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I was trying to prove that set A is equal to set B (A=B) if there is given a statement:

  1. $A \cup C$ = $B \cup C$

I tried to prove it by showing that ($A \cup C$) $\subseteq$ ($B \cup C$) and vice versa but i failed because i can't relate those three sets to get a relation between A and B. I was trying to use membership table too but i faced same problem, i couldn't relate A and B.

It seems obvious that it is not clear wheter set A is equal to set B or not if we assumed all the elements that contained in all the sets (example, I assume that C is {1,2,3,4,5} and A is {1,4,6,7,8}, and A $\cup$ C is {1,2,3,4,5,6,7,8}, and according to statement 1, because elements of C are never change, so it force B to be set that contained elements {3,4,5}) , because A, or B, still have a multiple form that can fulfill statement 1, or A may be different with B.

But, can we prove it by sets identity or another mathematical method?

And can we determine if A = B, if we have a statement :

$A \cap C$ = $B \cap C$

or a statement

$A \cup C$ = $B \cup C$ and $A \cap C$ = $B \cap C$ ,

Please, someone help me to solve this problem

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  • $\begingroup$ Do you mean that the equality $A\cup C=B\cup C$ holds for every set $C$: Or for some set? $\endgroup$ – José Carlos Santos Sep 9 '18 at 14:20
  • $\begingroup$ for every set C, Mr. Santos $\endgroup$ – Dziban N Sep 9 '18 at 14:25
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In order to prove that $A=B$, take $C=\emptyset$. Then $A\cup C=B\cup C$. But$$A\cup C=B\cup C\iff A\cup\emptyset=B\cup\emptyset\iff A=B.$$

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Even if we don't have enough axioms to prove there is an empty set, suppose we have the reasonable-looking axioms $\forall X\;(X=X\cup X)$ and $\forall X, Y\;(X\cup Y=Y\cup X).$ Then if $\forall C\;(A\cup C=B\cup C),$ let $C_1=A$ and $C_2=B,$ and we have $$A=A\cup A=A\cup C_1=B\cup C_1 = B\cup A$$ $$B=B\cup B=B\cup C_2=A\cup C_2=A\cup B$$ Without regard to what $\cup$ means.

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