How would I prove this
A⊆B ^ |A|=|B| -> A=B
Is this proof ok and formal?
Suppose ¬(A⊆B ^ |A|=|B|)
Then |A|≠|B|
Therefore A⊆B ^ |A|=|B| -> A=B
I forgot to mention that the sets were finite.
So if A≠B then since A⊆B, then |A|<|B|
therefore A⊆B ^ |A|=|B| -> A=B
is this better?
why wont it work for non-finite sets?
Is it because then you cant say
A⊆B ^ |A|=|B| ?
No, it means that E is a proper subset of Z^+, i.e., it does not equal the whole Z^+.does the symbol of the C with the 'not equal sign' under mean not a proper subset of ?
There is still a question whether enough details are provided to justify this statement: "if A≠B then since A⊆B, then |A|<|B|." This statement is true and sufficiently obvious, but whether more needs to be said depends on the course requirements and the definition of |A|.And also is my proof ok now?
(1) As has been pointed out, we need to assume that either A or B is finite.
(2) Your logic is wrong anyway. You don't prove P -> Q by assuming ~P and showing ~Q. You have it backwards. What you want here is to assume ~Q and show ~P.
So suppose ~A=B. Show ~(A subset of B & |A|=|B|). Suppose A subset of B. Then, since ~A=B, we have A proper subset of B. But no finite set has the same cardinality as a proper subset of itself (this is called 'the pigeonhole principle').
However, if you haven't already proven the pigeonhole principle, then you have to prove it, and that is a longer, more involved proof.