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