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 ) 𝑃𝐹 ( Cycles ‘ 𝐺 ) 𝑃 ) )

Proof

Step Hyp Ref Expression
1 pthisspthorcycl ( 𝐹 ( Paths ‘ 𝐺 ) 𝑃 → ( 𝐹 ( SPaths ‘ 𝐺 ) 𝑃𝐹 ( Cycles ‘ 𝐺 ) 𝑃 ) )
2 spthispth ( 𝐹 ( SPaths ‘ 𝐺 ) 𝑃𝐹 ( Paths ‘ 𝐺 ) 𝑃 )
3 cyclispth ( 𝐹 ( Cycles ‘ 𝐺 ) 𝑃𝐹 ( Paths ‘ 𝐺 ) 𝑃 )
4 2 3 jaoi ( ( 𝐹 ( SPaths ‘ 𝐺 ) 𝑃𝐹 ( Cycles ‘ 𝐺 ) 𝑃 ) → 𝐹 ( Paths ‘ 𝐺 ) 𝑃 )
5 1 4 impbii ( 𝐹 ( Paths ‘ 𝐺 ) 𝑃 ↔ ( 𝐹 ( SPaths ‘ 𝐺 ) 𝑃𝐹 ( Cycles ‘ 𝐺 ) 𝑃 ) )