# Existential import.
七、請說明什麼是「存在預設(existential import)」?此概念對有效性定義有何影響?(10%) (Please explain the significance of "existential import" and show how the definition of validity is affected by the concept.)
(105)
(unit 10)
答.
存在預設是在亞里斯多德的傳統三段論法中對"全稱肯定" (A 句型) 及"全稱否定" (E 句型) 的一個隱藏假設.反而如果是"特稱肯定" (I 句型) 或"特稱否定" (O 句型), 這種存在預設是明確的. 例如, 命題 "所有的天狗都不會吃草" (E句型), 在亞里斯多德邏輯中, 隱含存在預設，即假設 "天狗存在". 這影響判斷論證的結論. 例如:
$$
\begin{array}{ }
所有的羊都會吃草 \\
所有的天狗都不會吃草 \\ \hline
有些天狗不是羊
\end{array}
$$
這裡$S=天狗$, $M=會吃草$, $P=羊$. 所以我們有 $$
\begin{array}{ }
P^{\circ} & A & M^{\times} \\
S ^{\circ}& E & M^{\circ} \\ \hline
S^{\times} & O & P^{\circ}
\end{array}
$$
用亞里斯多德邏輯的檢測:
(1) 結論為否定(O)句, 且前論為一肯一否. 這個通過 $\checkmark$
(2) 畫上周延符號, 中項至少一次周延. 這個通過 $\checkmark$
(3) 結論有周延的語詞, 前論也周延. 這個通過 $\checkmark$
在亞里斯多德邏輯中 (假設存在預設), 這個論證是有效的.
但在現代邏輯 (利用范恩圖分析) 下:
![[basic-logic/---attachments/logic-problem-7-e-2025-11-13-22-22-45.svg]]
但 $S$ 集合裡不一定有天狗! 這個論證是無效的.
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Existential import asserts that any categorical description of things must exists one such witness. This is an Aristotelian point of view. For example, when we say the statement "all birds can fly", we already assumes the category of birds contains at least one example of such birds. At the same time, when we say "all Tengus do not eat grass", this also asserts the existence of one such Tengus. However, this assertion of existence affects validity of proofs, for example validity of syllogisms. Consider the following syllogism,
All goats eat grass
All Tengus do not eat grass
Some Tengus are not goats
In Aristotelian that assumes the existential import, this is a valid argument. However, with Venn diagram approach (which does not assume existential import), one can see this is not a valid argument.