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