Metamath Proof Explorer


Theorem gpg5order

Description: The order of a generalized Petersen graph G(5,K), which is either the Petersen graph G(5,2) or the 5-prism G(5,1), is 10. (Contributed by AV, 26-Aug-2025)

Ref Expression
Assertion gpg5order ( 𝐾 ∈ ( 1 ... 2 ) → ( ♯ ‘ ( Vtx ‘ ( 5 gPetersenGr 𝐾 ) ) ) = 1 0 )

Proof

Step Hyp Ref Expression
1 5nn ⊢ 5 ∈ ℕ
2 2z ⊢ 2 ∈ ℤ
3 fzval3 ⊢ ( 2 ∈ ℤ → ( 1 ... 2 ) = ( 1 ..^ ( 2 + 1 ) ) )
4 2 3 ax-mp ⊢ ( 1 ... 2 ) = ( 1 ..^ ( 2 + 1 ) )
5 2p1e3 ⊢ ( 2 + 1 ) = 3
6 ceil5half3 ⊢ ( ⌈ ‘ ( 5 / 2 ) ) = 3
7 5 6 eqtr4i ⊢ ( 2 + 1 ) = ( ⌈ ‘ ( 5 / 2 ) )
8 7 oveq2i ⊢ ( 1 ..^ ( 2 + 1 ) ) = ( 1 ..^ ( ⌈ ‘ ( 5 / 2 ) ) )
9 4 8 eqtri ⊢ ( 1 ... 2 ) = ( 1 ..^ ( ⌈ ‘ ( 5 / 2 ) ) )
10 9 eleq2i ⊢ ( 𝐾 ∈ ( 1 ... 2 ) ↔ 𝐾 ∈ ( 1 ..^ ( ⌈ ‘ ( 5 / 2 ) ) ) )
11 10 biimpi ⊢ ( 𝐾 ∈ ( 1 ... 2 ) → 𝐾 ∈ ( 1 ..^ ( ⌈ ‘ ( 5 / 2 ) ) ) )
12 eqid ⊢ ( 1 ..^ ( ⌈ ‘ ( 5 / 2 ) ) ) = ( 1 ..^ ( ⌈ ‘ ( 5 / 2 ) ) )
13 12 gpgorder ⊢ ( ( 5 ∈ ℕ ∧ 𝐾 ∈ ( 1 ..^ ( ⌈ ‘ ( 5 / 2 ) ) ) ) → ( ♯ ‘ ( Vtx ‘ ( 5 gPetersenGr 𝐾 ) ) ) = ( 2 · 5 ) )
14 1 11 13 sylancr ⊢ ( 𝐾 ∈ ( 1 ... 2 ) → ( ♯ ‘ ( Vtx ‘ ( 5 gPetersenGr 𝐾 ) ) ) = ( 2 · 5 ) )
15 5cn ⊢ 5 ∈ ℂ
16 2cn ⊢ 2 ∈ ℂ
17 5t2e10 ⊢ ( 5 · 2 ) = 1 0
18 15 16 17 mulcomli ⊢ ( 2 · 5 ) = 1 0
19 14 18 eqtrdi ⊢ ( 𝐾 ∈ ( 1 ... 2 ) → ( ♯ ‘ ( Vtx ‘ ( 5 gPetersenGr 𝐾 ) ) ) = 1 0 )