Metamath Proof Explorer


Theorem eucrct2eupth1

Description: Removing one edge ( I( FN ) ) from a nonempty graph G with an Eulerian circuit <. F , P >. results in a graph S with an Eulerian path <. H , Q >. . This is the special case of eucrct2eupth (with J = ( N - 1 ) ) where the last segment/edge of the circuit is removed. (Contributed by AV, 11-Mar-2021) Hypothesis revised using the prefix operation. (Revised by AV, 30-Nov-2022)

Ref Expression
Hypotheses eucrct2eupth1.v ⊢ V = Vtx ⁡ G
eucrct2eupth1.i ⊢ I = iEdg ⁡ G
eucrct2eupth1.d ⊢ φ → F EulerPaths ⁡ G P
eucrct2eupth1.c ⊢ φ → F Circuits ⁡ G P
eucrct2eupth1.s ⊢ Vtx ⁡ S = V
eucrct2eupth1.g ⊢ φ → 0 < F
eucrct2eupth1.n ⊢ φ → N = F − 1
eucrct2eupth1.e ⊢ φ → iEdg ⁡ S = I ↾ F 0 ..^ N
eucrct2eupth1.h ⊢ H = F prefix N
eucrct2eupth1.q ⊢ Q = P ↾ 0 … N
Assertion eucrct2eupth1 ⊢ φ → H EulerPaths ⁡ S Q

Proof

Step Hyp Ref Expression
1 eucrct2eupth1.v ⊢ V = Vtx ⁡ G
2 eucrct2eupth1.i ⊢ I = iEdg ⁡ G
3 eucrct2eupth1.d ⊢ φ → F EulerPaths ⁡ G P
4 eucrct2eupth1.c ⊢ φ → F Circuits ⁡ G P
5 eucrct2eupth1.s ⊢ Vtx ⁡ S = V
6 eucrct2eupth1.g ⊢ φ → 0 < F
7 eucrct2eupth1.n ⊢ φ → N = F − 1
8 eucrct2eupth1.e ⊢ φ → iEdg ⁡ S = I ↾ F 0 ..^ N
9 eucrct2eupth1.h ⊢ H = F prefix N
10 eucrct2eupth1.q ⊢ Q = P ↾ 0 … N
11 eupthiswlk ⊢ F EulerPaths ⁡ G P → F Walks ⁡ G P
12 wlkcl ⊢ F Walks ⁡ G P → F ∈ ℕ 0
13 nn0z ⊢ F ∈ ℕ 0 → F ∈ ℤ
14 13 anim1i ⊢ F ∈ ℕ 0 ∧ 0 < F → F ∈ ℤ ∧ 0 < F
15 elnnz ⊢ F ∈ ℕ ↔ F ∈ ℤ ∧ 0 < F
16 14 15 sylibr ⊢ F ∈ ℕ 0 ∧ 0 < F → F ∈ ℕ
17 16 ex ⊢ F ∈ ℕ 0 → 0 < F → F ∈ ℕ
18 3 11 12 17 4syl ⊢ φ → 0 < F → F ∈ ℕ
19 6 18 mpd ⊢ φ → F ∈ ℕ
20 fzo0end ⊢ F ∈ ℕ → F − 1 ∈ 0 ..^ F
21 19 20 syl ⊢ φ → F − 1 ∈ 0 ..^ F
22 7 21 eqeltrd ⊢ φ → N ∈ 0 ..^ F
23 1 2 3 22 8 9 10 5 eupthres ⊢ φ → H EulerPaths ⁡ S Q