Metamath Proof Explorer


Theorem cycliscrct

Description: A cycle is a circuit. (Contributed by Alexander van der Vekens, 30-Oct-2017) (Revised by AV, 31-Jan-2021) (Proof shortened by AV, 30-Oct-2021)

Ref Expression
Assertion cycliscrct ( 𝐹 ( Cycles ‘ 𝐺 ) 𝑃 → 𝐹 ( Circuits ‘ 𝐺 ) 𝑃 )

Proof

Step Hyp Ref Expression
1 pthistrl ⊢ ( 𝐹 ( Paths ‘ 𝐺 ) 𝑃 → 𝐹 ( Trails ‘ 𝐺 ) 𝑃 )
2 1 anim1i ⊢ ( ( 𝐹 ( Paths ‘ 𝐺 ) 𝑃 ∧ ( 𝑃 ‘ 0 ) = ( 𝑃 ‘ ( ♯ ‘ 𝐹 ) ) ) → ( 𝐹 ( Trails ‘ 𝐺 ) 𝑃 ∧ ( 𝑃 ‘ 0 ) = ( 𝑃 ‘ ( ♯ ‘ 𝐹 ) ) ) )
3 iscycl ⊢ ( 𝐹 ( Cycles ‘ 𝐺 ) 𝑃 ↔ ( 𝐹 ( Paths ‘ 𝐺 ) 𝑃 ∧ ( 𝑃 ‘ 0 ) = ( 𝑃 ‘ ( ♯ ‘ 𝐹 ) ) ) )
4 iscrct ⊢ ( 𝐹 ( Circuits ‘ 𝐺 ) 𝑃 ↔ ( 𝐹 ( Trails ‘ 𝐺 ) 𝑃 ∧ ( 𝑃 ‘ 0 ) = ( 𝑃 ‘ ( ♯ ‘ 𝐹 ) ) ) )
5 2 3 4 3imtr4i ⊢ ( 𝐹 ( Cycles ‘ 𝐺 ) 𝑃 → 𝐹 ( Circuits ‘ 𝐺 ) 𝑃 )