If suppose we know "All A's are B's" then can we say "some B's are A's".
I know that
"All A's are B's" --> "Some B's are not A's"
But can we say
"Some B's are not A's" -->"Some B's are A's"
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Sign up to join this communityIf suppose we know "All A's are B's" then can we say "some B's are A's".
I know that
"All A's are B's" --> "Some B's are not A's"
But can we say
"Some B's are not A's" -->"Some B's are A's"
Here is a simple counterexample:
All unicorns are horse-like creatures — true, by the definition of a unicorn;
Some horse-like creatures are unicorns — false, because there happens not to be any unicorns.
Every case where your syllogism fails, it will be because the class A is in fact the empty set. The empty set is a subset of all sets; but that doesn't mean that every set contains elements which belong to the empty set. In fact, none do.
A historical footnote to Niel's already complete answer.
In Aristotle's syllogistic the (all → some) inference is valid. The argument given for it in the body of the question is not. In modern axiomatizations of Aristotle's assertoric syllogistic (e.g., Corcoran's), the inference is captured as a rule called "a-i conversion." Modern logics don't validate the move because whenever the plurality (≈ extension) corresponding to A is empty, the universal affirmation will vacuously be true, but since there won't be any As, the particular affirmation will be false.
The question states:
∀ x . A ( x ) → B ( x ) ⊢ ∃ y . B ( y ) → ¬ A ( y )This is false.
Then it asks:
∃ x . B ( x ) → ¬ A ( x ) ⊢ ∃ y . B ( y ) → A ( y )This is false as well.
The title says:
∀ x . A ( x ) → B ( x ) ⊢ ∃ y . B ( y ) → A ( y )This is false as well.
We cannot derive those conclusions. They are possible, but not necessary.
You simple ask if set A is included in set B will set B contains elements of set A. so lets give an example if A = {1,2,3} and B = {1,2,3,4} ,yes A is included in B and yes there are elements from B which are equal to elements from A