Metamath Proof Explorer


Theorem isspth

Description: Conditions for a pair of classes/functions to be a simple path (in an undirected graph). (Contributed by Alexander van der Vekens, 21-Oct-2017) (Revised by AV, 9-Jan-2021) (Revised by AV, 29-Oct-2021)

Ref Expression
Assertion isspth ( 𝐹 ( SPaths ‘ 𝐺 ) 𝑃 ↔ ( 𝐹 ( Trails ‘ 𝐺 ) 𝑃 ∧ Fun 𝑃 ) )

Proof

Step Hyp Ref Expression
1 spthsfval ( SPaths ‘ 𝐺 ) = { ⟨ 𝑓 , 𝑝 ⟩ ∣ ( 𝑓 ( Trails ‘ 𝐺 ) 𝑝 ∧ Fun 𝑝 ) }
2 cnveq ( 𝑝 = 𝑃 𝑝 = 𝑃 )
3 2 funeqd ( 𝑝 = 𝑃 → ( Fun 𝑝 ↔ Fun 𝑃 ) )
4 3 adantl ( ( 𝑓 = 𝐹𝑝 = 𝑃 ) → ( Fun 𝑝 ↔ Fun 𝑃 ) )
5 reltrls Rel ( Trails ‘ 𝐺 )
6 1 4 5 brfvopabrbr ( 𝐹 ( SPaths ‘ 𝐺 ) 𝑃 ↔ ( 𝐹 ( Trails ‘ 𝐺 ) 𝑃 ∧ Fun 𝑃 ) )