Metamath Proof Explorer


Theorem wksfval

Description: The set of walks (in an undirected graph). (Contributed by AV, 30-Dec-2020)

Ref Expression
Hypotheses wksfval.v ⊢ V = Vtx ⁡ G
wksfval.i ⊢ I = iEdg ⁡ G
Assertion wksfval ⊢ G ∈ W → Walks ⁡ G = f p | f ∈ Word dom ⁡ I ∧ p : 0 … f ⟶ V ∧ ∀ k ∈ 0 ..^ f if- p ⁡ k = p ⁡ k + 1 I ⁡ f ⁡ k = p ⁡ k p ⁡ k p ⁡ k + 1 ⊆ I ⁡ f ⁡ k

Proof

Step Hyp Ref Expression
1 wksfval.v ⊢ V = Vtx ⁡ G
2 wksfval.i ⊢ I = iEdg ⁡ G
3 df-wlks ⊢ Walks = g ∈ V ⟼ f p | f ∈ Word dom ⁡ iEdg ⁡ g ∧ p : 0 … f ⟶ Vtx ⁡ g ∧ ∀ k ∈ 0 ..^ f if- p ⁡ k = p ⁡ k + 1 iEdg ⁡ g ⁡ f ⁡ k = p ⁡ k p ⁡ k p ⁡ k + 1 ⊆ iEdg ⁡ g ⁡ f ⁡ k
4 fveq2 ⊢ g = G → iEdg ⁡ g = iEdg ⁡ G
5 4 2 eqtr4di ⊢ g = G → iEdg ⁡ g = I
6 5 dmeqd ⊢ g = G → dom ⁡ iEdg ⁡ g = dom ⁡ I
7 wrdeq ⊢ dom ⁡ iEdg ⁡ g = dom ⁡ I → Word dom ⁡ iEdg ⁡ g = Word dom ⁡ I
8 6 7 syl ⊢ g = G → Word dom ⁡ iEdg ⁡ g = Word dom ⁡ I
9 8 eleq2d ⊢ g = G → f ∈ Word dom ⁡ iEdg ⁡ g ↔ f ∈ Word dom ⁡ I
10 fveq2 ⊢ g = G → Vtx ⁡ g = Vtx ⁡ G
11 10 1 eqtr4di ⊢ g = G → Vtx ⁡ g = V
12 11 feq3d ⊢ g = G → p : 0 … f ⟶ Vtx ⁡ g ↔ p : 0 … f ⟶ V
13 5 fveq1d ⊢ g = G → iEdg ⁡ g ⁡ f ⁡ k = I ⁡ f ⁡ k
14 13 eqeq1d ⊢ g = G → iEdg ⁡ g ⁡ f ⁡ k = p ⁡ k ↔ I ⁡ f ⁡ k = p ⁡ k
15 13 sseq2d ⊢ g = G → p ⁡ k p ⁡ k + 1 ⊆ iEdg ⁡ g ⁡ f ⁡ k ↔ p ⁡ k p ⁡ k + 1 ⊆ I ⁡ f ⁡ k
16 14 15 ifpbi23d ⊢ g = G → if- p ⁡ k = p ⁡ k + 1 iEdg ⁡ g ⁡ f ⁡ k = p ⁡ k p ⁡ k p ⁡ k + 1 ⊆ iEdg ⁡ g ⁡ f ⁡ k ↔ if- p ⁡ k = p ⁡ k + 1 I ⁡ f ⁡ k = p ⁡ k p ⁡ k p ⁡ k + 1 ⊆ I ⁡ f ⁡ k
17 16 ralbidv ⊢ g = G → ∀ k ∈ 0 ..^ f if- p ⁡ k = p ⁡ k + 1 iEdg ⁡ g ⁡ f ⁡ k = p ⁡ k p ⁡ k p ⁡ k + 1 ⊆ iEdg ⁡ g ⁡ f ⁡ k ↔ ∀ k ∈ 0 ..^ f if- p ⁡ k = p ⁡ k + 1 I ⁡ f ⁡ k = p ⁡ k p ⁡ k p ⁡ k + 1 ⊆ I ⁡ f ⁡ k
18 9 12 17 3anbi123d ⊢ g = G → f ∈ Word dom ⁡ iEdg ⁡ g ∧ p : 0 … f ⟶ Vtx ⁡ g ∧ ∀ k ∈ 0 ..^ f if- p ⁡ k = p ⁡ k + 1 iEdg ⁡ g ⁡ f ⁡ k = p ⁡ k p ⁡ k p ⁡ k + 1 ⊆ iEdg ⁡ g ⁡ f ⁡ k ↔ f ∈ Word dom ⁡ I ∧ p : 0 … f ⟶ V ∧ ∀ k ∈ 0 ..^ f if- p ⁡ k = p ⁡ k + 1 I ⁡ f ⁡ k = p ⁡ k p ⁡ k p ⁡ k + 1 ⊆ I ⁡ f ⁡ k
