Metamath Proof Explorer


Theorem frgrwopreglem3

Description: Lemma 3 for frgrwopreg . The vertices in the sets A and B have different degrees. (Contributed by Alexander van der Vekens, 30-Dec-2017) (Revised by AV, 10-May-2021) (Proof shortened by AV, 2-Jan-2022)

Ref Expression
Hypotheses frgrwopreg.v ⊢ V = Vtx ⁡ G
frgrwopreg.d ⊢ D = VtxDeg ⁡ G
frgrwopreg.a ⊢ A = x ∈ V | D ⁡ x = K
frgrwopreg.b ⊢ B = V ∖ A
Assertion frgrwopreglem3 ⊢ X ∈ A ∧ Y ∈ B → D ⁡ X ≠ D ⁡ Y

Proof

Step Hyp Ref Expression
1 frgrwopreg.v ⊢ V = Vtx ⁡ G
2 frgrwopreg.d ⊢ D = VtxDeg ⁡ G
3 frgrwopreg.a ⊢ A = x ∈ V | D ⁡ x = K
4 frgrwopreg.b ⊢ B = V ∖ A
5 fveqeq2 ⊢ x = Y → D ⁡ x = K ↔ D ⁡ Y = K
6 5 notbid ⊢ x = Y → ¬ D ⁡ x = K ↔ ¬ D ⁡ Y = K
7 3 difeq2i ⊢ V ∖ A = V ∖ x ∈ V | D ⁡ x = K
8 notrab ⊢ V ∖ x ∈ V | D ⁡ x = K = x ∈ V | ¬ D ⁡ x = K
9 4 7 8 3eqtri ⊢ B = x ∈ V | ¬ D ⁡ x = K
10 6 9 elrab2 ⊢ Y ∈ B ↔ Y ∈ V ∧ ¬ D ⁡ Y = K
11 fveqeq2 ⊢ x = X → D ⁡ x = K ↔ D ⁡ X = K
12 11 3 elrab2 ⊢ X ∈ A ↔ X ∈ V ∧ D ⁡ X = K
13 eqeq2 ⊢ D ⁡ X = K → D ⁡ Y = D ⁡ X ↔ D ⁡ Y = K
14 13 notbid ⊢ D ⁡ X = K → ¬ D ⁡ Y = D ⁡ X ↔ ¬ D ⁡ Y = K
15 neqne ⊢ ¬ D ⁡ Y = D ⁡ X → D ⁡ Y ≠ D ⁡ X
16 15 necomd ⊢ ¬ D ⁡ Y = D ⁡ X → D ⁡ X ≠ D ⁡ Y
17 14 16 biimtrrdi ⊢ D ⁡ X = K → ¬ D ⁡ Y = K → D ⁡ X ≠ D ⁡ Y
18 12 17 simplbiim ⊢ X ∈ A → ¬ D ⁡ Y = K → D ⁡ X ≠ D ⁡ Y
19 18 com12 ⊢ ¬ D ⁡ Y = K → X ∈ A → D ⁡ X ≠ D ⁡ Y
20 10 19 simplbiim ⊢ Y ∈ B → X ∈ A → D ⁡ X ≠ D ⁡ Y
21 20 impcom ⊢ X ∈ A ∧ Y ∈ B → D ⁡ X ≠ D ⁡ Y