Metamath Proof Explorer


Theorem pthspthcyc

Description: A pair <. F , P >. represents a path if it represents either a simple path or a cycle. The exclusivity only holds for non-trivial paths ( F =/= (/) ), see cyclnspth . (Contributed by AV, 2-Oct-2025)

Ref Expression
Assertion pthspthcyc
|- ( F ( Paths ` G ) P <-> ( F ( SPaths ` G ) P \/ F ( Cycles ` G ) P ) )

Proof

Step Hyp Ref Expression
1 pthisspthorcycl
 |-  ( F ( Paths ` G ) P -> ( F ( SPaths ` G ) P \/ F ( Cycles ` G ) P ) )
2 spthispth
 |-  ( F ( SPaths ` G ) P -> F ( Paths ` G ) P )
3 cyclispth
 |-  ( F ( Cycles ` G ) P -> F ( Paths ` G ) P )
4 2 3 jaoi
 |-  ( ( F ( SPaths ` G ) P \/ F ( Cycles ` G ) P ) -> F ( Paths ` G ) P )
5 1 4 impbii
 |-  ( F ( Paths ` G ) P <-> ( F ( SPaths ` G ) P \/ F ( Cycles ` G ) P ) )