19 18 opabbidv ⊢ g = G → f p | f ∈ Word dom ⁡ iEdg ⁡ g ∧ p : 0 … f ⟶ Vtx ⁡ g ∧ ∀ k ∈ 0 ..^ f if- p ⁡ k = p ⁡ k + 1 iEdg ⁡ g ⁡ f ⁡ k = p ⁡ k p ⁡ k p ⁡ k + 1 ⊆ iEdg ⁡ g ⁡ f ⁡ k = f p | f ∈ Word dom ⁡ I ∧ p : 0 … f ⟶ V ∧ ∀ k ∈ 0 ..^ f if- p ⁡ k = p ⁡ k + 1 I ⁡ f ⁡ k = p ⁡ k p ⁡ k p ⁡ k + 1 ⊆ I ⁡ f ⁡ k
20 elex ⊢ G ∈ W → G ∈ V
21 3anass ⊢ f ∈ Word dom ⁡ I ∧ p : 0 … f ⟶ V ∧ ∀ k ∈ 0 ..^ f if- p ⁡ k = p ⁡ k + 1 I ⁡ f ⁡ k = p ⁡ k p ⁡ k p ⁡ k + 1 ⊆ I ⁡ f ⁡ k ↔ f ∈ Word dom ⁡ I ∧ p : 0 … f ⟶ V ∧ ∀ k ∈ 0 ..^ f if- p ⁡ k = p ⁡ k + 1 I ⁡ f ⁡ k = p ⁡ k p ⁡ k p ⁡ k + 1 ⊆ I ⁡ f ⁡ k
22 21 opabbii ⊢ f p | f ∈ Word dom ⁡ I ∧ p : 0 … f ⟶ V ∧ ∀ k ∈ 0 ..^ f if- p ⁡ k = p ⁡ k + 1 I ⁡ f ⁡ k = p ⁡ k p ⁡ k p ⁡ k + 1 ⊆ I ⁡ f ⁡ k = f p | f ∈ Word dom ⁡ I ∧ p : 0 … f ⟶ V ∧ ∀ k ∈ 0 ..^ f if- p ⁡ k = p ⁡ k + 1 I ⁡ f ⁡ k = p ⁡ k p ⁡ k p ⁡ k + 1 ⊆ I ⁡ f ⁡ k
23 2 fvexi ⊢ I ∈ V
24 23 dmex ⊢ dom ⁡ I ∈ V
25 wrdexg ⊢ dom ⁡ I ∈ V → Word dom ⁡ I ∈ V
26 24 25 mp1i ⊢ G ∈ W → Word dom ⁡ I ∈ V
27 ovex ⊢ 0 … f ∈ V
28 1 fvexi ⊢ V ∈ V
29 28 a1i ⊢ G ∈ W ∧ f ∈ Word dom ⁡ I → V ∈ V
30 mapex ⊢ 0 … f ∈ V ∧ V ∈ V → p | p : 0 … f ⟶ V ∈ V
31 27 29 30 sylancr ⊢ G ∈ W ∧ f ∈ Word dom ⁡ I → p | p : 0 … f ⟶ V ∈ V
32 simpl ⊢ p : 0 … f ⟶ V ∧ ∀ k ∈ 0 ..^ f if- p ⁡ k = p ⁡ k + 1 I ⁡ f ⁡ k = p ⁡ k p ⁡ k p ⁡ k + 1 ⊆ I ⁡ f ⁡ k → p : 0 … f ⟶ V
33 32 ss2abi ⊢ p | p : 0 … f ⟶ V ∧ ∀ k ∈ 0 ..^ f if- p ⁡ k = p ⁡ k + 1 I ⁡ f ⁡ k = p ⁡ k p ⁡ k p ⁡ k + 1 ⊆ I ⁡ f ⁡ k ⊆ p | p : 0 … f ⟶ V
34 33 a1i ⊢ G ∈ W ∧ f ∈ Word dom ⁡ I → p | p : 0 … f ⟶ V ∧ ∀ k ∈ 0 ..^ f if- p ⁡ k = p ⁡ k + 1 I ⁡ f ⁡ k = p ⁡ k p ⁡ k p ⁡ k + 1 ⊆ I ⁡ f ⁡ k ⊆ p | p : 0 … f ⟶ V
35 31 34 ssexd ⊢ G ∈ W ∧ f ∈ Word dom ⁡ I → p | p : 0 … f ⟶ V ∧ ∀ k ∈ 0 ..^ f if- p ⁡ k = p ⁡ k + 1 I ⁡ f ⁡ k = p ⁡ k p ⁡ k p ⁡ k + 1 ⊆ I ⁡ f ⁡ k ∈ V
36 26 35 opabex3d ⊢ G ∈ W → f p | f ∈ Word dom ⁡ I ∧ p : 0 … f ⟶ V ∧ ∀ k ∈ 0 ..^ f if- p ⁡ k = p ⁡ k + 1 I ⁡ f ⁡ k = p ⁡ k p ⁡ k p ⁡ k + 1 ⊆ I ⁡ f ⁡ k ∈ V
37 22 36 eqeltrid ⊢ G ∈ W → f p | f ∈ Word dom ⁡ I ∧ p : 0 … f ⟶ V ∧ ∀ k ∈ 0 ..^ f if- p ⁡ k = p ⁡ k + 1 I ⁡ f ⁡ k = p ⁡ k p ⁡ k p ⁡ k + 1 ⊆ I ⁡ f ⁡ k ∈ V
38 3 19 20 37 fvmptd3 ⊢ G ∈ W → Walks ⁡ G = f p | f ∈ Word dom ⁡ I ∧ p : 0 … f ⟶ V ∧ ∀ k ∈ 0 ..^ f if- p ⁡ k = p ⁡ k + 1 I ⁡ f ⁡ k = p ⁡ k p ⁡ k p ⁡ k + 1 ⊆ I ⁡ f ⁡ k