Metamath Proof Explorer


Theorem uneqdifeq

Description: Two ways to say that A and B partition C (when A and B don't overlap and A is a part of C ). (Contributed by FL, 17-Nov-2008) (Proof shortened by JJ, 14-Jul-2021)

Ref Expression
Assertion uneqdifeq A C A B = A B = C C A = B

Proof

Step Hyp Ref Expression
1 uncom B A = A B
2 eqtr B A = A B A B = C B A = C
3 2 eqcomd B A = A B A B = C C = B A
4 difeq1 C = B A C A = B A A
5 difun2 B A A = B A
6 eqtr C A = B A A B A A = B A C A = B A
7 incom A B = B A
8 7 eqeq1i A B = B A =
9 disj3 B A = B = B A
10 8 9 bitri A B = B = B A
11 eqtr C A = B A B A = B C A = B
12 11 expcom B A = B C A = B A C A = B
13 12 eqcoms B = B A C A = B A C A = B
14 10 13 sylbi A B = C A = B A C A = B
15 6 14 syl5com C A = B A A B A A = B A A B = C A = B
16 4 5 15 sylancl C = B A A B = C A = B
17 3 16 syl B A = A B A B = C A B = C A = B
18 1 17 mpan A B = C A B = C A = B
19 18 com12 A B = A B = C C A = B
20 19 adantl A C A B = A B = C C A = B
21 simpl A C C A = B A C
22 difssd C A = B C A C
23 sseq1 C A = B C A C B C
24 22 23 mpbid C A = B B C
25 24 adantl A C C A = B B C
26 21 25 unssd A C C A = B A B C
27 eqimss C A = B C A B
28 ssundif C A B C A B
29 27 28 sylibr C A = B C A B
30 29 adantl A C C A = B C A B
31 26 30 eqssd A C C A = B A B = C
32 31 ex A C C A = B A B = C
33 32 adantr A C A B = C A = B A B = C
34 20 33 impbid A C A B = A B = C C A = B