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發表於 2006-10-20 12:48 AM
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Modus ponendo ponens
. e0 ~0 G& _2 ^& D7 ^! iP --> Q, 因為 P 所以 Q
7 h6 L; J, q2 [. I* G
/ x4 [' K9 H( c; ?1 k! rModus tollendo tollens
5 Y$ ~8 @! r, D% WP --> Q, 取得 ~P 所以 ~Q
+ W6 j7 N, {1 p5 ~; |% d9 L
0 V4 r( _2 h lDisjunctive syllogism1 q) b* I7 f' q8 @- s9 Y
P or Q, 取得 ~P 所以 Q
7 D3 c' V) g NP or Q, 取得 ~Q 所以 P
( I& }5 A( Z. x1 i# k$ q% z+ Y1 M! q6 z+ j
Conjunctive syllogism' K, }' R/ N; v% }% ]3 ^
~(P & Q), 取得 P 所以 ~Q
2 I: p& X' F" t7 {~(P & Q), 取得 Q 所以 ~P
8 W3 B, M4 U; d) j7 D( P4 J7 \4 J
7 D( V$ L0 I5 B/ K- `7 IHypothetical syllogism0 R3 A+ E( B2 A4 P! r! w
P --> Q, 取得 Q --> R 所以 P --> R
6 j) G7 N* D M- X$ G' L- ?
) t1 I5 s8 @ U2 `De Morgan's theorem- P. W, S+ c/ L6 p1 B, Y5 ~
~(P & Q) 所以 ~P or ~Q8 y3 R. ~- z% g2 J
~(P or Q) 所以 ~P & ~Q |
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