Prove set

I'd start by making a Venn diagram to check whether it's even true. I assume this is supposed to be true for all sets A and B?
 
I assume this is supposed to be true for all sets A and B?
Yep.
I have doubts in how to approach the resolution..
In a similar exercise i was able to prove:
? - ? = ? ∩ ?' ( ' = complement)
? - ? ⊂ ? ∩ ?'
? ∈ ? - ? ⇒ ? ∈ ? ∧ ? ∉ ?
⇒ ? ∈ ? ∧ ? ∈ ?'
⇒ ? ∈ ? ∩ ?'
∴ ? - ? ⊂ ? ∩ ?'

? ∩ ?' ⊂ ? - ?
? ∈ ? ∩ ?' ⇒ ? ∈ ? ∧ ? ∈ ?'
⇒ ? ∈ ? ∧ ? ∉ ?
⇒ ? ∈ ? - ?
∴ ? ∩ ?' ⊂ ? - ?

So: ? - ? = ? ∩ ?'
 
prove ? ∩ ? ∪ ? − ? = A
Surely you must mean \((A\cap B)\cup(A\setminus B)=A\)
Please learn to post the correct question!
If \(t\in(A\cap B)\cup(A\setminus B)\) then what does that mean?
Then \(t\in(A\cap B)\vee t\in(A\cap B^c)\)
That means \((t\in A)\vee[(t\in B)\wedge(t\notin B)]\)
Is that qed?:
 
Can you show us what "order of operations" or "precedence" you have been taught for set operations? I understand that some authors give different rules, but here is one source by which your statement is false (and which I followed in reading it):

Sometimes we want to combine more than two sets and more than one operation to create a more compound expression. But in order to do this we have to establish some set of rules so that we know in what order to do each operation. Just like with numbers, we use parentheses if we want an operation to be done first.​

  1. Just like with numbers, we always do anything in parentheses first. If there is more than one set of parentheses, we work from the inside out.​
  2. We do complements first.​
  3. Union , intersection, and difference operations are all equal in the order. So if we have more than one of these at a time, we have to use parentheses to indicate which of these operations should be done first. For example, the expression A∪B-C doesn’t make any sense because we don’t know which operation we should do first: should we take the union first, and then the difference, or should we take the difference first and then the union? In order to make this clear, we need to either write (A∪B)-C or A∪(B-C).​
 
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