请教一道基础的逻辑关系

fuxinwang

If A, then B and C; If B or C, then D
能否推出：If A, then D?
因为 B or C 包括了 B and C 的可能性，所以应该可以成立。
但实际上的逻辑关系到底应该是：
A--> B and C --> D
或者是：
A--> B or C --> D 这样表达。
之所以要确定，是因为在做contrapositive的时候，以上两个式子表达的方式是不一样的：
A--> B and C --> D ==> NOT D --> NOT B or C --> NOT A
A--> B or C --> D ==> NOT D --> NOT B and C --> NOT A
感谢万分。

我做的原题如下，有兴趣的同学可以感受一下：
A music store carries exactly ten types of CDs - both new and used of each of Jazz, Opera, Pop, Rap, and Soul. The store is having a sale on some of these types of CDs. THe following conditions must apply:

Used Pop is on sale; New Opera is not
If both types of Pop are on sale, then all Soul is
If both types of jazz are on sale, then no rap is
If neither type of jazz is on sale, then new pop is
If either type of rap is on sale, then no soul is


QUESTION 10: Which of the following CANNOT be true?

A) Neither type of Opera and neither type of rap is on sale
B) Neither type of Jazz and neither type of opera is on sale
C) Neither type of opera and neither type of sould is on sale
D) Neither type of jazz and neither type of sould is on sale
E) Neither type of jazz and neither type of rap is on sale


QUESTION 11: IF neither type of jazz is on sale, then the following must be true EXCEPT:

A) Used opera is on sale
B) New rap is not on sale
C) Used rap is not on sale
D) New soul is on sale
E) Used soul is on sale


QUESTION 12: If new sould is the only type of new CD on sale, then which one of the following CANNOT be true?

A) Used jazz is not on sale
B) Used opera is not on sale
C) Used rap is not on sale
D) Used soul is on sale
E) Used soul is not on sale


QUESTION 13: IF exactly four of the five types of used CDs are the only CDs on sale, which of the following could be true?

A) Used jazz is not on sale
B) Used opera is not on sale
C) Used rap is not on sale
D) Neither type of jazz is on sale
E) Neither type of rap and neither type of soul is on sale

cdwayne

你這是LSAT的game吧

A --> B and C  --> D
          ↘
             D


~D --> ~B and ~C --> ~A
               ↘
                 ~A

fuxinwang

你這是LSAT的game吧

A --> B and C  --> D
          ↘
             D


~D --> ~B and ~C --> ~A
               ↘
                 ~A

-- by 会员 cdwayne (2011/5/10 1:27:33)

感谢回复，不过这个表达式A --> B and C  --> D
                                                    ↘
                                                      D
是说A--> B and C OR D --> D这个意思吗？这个没太明白。
还有，and 的contrapositive不应该是 or吗？
恩，是LSAT的game题。再次感谢你的回复。。。

fuxinwang

最后还是自己做出来了，大家帮我检查一下看正确与否。
Initial rules:
Since UP is in already, rule 1 can also be written as
1. NP->US+NS=>~US/~NS->~NP
2. UJ+NJ->~UR+~NR=>UR+NR->~UJ/~NJ
3. ~UJ+~NJ->NP=>~NP->UJ/NJ
4. UR/NR->~US+~NS=>US/NS->~UR+~NR
By combining link 3 ~UJ+~NJ->NP with link 1 NP->US+NS and then since we know US+NS is sufficient to US/NS, we can further add link 4 US/NS->~UR+~NR into the chain.
Finally we have: ~UJ+~NJ->NP->US+NS->~UR+~NR with its contrapositive UR/NR->~US/~NS->~NP->UJ/NJ
Thus, we have included all elements except UO, which is not mentioned in the question.
Finally, I will solve one question using the logic chain I derived above.
11. If neither type of j is on sale, then each of the following must be true EXCEPT:
A. Used o is on sale.
B. New r is not on sale.
C. Used r is not on sale.
D. New s is on sale.
E. Used s is on sale.

The initial in/out diagram:
Sale      Not on sale
UP        NO

If neither j is on sale, then:
Sale      Not on sale
UP        NO
NJ
UJ
and from the chain derived above, ~UJ+~NJ->NP->US+NS->~UR+~NR, we know that
Sale      Not on sale
UP        NO
NP        NJ
US        UJ
NS        UR
NR
Everything is known except UO, basically a reiteration of the chain.