a ∈ {1, 3}. {1, 3} ∈ F.
S. Choi (KAIST)
Logic and set theory
October 7, 2012
26 / 26
