Metamath Proof Explorer


Theorem pg4cyclnex

Description: In the Petersen graph G(5,2), there is no cycle of length 4. (Contributed by AV, 22-Nov-2025)

Ref Expression
Assertion pg4cyclnex ¬ ∃ 𝑝 ∃ 𝑓 ( 𝑓 ( Cycles ‘ ( 5 gPetersenGr 2 ) ) 𝑝 ∧ ( ♯ ‘ 𝑓 ) = 4 )

Proof

Step Hyp Ref Expression
1 eqid ⊢ ( 5 gPetersenGr 2 ) = ( 5 gPetersenGr 2 )
2 1 pgn4cyclex ⊢ ( 𝑓 ( Cycles ‘ ( 5 gPetersenGr 2 ) ) 𝑝 → ( ♯ ‘ 𝑓 ) ≠ 4 )
3 2 imori ⊢ ( ¬ 𝑓 ( Cycles ‘ ( 5 gPetersenGr 2 ) ) 𝑝 ∨ ( ♯ ‘ 𝑓 ) ≠ 4 )
4 3 gen2 ⊢ ∀ 𝑝 ∀ 𝑓 ( ¬ 𝑓 ( Cycles ‘ ( 5 gPetersenGr 2 ) ) 𝑝 ∨ ( ♯ ‘ 𝑓 ) ≠ 4 )
5 2nexaln ⊢ ( ¬ ∃ 𝑝 ∃ 𝑓 ( 𝑓 ( Cycles ‘ ( 5 gPetersenGr 2 ) ) 𝑝 ∧ ( ♯ ‘ 𝑓 ) = 4 ) ↔ ∀ 𝑝 ∀ 𝑓 ¬ ( 𝑓 ( Cycles ‘ ( 5 gPetersenGr 2 ) ) 𝑝 ∧ ( ♯ ‘ 𝑓 ) = 4 ) )
6 ianor ⊢ ( ¬ ( 𝑓 ( Cycles ‘ ( 5 gPetersenGr 2 ) ) 𝑝 ∧ ( ♯ ‘ 𝑓 ) = 4 ) ↔ ( ¬ 𝑓 ( Cycles ‘ ( 5 gPetersenGr 2 ) ) 𝑝 ∨ ¬ ( ♯ ‘ 𝑓 ) = 4 ) )
7 df-ne ⊢ ( ( ♯ ‘ 𝑓 ) ≠ 4 ↔ ¬ ( ♯ ‘ 𝑓 ) = 4 )
8 7 bicomi ⊢ ( ¬ ( ♯ ‘ 𝑓 ) = 4 ↔ ( ♯ ‘ 𝑓 ) ≠ 4 )
9 8 orbi2i ⊢ ( ( ¬ 𝑓 ( Cycles ‘ ( 5 gPetersenGr 2 ) ) 𝑝 ∨ ¬ ( ♯ ‘ 𝑓 ) = 4 ) ↔ ( ¬ 𝑓 ( Cycles ‘ ( 5 gPetersenGr 2 ) ) 𝑝 ∨ ( ♯ ‘ 𝑓 ) ≠ 4 ) )
10 6 9 bitri ⊢ ( ¬ ( 𝑓 ( Cycles ‘ ( 5 gPetersenGr 2 ) ) 𝑝 ∧ ( ♯ ‘ 𝑓 ) = 4 ) ↔ ( ¬ 𝑓 ( Cycles ‘ ( 5 gPetersenGr 2 ) ) 𝑝 ∨ ( ♯ ‘ 𝑓 ) ≠ 4 ) )
11 10 2albii ⊢ ( ∀ 𝑝 ∀ 𝑓 ¬ ( 𝑓 ( Cycles ‘ ( 5 gPetersenGr 2 ) ) 𝑝 ∧ ( ♯ ‘ 𝑓 ) = 4 ) ↔ ∀ 𝑝 ∀ 𝑓 ( ¬ 𝑓 ( Cycles ‘ ( 5 gPetersenGr 2 ) ) 𝑝 ∨ ( ♯ ‘ 𝑓 ) ≠ 4 ) )
12 5 11 bitri ⊢ ( ¬ ∃ 𝑝 ∃ 𝑓 ( 𝑓 ( Cycles ‘ ( 5 gPetersenGr 2 ) ) 𝑝 ∧ ( ♯ ‘ 𝑓 ) = 4 ) ↔ ∀ 𝑝 ∀ 𝑓 ( ¬ 𝑓 ( Cycles ‘ ( 5 gPetersenGr 2 ) ) 𝑝 ∨ ( ♯ ‘ 𝑓 ) ≠ 4 ) )
13 4 12 mpbir ⊢ ¬ ∃ 𝑝 ∃ 𝑓 ( 𝑓 ( Cycles ‘ ( 5 gPetersenGr 2 ) ) 𝑝 ∧ ( ♯ ‘ 𝑓 ) = 4 )