Metamath Proof Explorer


Theorem uneqin

Description: Equality of union and intersection implies equality of their arguments. (Contributed by NM, 16-Apr-2006) (Proof shortened by Andrew Salmon, 26-Jun-2011)

Ref Expression
Assertion uneqin A B = A B A = B

Proof

Step Hyp Ref Expression
1 eqimss A B = A B A B A B
2 unss A A B B A B A B A B
3 ssin A A A B A A B
4 sstr A A A B A B
5 3 4 sylbir A A B A B
6 ssin B A B B B A B
7 simpl B A B B B A
8 6 7 sylbir B A B B A
9 5 8 anim12i A A B B A B A B B A
10 2 9 sylbir A B A B A B B A
11 1 10 syl A B = A B A B B A
12 eqss A = B A B B A
13 11 12 sylibr A B = A B A = B
14 unidm A A = A
15 inidm A A = A
16 14 15 eqtr4i A A = A A
17 uneq2 A = B A A = A B
18 ineq2 A = B A A = A B
19 16 17 18 3eqtr3a A = B A B = A B
20 13 19 impbii A B = A B A = B