最近看到一些簡單的邏輯 (Logic) 英詞名詞，並不是很清楚，所以，找出來再複習一下。
其實大部份是很簡單的，只是換成英文就不熟了.
Rules of Inference
| Modus Ponens [tex] \displaystyle\begin{array}{l} p\rightarrow q \\ p \\ \therefore q \end{array} [/tex] | Modus Tollens [tex] \begin{array}{l} p\rightarrow q \\ \sim q \\ \therefore \,\sim p \end{array} [/tex] | Hypothetical Syllogism [tex] \begin{array}{l} p\rightarrow q \\ q\rightarrow r \\ \therefore p\rightarrow r \end{array} [/tex] |
| Disjunctive Syllogism [tex] \begin{array}{l} p\vee q \\ \sim p \\ \therefore q \end{array} [/tex] | Constructive Dilemma [tex] \begin{array}{l} (p\rightarrow q) \& (r\rightarrow s) \\ p\vee r \\ \therefore q\vee s \end{array} [/tex]
| Absorption [tex] \begin{array}{l} \,\\ p\rightarrow q \\ \therefore p\rightarrow (p\,\&\,q) \end{array} [/tex] |
| Simplification [tex] \begin{array}{l} \\ p\,\&\,q \\ \therefore p \end{array} [/tex] | Conjunction [tex] \begin{array}{l} p \\ q \\ \therefore p\,\&\,q \end{array} [/tex] | Addition [tex] \begin{array}{l} \\ p \\ \therefore p \vee q \end{array} [/tex] |
Rules of Replacement
| Double Negation | [tex] p\leftrightarrow \,\sim\sim q[/tex] |
| Commutation | [tex] \begin{array}{l} \\ (p\vee q)\leftrightarrow (q\vee p) \\ (p\,\&\,q)\leftrightarrow (q\,\&\,p) \\ \end{array} [/tex] |
| Tautology | [tex] \begin{array}{l} \\ p\leftrightarrow (p\vee p) \\ p\leftrightarrow (p\,\&\,p) \\ \end{array} [/tex] |
| Association | [tex] \begin{array}{l} \\ \left[p\vee (q\vee r)\right] \leftrightarrow \left[(p\vee q)\vee r\right] \\ \left[p\,\&\, (q\,\&\, r)\right] \leftrightarrow \left[(p\,\&\, q)\,\&\, r\right] \\ \end{array} [/tex] |
| Transposition | [tex] \begin{array}{l} \\ (p\rightarrow q) \leftrightarrow (\sim q\rightarrow \sim p) \\ \end{array} [/tex] |
| Material Implication | [tex] \begin{array}{l} \\ (p\rightarrow q) \leftrightarrow (\sim p\vee q) \\ \end{array} [/tex] |
| Exportation | [tex] \begin{array}{l} \\ \left[(p\,\&\, q)\rightarrow r }\right] \leftrightarrow \left[p\rightarrow (q\rightarrow r)\right] \\ \end{array} [/tex] |
| Material Equivalence | [tex] \begin{array}{l} \\ (p\leftrightarrow q)\leftrightarrow \left[(p\rightarrow q) \,\&\, (q\rightarrow p)\right] \\ (p\leftrightarrow q)\leftrightarrow \left[(p\,\&\, q) \vee (\sim p\,\&\, \sim q)\right] \\ \end{array} [/tex] |
| Distribution | [tex] \begin{array}{l} \\ \left[p \,\&\, (q\vee r)\left] \leftrightarrow \left[(p\,\&\, q)\vee (p\,\&\, r)\right] \\ \left[p \vee (q\,\&\, r)\left] \leftrightarrow \left[(p\vee q)\,\&\, (p\vee r)\right] \\ \end{array} [/tex] |
| De Morgan's Theorems | [tex] \begin{array}{l} \\ \sim(p \,\&\, q)\leftrightarrow (\sim p \vee \sim q) \\ \sim(p \vee q)\leftrightarrow (\sim p \,\&\, \sim q) \\ \end{array} [/tex] |
Bi-conditionals Logical Equivalence
[tex] (\forall x)(\psi x\rightarrow \varphi x) \leftrightarrow \,\sim(\exists x)(\psi x \,\&\, \sim\varphi x) [/tex]
"Everything in the lake is wet."
is logically equivalent to
"There isn't anything in the lake which is not wet."
[tex] (\exists x)(\psi x\,\&\, \varphi x) \leftrightarrow \,\sim(\forall x)(\psi x \rightarrow \sim\varphi x) [/tex]
"There exists at least one individual who is both a native of Boston and of Irish descent."
is logically equivalent to
"It's not true that no natives of Boston are of Irish descent."
[tex] (\forall x)(\psi x\rightarrow \sim\varphi x) \leftrightarrow \,\sim(\exists x)(\psi x \,\&\, \varphi x) [/tex]
"No residents of Boston are Irish."
is logically equivalent to
"It's not true that some residents of Boston are Irish."
[tex] (\exists x)(\psi x\,\&\, \sim\varphi x) \leftrightarrow \,\sim(\forall x)(\psi x \rightarrow \varphi x) [/tex]
"Some residents of Boston are not Irish."
is logically equivalent to
"Not all residents of Boston are Irish."
Orignal From: 基本邏輯名詞
