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 )