Metamath Proof Explorer


Theorem 3cycld

Description: Construction of a 3-cycle from three given edges in a graph. (Contributed by Alexander van der Vekens, 13-Nov-2017) (Revised by AV, 10-Feb-2021) (Revised by AV, 24-Mar-2021) (Proof shortened by AV, 30-Oct-2021)

Ref Expression
Hypotheses 3wlkd.p ⊢ P = ⟨“ ABCD ”⟩
3wlkd.f ⊢ F = ⟨“ JKL ”⟩
3wlkd.s ⊢ φ → A ∈ V ∧ B ∈ V ∧ C ∈ V ∧ D ∈ V
3wlkd.n ⊢ φ → A ≠ B ∧ A ≠ C ∧ B ≠ C ∧ B ≠ D ∧ C ≠ D
3wlkd.e ⊢ φ → A B ⊆ I ⁡ J ∧ B C ⊆ I ⁡ K ∧ C D ⊆ I ⁡ L
3wlkd.v ⊢ V = Vtx ⁡ G
3wlkd.i ⊢ I = iEdg ⁡ G
3trld.n ⊢ φ → J ≠ K ∧ J ≠ L ∧ K ≠ L
3cycld.e ⊢ φ → A = D
Assertion 3cycld ⊢ φ → F Cycles ⁡ G P

Proof

Step Hyp Ref Expression
1 3wlkd.p ⊢ P = ⟨“ ABCD ”⟩
2 3wlkd.f ⊢ F = ⟨“ JKL ”⟩
3 3wlkd.s ⊢ φ → A ∈ V ∧ B ∈ V ∧ C ∈ V ∧ D ∈ V
4 3wlkd.n ⊢ φ → A ≠ B ∧ A ≠ C ∧ B ≠ C ∧ B ≠ D ∧ C ≠ D
5 3wlkd.e ⊢ φ → A B ⊆ I ⁡ J ∧ B C ⊆ I ⁡ K ∧ C D ⊆ I ⁡ L
6 3wlkd.v ⊢ V = Vtx ⁡ G
7 3wlkd.i ⊢ I = iEdg ⁡ G
8 3trld.n ⊢ φ → J ≠ K ∧ J ≠ L ∧ K ≠ L
9 3cycld.e ⊢ φ → A = D
10 1 2 3 4 5 6 7 8 3pthd ⊢ φ → F Paths ⁡ G P
11 1 fveq1i ⊢ P ⁡ 0 = ⟨“ ABCD ”⟩ ⁡ 0
12 s4fv0 ⊢ A ∈ V → ⟨“ ABCD ”⟩ ⁡ 0 = A
13 11 12 eqtrid ⊢ A ∈ V → P ⁡ 0 = A
14 13 ad3antrrr ⊢ A ∈ V ∧ B ∈ V ∧ C ∈ V ∧ D ∈ V ∧ A = D → P ⁡ 0 = A
15 simpr ⊢ A ∈ V ∧ B ∈ V ∧ C ∈ V ∧ D ∈ V ∧ A = D → A = D
16 2 fveq2i ⊢ F = ⟨“ JKL ”⟩
17 s3len ⊢ ⟨“ JKL ”⟩ = 3
18 16 17 eqtri ⊢ F = 3
19 1 18 fveq12i ⊢ P ⁡ F = ⟨“ ABCD ”⟩ ⁡ 3
20 s4fv3 ⊢ D ∈ V → ⟨“ ABCD ”⟩ ⁡ 3 = D
21 19 20 eqtr2id ⊢ D ∈ V → D = P ⁡ F
22 21 adantl ⊢ C ∈ V ∧ D ∈ V → D = P ⁡ F
23 22 ad2antlr ⊢ A ∈ V ∧ B ∈ V ∧ C ∈ V ∧ D ∈ V ∧ A = D → D = P ⁡ F
24 14 15 23 3eqtrd ⊢ A ∈ V ∧ B ∈ V ∧ C ∈ V ∧ D ∈ V ∧ A = D → P ⁡ 0 = P ⁡ F
25 3 9 24 syl2anc ⊢ φ → P ⁡ 0 = P ⁡ F
26 iscycl ⊢ F Cycles ⁡ G P ↔ F Paths ⁡ G P ∧ P ⁡ 0 = P ⁡ F
27 10 25 26 sylanbrc ⊢ φ → F Cycles ⁡ G P