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# Sufficient condition set theory ?

Alum Member
71 karma

I want to challenge in the following example ( https://7sage.com/lesson/basic-translation-group-1-flashcards )

Each and every member of Q is omnipotent. Lecture tells us M is SC and O is NC.

However, if we use venn diagram (from set theory) and sorry for the long explanation but I cannot post pictures or draw here.
it raises questions.

Q1: Out of M or O, which "IS" the bigger set ? A1: we do not know
Q2: Out of M or O, which "cannot" be bigger than the other ? A2: M cannot be bigger than O because .... "Each and every member of Q is omnipotent" but are all omnipotent set members members of Q ? Obviously, not necessarily .

Well if Q2, A2 is right than why is O not the SC and M the NC ( not by logic but because whatever is left should fit in the remaining slot )

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• Member
640 karma

Hi. I'm unsure what you mean by "whatever is left should fit in the remaining slot"

In Q2 A2, you wrote:
Each and every member of Q is omnipotent" but are all omnipotent set members members of Q ? Obviously, not necessarily

If it is not necessary that all O are Members of Q, O cannot be the SC.

If O were SC for M, it must be necessary that all O are Members of Q.

• Member
edited January 2019 640 karma

For example:
All strawberries (S) are red (R).

The "strawberries" set cannot be the bigger set than the "red" set.
If there are five strawberries, there cannot be four red things. There must be at least five red things because all strawberries are red.

But this doesn't lead us to conclude that (R) must be the SC for (S).

Although the (R) set must be at least as big as the (S) set, it doesn't mean necessarily mean all red things are strawberries. Some red things may be apples or cars, not strawberries.

It is possible for the (S) set to be equal to (R) set. This is a biconditial relationship in which all S are R and all R are S. Both sets are sufficient and necessary conditions for each other. In this case, the size of the sets is the same. But unless we are told that S and R exist in a biconditional relationship we cannot conclude that all R are S from the statement all S are R.

• Alum Member
2186 karma

What is this set theory you're using? I feel like for these sorts of translations it will probably do more harm than good. The translation should be almost instinctual.

"Each and every member of Q is omnipotent" is the same as saying "All members of Q are Omnipotent". Aka if M then O. Being a Member is sufficient for Omnipotence.

It seems you got to this point when you said, "Each and every member of Q is omnipotent" but are all omnipotent set members members of Q ? Obviously, not necessarily." What else is there to figure out?

• Alum Member
71 karma

Point 1: size of M <= size of O
Point 2: all M members are also all O member but NOT vice versa

therefore, M is the NC and O is the SC, right ? (Opposite to what the answer is in the lecture).

Any thoughts ?

• Member
edited January 2019 640 karma

Please correct me if I misunderstood you.

All members of M are also O members, but not vice versa.

Yes, it is true that the O set contains M members. M members are completely subsumed under the O set. But that does not make O the SC and M the NC.
Although M belongs to the O set, M is not the necessary condition for O.

Let's say, John is a Member of Team O.

If you were asked, what team is John on? The answer is clear that John is a Member of Team O.

John (SC) --> Team O (NC)

John is the sufficient condition for being a member of Team O.

But if you were asked, if someone is a member of Team O, who must it be? You cannot say for certain that it is John.

That is,

Team O is NOT the sufficient condition for John.

This is because there is a possibility that other members exist in Team O. Merely knowing that someone belongs to Team O does not allow us to conclude that that someone must be John.