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The Foundations: Logic and Proofs (2)수학/이산수학 2020. 4. 6. 12:10
Predicates
* 속성과 관계를 나타낸다. 명제와 달리, 주어와 술어를 구분하여 참 거짓 판단한다.
- x is greater than 3
- x: variable
- is greater than 3: predicate P
- P(x) :P(x)자체가 명제는 아니고 변수 x가 적절한 값으로 치환되면 결과적으로 명제가 된다.
- Let Q(x,y) denote the statement “x=y+3”
- Q(1,2) is false but Q(3,0) is true 하나 이상의 변수가 존재할 수 있다.
- n-place predicate (or n-ary predicate)
- P(x1 , x2 , … , xn )
-> 그 자체로는 바로 T, F 인지 구분 불가능.
Quantifiers
˙Universal quantification: ∀x P(x) 모든 x 에 대하여
- A predicate is true for every element
- If P(x) is x<2, is ∀x P(x) true for the domain of integers?
˙ Existential quantification: ∃x P(x) 어떤 x 에 대하여
- A predicate is true for one or more elements
- If P(x) is x<2, is ∃x P(x) true for the domain of integers?
˙ Precedence of Quantifiers
- Quantifiers ∀ and ∃have higher precedence then all logical operators : 제일 높은 우선순위를 가짐.
- ∀x P(x) ∨ Q(x) means (∀x P(x)) ∨ Q(x)
-> predicates에 quantifiers를 붙이면, T,F를 바로 판단 가능하다.
이때, universal(∀)또는 existential(∃)를 붙이냐 2가지 방법이 있다.
Logical Equivalences Involving Quantifiers
Nested Quantifiers
˙∀x ∃y(x+y=0) -> T
- ∀x Q(x) where Q(x) is ∃y(x+y=0)
- For every real number x, there is a real number y such that x+y=0
- ∀x ∃y(x*y=0) -> T
˙∃y∀x(x+y=0) -> F
- ∃y Q(y) where Q(y) is ∀x(x+y=0)
- There is a real number y such that for every real number x, x+y=0
- ∃y∀x(x*y=0) -> T
˙Negating nested quantifiers
- ~∀x∃y(xy=1) ≡ ∃x ~∃y(xy=1) ≡ ∃x∀y~(xy=1) ≡ ∃x∀y(xy≠1)
Valid Arguments
˙ Argument
- Sequence of propositions (p1 , p2 , …, pn )
- Premises (p1 , p2 , …, pn-1 )
: All but the final proposition in the argument
˙ Conclusion (pn )
- The final proposition
˙ The argument is valid if the truth of all its premises implies that the conclusion is true
- Tautology: (p1 ∧ p2 ∧ … ∧ pn-1 ) -> pn
: 주어진 전제가 모두,동시에 true이면 결론은 무조건 true이다.
If you have a password, then you can log onto the network ---------------premises
You have a password ---------------premises
Therefore, You can log onto the network --------------------------- conclusion
Rules of Inference
example 1.
˙ Hypotheses
- Jasmine is skiing or it is not snowing
- It is snowing or Bart is playing hockey
˙ Conclusion
- Jasmine is skiing or Bart is playing hockey
˙ p: it is snowing /q: Jasmine is skiing /r: Bart is playing hokey
1. ~p∨q Hypothesis T
2. p∨r Hypothesis T
3. q∨r Resolution (1) and (2) T
example 2.
˙ Show that the hypotheses (p∧ q)∨r and r -> s imply the conclusion p∨s
1. (p∧ q)∨r Hypothesis
2. (p∨r)∧ (p∨q) Distributive law
3. p∨r Simplification
4. r -> s Hypothesis
5. ~r∨s
6. p∨s Resolution (3) and (5)
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