[Math] Logic
Propositions
Conditions
variable에 따라 True, False 가 달라지는 expression.
\[x^2 = 4\]- $x= \pm 2$ 이면 True.
- otherwise: False
비교: true/false propositions
- true proposition: 항상 true.
- false proposition: 항상 false.
Truth Sets
어떤 condition을 만족하는 element들로 구성된 set.
condition이 정해지면, truth set이 지정됨.
condition을 통해 set이 정해짐.
Logical Operations
Unary Operations
- $\neg$ : Logical Negation
Binary Operations
- $\land$ : Logical Conjunction (=
and) - $\lor$ : Logical Disjunction (=
or) - $\implies$ : Logical Implication
- $\oplus$ : Logical Exclusive Disjunction (=
xor)
참고: Truth Table
Logical Implications
Truth Table of Logical Implication
\[p \implies q\]cause $p$가 False 경우 무조건 True임.
무죄 추정의 원칙?
| $p$ | $q$ | |
|---|---|---|
| False | False | True |
| False | True | True |
| True | False | False |
| True | True | True |
Sufficiency and Necessity
Converse, Contrapositive, Inverse
\[p \implies q\]- converse (역): $q\implies p$
- inverse (부정): $\neg p \implies \neg q$
- contrapositive (대우): $\neg q \implies \neg q$
All and Existent
For All $\forall$
$\forall x, p$: 모든 $x$ 에 대해 $p$가 만족함.
- $\forall x, 0x = 0$
- $\forall x \text{ except }x=0, x\frac{1}{x}=1$
There Exists $\exists$
$\exists x, q$: 최소한 하나 이상의 $x$에서 $q$가 만족함.
- $\exists x, x+3 = 0$
- $\exists x, x^2=1$
Identities and Inverses
Additive Identities and Inverse
\[\begin{aligned} \forall x, \exists a, x+a =x & \implies a=0 \\ \forall x, \exists a, x+a = 0 &\implies a=-x \end{aligned}\]Multiplicative Identities and Inverse
\[\begin{aligned} \forall x \text{ except } x=0, \exists a, xa=x &\implies a=1 \\ \forall x \text{ except } x=0, \exists a, xa=1 &\implies a=x^{-1} \end{aligned}\]Logical Computations
Commutativity
교환법칙
- $p \land q \iff q \land p$
- $p \lor q \iff q \land p$
Associativity
결합법칙
- $(p \land q) \land r \iff p \land (q \land r)$
- $(p \lor q) \lor r \iff p \lor (q \lor r)$
Distributivity
분배접칙
- $p \land (q \lor r) \iff (p \land q) \lor (p \land r)$
-
$p \lor (q \land r) \iff (p \lor q) \land (p \lor r)$
- $(p \implies q) \iff (\neg q \implies \neg p)$
- $(p \oplus q) \iff [{p \land (\neg q)} \lor { (\neg p) \land q } ]$