All A's are B's.
No B's are C's.
Which conclusion is necessarily true?
A) No A's are C's.
B) All C's are A's.
C) Some B's are A's.
D) No C exists.
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D) No C exists
If all A’s are B’s and No B’s are C’s than it would mean “all” is either A or B so the lack B “No B’s are C’s” would mean that No C exists.
A) No A's are C's.
No A is C.
Here's the step-by-step reasoning:
All A's are B's: This means that the set of "A's" is completely contained within the set of "B's". If something is an A, it must necessarily be a B.
No B's are C's: This means that the sets of "B's" and "C's" have no intersection; they are completely separate.
If you take any element that is an A, you'll first know that it's a B (by the first rule). But since it's a B, and no B can be a C (by the second rule), then that element cannot be a C.
Logical conclusion: The intersection of A and C is empty.
No A's are C's
A) No A's are C's.
Please justify your answer
if all A’s are B’s they cannot be C’s because that would mean a member of B was also a member of C and the second statement disallows that condition.
Not sure if I get this right but here is how I see it.
All elements of the set A belong to B. No elements of the set B belong to C.
Hence No elements of A are in C seems correct to me. Answer A seems right.
All elements of C belong to A contradicts the second statement. Answer B seems false.
Some elements of B are A seems right to me as the first statement says that all elements of A are in B.
I am not a native English speaker but I understand the last statement as the set C does not exist, which seems false fo me given the second statement.
My answer is A is necessarily true.
A) No A's are C's
Please justify your answer
a.