逻辑学中三段论规则7的论证

2024-12-03 01:34:10
推荐回答(1个)
回答1:

三段论规则7内容:有一个特称的前提,则结论是特称的。
这个规则是个导出规则,即可以从前面的五条基本规则中引申出来。其证明比较容易,通常教科书里都讲过:
只有一个特称前提的情况不外是这样几种情况:1)AI;2)AO;3)EI;4)EO;5)IA;6)OA;7)IE;8)OE。
①情形1)和5),只有A命题主项是周延的,根据规则1,中项M必须周延一次,这周延的机会必须让M得以周延,大词和小词都没有机会周延,故,结论是特称肯定命题。
②情形2)和6),情形3)和7)可以合并一起讨论。它们都有两个周延词项的机会,根据规则1,周延的一个机会要给予中项(词)M,另外,它们都有一个否定的前提,根据规则4,有一个否定的前提结论就是否定的。既然结论是否定的,大词P在否定命题结论(谓项)中就是周延。根据规则3,前提中不周延的词项,在结论中不得周延。因此要求大前提中的大词P也要周延。这样第二个周延的机会就被大词占用了。因此,小词同样得不到周延,所以结论是特称否定命题。
③情形4)和8),EO和OE,都是否定命题,违背三段论规则3,两个否定的前提退不出结论。
从①②③可以证明规则7。

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