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 ( ( 𝐴𝐶 ∧ ( 𝐴𝐵 ) = ∅ ) → ( ( 𝐴𝐵 ) = 𝐶 ↔ ( 𝐶𝐴 ) = 𝐵 ) )

Proof

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