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March 29, 2015

Posted by **Deb** on Tuesday, April 27, 2010 at 12:24am.

(a)No A is B (b) All B is A

Some C is A All C is A

____________ _____________

Some C is not B All C is B

- Math -
**alec**, Monday, May 24, 2010 at 4:54pmHi,

1)

The area is part of C and not part of B hence the statement some C is not B is valid

Let r be "The object belongs to set C"

(p=>~q)Λ(pΛr) => (rΛ~q)

The statement is valid because in all cases the statment is true (note that all Ts in the final column)

2)

The diagram shows the arguement is not valid since the diagram satisfies all B is A and all C is A without the conclusion all C is B (in fact you couldn't concluded even some C is B as the diagram shows that it is possible that no C is B)

In statement form we have

(q=>p)Λ(r=>p) => (r=>q)

graphic

View Full Image

As you can see there is a case when the staement isn't true (counterexample I guess). This is when an element was a member of A and C but not B. This case didn't contradict either of the statements All B is A and All C is A but it did contradict the conclusion that all C is B.

I hope this has helped. Any questions then please ask. All the best,

justanswer dot com/questions/1s1wr-truth-tables-are-related-to-euler-circles-arguments-in-the#ixzz0osmCruQD

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