Metamath Proof Explorer

Theorem equncomVD

Description: If a class equals the union of two other classes, then it equals the union of those two classes commuted. The following User's Proof is a Virtual Deduction proof completed automatically by the tools program completeusersproof.cmd, which invokes Mel L. O'Cat's mmj2 and Norm Megill's Metamath Proof Assistant. equncom is equncomVD without virtual deductions and was automatically derived from equncomVD .

1:: |- (. A = ( B u. C ) ->. A = ( B u. C ) ).
2:: |- ( B u. C ) = ( C u. B )
3:1,2: |- (. A = ( B u. C ) ->. A = ( C u. B ) ).
4:3: |- ( A = ( B u. C ) -> A = ( C u. B ) )
5:: |- (. A = ( C u. B ) ->. A = ( C u. B ) ).
6:5,2: |- (. A = ( C u. B ) ->. A = ( B u. C ) ).
7:6: |- ( A = ( C u. B ) -> A = ( B u. C ) )
8:4,7: |- ( A = ( B u. C ) <-> A = ( C u. B ) )
(Contributed by Alan Sare, 17-Feb-2012) (Proof modification is discouraged.) (New usage is discouraged.)

Ref Expression
Assertion equncomVD ( 𝐴 = ( 𝐵 ∪ 𝐶 ) ↔ 𝐴 = ( 𝐶 ∪ 𝐵 ) )

Proof

Step Hyp Ref Expression
1 idn1 ⊢ (    𝐴 = ( 𝐵 ∪ 𝐶 )    ▶    𝐴 = ( 𝐵 ∪ 𝐶 )    )
2 uncom ⊢ ( 𝐵 ∪ 𝐶 ) = ( 𝐶 ∪ 𝐵 )
3 eqeq1 ⊢ ( 𝐴 = ( 𝐵 ∪ 𝐶 ) → ( 𝐴 = ( 𝐶 ∪ 𝐵 ) ↔ ( 𝐵 ∪ 𝐶 ) = ( 𝐶 ∪ 𝐵 ) ) )
4 3 biimprd ⊢ ( 𝐴 = ( 𝐵 ∪ 𝐶 ) → ( ( 𝐵 ∪ 𝐶 ) = ( 𝐶 ∪ 𝐵 ) → 𝐴 = ( 𝐶 ∪ 𝐵 ) ) )
5 1 2 4 e10 ⊢ (    𝐴 = ( 𝐵 ∪ 𝐶 )    ▶    𝐴 = ( 𝐶 ∪ 𝐵 )    )
6 5 in1 ⊢ ( 𝐴 = ( 𝐵 ∪ 𝐶 ) → 𝐴 = ( 𝐶 ∪ 𝐵 ) )
7 idn1 ⊢ (    𝐴 = ( 𝐶 ∪ 𝐵 )    ▶    𝐴 = ( 𝐶 ∪ 𝐵 )    )
8 eqeq2 ⊢ ( ( 𝐵 ∪ 𝐶 ) = ( 𝐶 ∪ 𝐵 ) → ( 𝐴 = ( 𝐵 ∪ 𝐶 ) ↔ 𝐴 = ( 𝐶 ∪ 𝐵 ) ) )
9 8 biimprcd ⊢ ( 𝐴 = ( 𝐶 ∪ 𝐵 ) → ( ( 𝐵 ∪ 𝐶 ) = ( 𝐶 ∪ 𝐵 ) → 𝐴 = ( 𝐵 ∪ 𝐶 ) ) )
10 7 2 9 e10 ⊢ (    𝐴 = ( 𝐶 ∪ 𝐵 )    ▶    𝐴 = ( 𝐵 ∪ 𝐶 )    )
11 10 in1 ⊢ ( 𝐴 = ( 𝐶 ∪ 𝐵 ) → 𝐴 = ( 𝐵 ∪ 𝐶 ) )
12 6 11 impbii ⊢ ( 𝐴 = ( 𝐵 ∪ 𝐶 ) ↔ 𝐴 = ( 𝐶 ∪ 𝐵 ) )