How you are understanding the question is that the circles overlap which called a Venn diagram. In which case, there are three possible drawings:
- The two circles DO NOT overlap
- They PARTIALLY overlap
- They COMPLETELY overlap.
(1) If they DO NOT overlap, all that belong to A do not belong to B and vice versa, or alternatively something can belong to A or B but not both. (∃x:x∈A⊻x∈B)
(3) If they COMPLETELY overlap, all that belong to A belong to B, or alternatively all that belong to B belong to A, or alternatively all that belong to either belong to both. (∃x:x∈A∧x∈B)
and to your question:
(2) If they PARTIALLY overlap, some that belong to A belong to B and some do not, and some that belong to B belong to A and some do not, or alternatively, some belong to only A, only B, or only to both A and B. (∃x:(x∈A∧x∉B)⊻(x∉A∧x∈B)⊻(x∈A∧x∈B))
You believe that the last is the case implied by the question, however, it is not the only case!
- If two circles PARTIALLY overlap, either neither contains the other since both partially contain the other but not COMPLETELY or one contains the other completely. If one contains the other AND they overlap but not COMPLETELY, then one PROPERLY CONTAINS the other.
Thus, instead of drawing two circles overlapping delineating three regions since the two circles proper intersect at two points, you can draw two circles so their regions overlap by having one circle entirely inside the region of another with no intersection of the circles proper at all. In plainspeak just draw one circle entirely inside the other. Two concentric circles of different diameters would be an example.
'If A does not belong to some B, it is not necessary that B should not belong to some A'
This can be rephrased as 'some A are not B, but all B are A'. It can be a special case that there exists A in B, but no B outside of A because ALL B are members of A and B does not exhaust A. This is called a proper subset. (∃x:(x∈A∧x∉B)⊻(x∈A∧x∈B))
RESPONSE TO COMMENT, 2020-02-11
How come for "if A does not belongs [sic] to some B, it is not necessary that B should not belong to some A' one circle contains the other, whereas for 'if A belongs to some B, it is necessary for B to belong to some A' the circles partially overlap? Why is one a circle within another circle and one a partially overlapping circle, when the language is so similar?
Okay, I should have gone with a pic the first time, but I'm lazy. The language is difficult but gleeful for logicians because we don't use the double negative in English normally. The type of situation that makes ordinary language philosophers groan. Let's unpack the image.
"If A does not belong to some B", then Case 1 is NOT possible. So we can rule it out. That leaves three cases. In Case 4, "If A does not belong to some B", it IS necessary that B should not belong to some A. So far, so good. But, it is ALSO true, that "it is NOT necessary that B should not belong to some A" because we have Cases 2 and 3. In fact, with Case 2 we can say if "A does not belong to some B, it is not necessary that B should not belong to some A" because it is POSSIBLE by S3 that A can B. But that might be confusing, so let's look at Case 3. This is the definitive case because if "A does not belong to some B, it is not necessary that B should not belong to some A" because it is IMPOSSIBLE that B does not belong to A.
So to review, "if A does not belong to some B" (S2, S5, S7; x∈A∧x∉B) such as in Cases 2, 3, and 4, we have three possibilities:
- Either "it IS necessary that B should not belong to some A" (x∈B∧x∉A) which is only Case 4 since any B can't be A.
- Or B might or might not belong to A (S3⊻S4; (x∈B∧x∉A)⊻(x∈B∧x∈A)) as in Case 2 because the B has regions in and out of A
- Or it is necessary that B SHOULD belong to some A" (S6; x∈B∧x∈A) which is Case 3.
Ergo, "If A does not belong to some B, it is not necessary that B should not belong to some A".
The key is working back from the non-double negative version of the statement which makes this statement almost incomprehensible!