(述語論理の応用)
http://www.sw.it.aoyama.ac.jp/2016/Math1/lecture7.html
© 2005-16 Martin J. Dürst Aoyama Gakuin University
The examples all are about the students taking Discrete Mathematics I.
Predicates and functions used:
age(s): A student's age (fully completed years)
college(s): A student's faculty or college (example: College of Science and Engineering)
female(s), male(s): Predicates for a student's gender.
native(s, k): True if student s is a native of prefecture k (using "abroad" for students from outside Japan)
Law 5:
(∃x: P(x)) ∧ Q(y) = ∃x:
(P(x)∧Q(y))
Concrete example:
There is a student from Shizuoka, and student y is older than 20 means that there is a case where there is a student from Shizuoka and student y is older than 20, and the other way round.
(∃s: native(s, Shizuoka)) ∧ age(y)>20 = ∃s: (native(s, Shizuoka)∧age(y)>20)
Law 8:
(∀x: P(x)) ∧ (∀x: R(x)) =
∀x: P(x)∧R(x)
Concrete example:
All students are less that 30 years old, and all students belong to the College
of Science and Engineering. = All students are less than 30 years old and
belong to the College of Science and Engineering.
(∀s: age(s)≤30) ∧ (∀t: college(t)=CSE) = ∀u: (age(u)≤30 ∧ college(u)=CSE)
Law 11:
(∃x: P(x)) ∧ (∃x: R(x)) ←
∃x: P(x) ∧ R(x)
Example of how the left side follows from the right:
There is a student who is a native of Hiroshima and who is male. → There is a
student who is a native of Hiroshima, and there is a student who is male.
(∃s: native(s, Hiroshima)) ∧ (∃s:
male(s)) ← ∃s: (native(s, Hiroshima) ∧
male(s))
Example of how the right side does not follow from the left side:
There is a student who is a native from Hokkaido, and there is a student who is
female. However, this does not imply that there is a student who is a native
from Hokkaido and is female.
(∃s: native(s, Hokkaido)) ∧ (∃s:
female(s)) ↛ ∃s: (native(s, Hokkaido) ∧
female(s))
Law 12:
(∃y: ∀x: P(x, y)) →
(∀x: ∃y: P(x, y))
Example of how the left side follows from the right:
There is an age y (e.g. 30) so that for all students,
their age is below y. From this follows that for each
student, there is an age for which the student's age is lower.
(∃y: ∀s:
age(s)<y) → (∀s:
∃y: age(s)<y)
Example of how the right side doesn't follow from the left side:
All students are native of some prefecture (or abroad). But this does not mean
that there is a single prefecture of which all students are native.
(∀s: ∃y: native(s, y)) ↛ (∃y: ∀s: native(s, y))
(∃y: ∀x: P(x, y)) → (∀x: ∃y: P(x, y))
(∀x: ∃y: P(x, y)) ↛ (∃y: ∀x: P(x, y))
The number of prime numbers is infinite.
(This means that whatever big number x we choose, there will always be a bigger prime number y.)
∀x: ∃y: (y > x ∧ prime(y))
Reversing the order of the quantifiers changes the meaning:
∃y: ∀x: (y > x ∧ prime(y))
(There is a prime number y that is bigger than any (natural number) x. This statement is obviously false.)
prime(t) ∨ ∃s: (z<s<t ∧ prime(s)) ⇒
∃y: (y > x ∧ prime(y))
∀s: age(s)≤30) ∧ (∀t: college(t)=CSE) = ∀u: (age(u)≤30∧college(u)=CSE)
is the same as
∀s: age(s)≤30) ∧ (∀s: college(s)=CSE) = ∀s: (age(s)≤30∧college(s)=CSE)
There are three different variables s in the last statement.
Advice:
Sum: ∑∞i=1 1/i2 = 1 + 1/4 + 1/9 + 1/16 + 1/25 + ...
Product: ∏∞i=1 1+1/(-2)i = ...
Universal quantification: (∀i ∈ℕ+: i>0) = ⋀∞i=1 i>0 = 1>0 ∧2>0 ∧3>0 ∧...
Existential quantification: (∃i ∈ℕ+: odd(i)) = ⋁∞i=1 odd(i) = odd(1)∨odd(2)∨odd(3)∨...
Quantification is a generalization of conjunction/disjunction to more than two operands in the same way that sum and product are a generalization of addition/multiplication to more than two operands.
| two operands | many operands | ||
|---|---|---|---|
| name | symbol | name | symbol(s) |
| addition | + | sum | ∑ |
| multiplication | * | product | ∏ |
| conjunction | ∧ | universal quantification | ∀/⋀ |
| disjunction | ∨ | existential quantification | ∃/⋁ |
∀i (i<0⋀i>5): odd(i) = T (because the unit element of conjunction is T)
∃i (i<0⋀i>5): odd(i) = F (because the unit element of disjunction is F)
Laws 1 and 2 introduced in the last lecture are generalizations of DeMorgans' laws:
Simplify ¬(∃x: P(x) → ∀y: ¬Q(y))
¬(∃x: P(x) → ∀y: ¬Q(y)) [removing implication]
= ¬(¬∃x: P(x) ∨ ∀y: ¬Q(y)) [deMorgan's law]
= ¬¬∃x: P(x) ∧ ¬∀y: ¬Q(y) [law 1 of last lecture]
= ¬¬∃x: P(x) ∧ ∃y: ¬¬Q(y) [double negation]
= ∃x: P(x) ∧ ∃y: Q(y)
Actual example:
Let P(x) mean "it is raining in x", and Q(y) "it is snowing y"
Then the original formula says "It's wrong that if it rains somewhere, then it snows nowhere". The final formula says "There is a place where it rains and there is a place where it snows".
∀s: (male(s) ∨ female(s)) [all students are either male or female]
∀s: ¬(male(s) ∧ female(s)) [no student is both male and female]
∀s∈S: (∃k∈K: native(s, k) ∧(∀h∈K: h=k ∨¬native(s, h))) [all students are native of exactly one prefecture]
Deadline: November 11, 2016 (Thursday), 19:00.
Format: A4 single page (using both sides is okay; NO cover page), easily readable handwriting (NO printouts), name (kanji and kana) and student number at the top right
Where to submit: Box in front of room O-529 (building O, 5th floor)
Using your high school books/materials or other sources, research the following terms related to sets, and write a definition and short explanation for each of them:
(高校の本・資料や他の情報を活用して、上記の集合に関連する概念を調査し、定義と簡単な説明を書きなさい。)