Existential Quantifier: (\exists x \in S)P(x) means "there exists an x in S such that P(x) holds." (某个)
(\exists x \in \mathbb{R})(x^2 \geq 8)
\bigcup_{i\in I}A_i=\{x\in U | (\exists i \in I)(x \in A_i)\}
\mathcal{E} = \{x\in \mathbb{Z}|(\exists y \in \mathbb{Z})(x=2y)\}
(\exists x, y \in \mathbb{Q})(x^2+y^2=0)=(\exists x \in \mathbb{Q})((\exists y \in \mathbb{Q})(x^2+y^2=0))=(\exists y \in \mathbb{Q})(\exists x \in \mathbb{Q})(x^2+y^2=0)
:warning: existential quantifiers itself can switch places
Universal Quantifier: (\forall x \in S)P(x) mean "for every x in S, P(x) holds true". (全部)
(\forall x \in \mathbb{R})(x^2 \geq 8)
\bigcap_{i\in I}A_i=\{x\in U | (\forall i \in I)(x\in A_i)\}
(\forall p \in \mathbb{Z}^+)(\forall q \in \mathbb{Z}^+)(\sqrt{2} \neq \frac{p}{q}) = (\forall p, q \in \mathbb{Z}^+)(\sqrt{2} \neq \frac{p}{q})
:warning: universal quantifiers itself can switch places
When both \exists and \forall in a statement, order matters. "All people loves some people"
Let P(x, y) denotes "x loves y"
Truth Table: a table with one row for each possible combination of truth values for the elementary propositions (sentence symbols) and columns containing the truth values for compound statements formed from them.
Negation(\lnot, \neg, \sim) Conjunction(\land, \wedge, \&) Disjunction(\lor, \vee, \parallel) Implication(\implies, \to, \Rightarrow, \rightarrow, \supset): P \implies Q \equiv \lnot P \lor Q
:warning: implies does not mean causality
Modus Ponens: "If P->Q, and P is True, then Q must be true"
P | Q | P->Q |
---|---|---|
T | T | T |
T | F | F |
F | T | T |
F | F | T |
> When P is F, it has nothing to do with implication |
When Q is True independently, result always True
When P is False independently, result always True
P->Q only if P=True and Q=False
Examples:
A\cap B = \{x\in U | x=A \land x\in B\}
(\forall x \in \mathbb{R})(x<0 \lor (\exists y \in \mathbb{R})(x=y^2))
(\exists x \in \mathbb{R})(x > 2 \implies x^2 > 4) - True
(\exists x \in \mathbb{R})(x^2 > 4 \implies x > 2) - False
(\forall x \in \mathbb{N})(x \geq 0) \implies (\forall x \in \mathbb{R})(x^2 \geq 0) - True(no causality)
(\forall x \in \mathbb{Z})((\exists n \in \mathbb{Z})(2x=2n+1)\implies x^2<0) - True (vacuously true when P is always false) (for any integer, 2 times integer can never be an odd number)
Homework due by noon Friday. TODO go to office hour
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