Theorem spthdifv

Description: The vertices of a simple path are distinct, so the vertex function is one-to-one. (Contributed by Alexander van der Vekens, 26-Jan-2018) (Revised by AV, 5-Jun-2021) (Proof shortened by AV, 30-Oct-2021)

	Ref	Expression
Assertion	spthdifv	⊢ ( 𝐹 ( SPaths ‘ 𝐺 ) 𝑃 → 𝑃 : ( 0 ... ( ♯ ‘ 𝐹 ) ) –1-1→ ( Vtx ‘ 𝐺 ) )

Proof

Step	Hyp	Ref	Expression
1		isspth	⊢ ( 𝐹 ( SPaths ‘ 𝐺 ) 𝑃 ↔ ( 𝐹 ( Trails ‘ 𝐺 ) 𝑃 ∧ Fun ◡ 𝑃 ) )
2		trliswlk	⊢ ( 𝐹 ( Trails ‘ 𝐺 ) 𝑃 → 𝐹 ( Walks ‘ 𝐺 ) 𝑃 )
3		eqid	⊢ ( Vtx ‘ 𝐺 ) = ( Vtx ‘ 𝐺 )
4	3	wlkp	⊢ ( 𝐹 ( Walks ‘ 𝐺 ) 𝑃 → 𝑃 : ( 0 ... ( ♯ ‘ 𝐹 ) ) ⟶ ( Vtx ‘ 𝐺 ) )
5		df-f1	⊢ ( 𝑃 : ( 0 ... ( ♯ ‘ 𝐹 ) ) –1-1→ ( Vtx ‘ 𝐺 ) ↔ ( 𝑃 : ( 0 ... ( ♯ ‘ 𝐹 ) ) ⟶ ( Vtx ‘ 𝐺 ) ∧ Fun ◡ 𝑃 ) )
6	5	simplbi2	⊢ ( 𝑃 : ( 0 ... ( ♯ ‘ 𝐹 ) ) ⟶ ( Vtx ‘ 𝐺 ) → ( Fun ◡ 𝑃 → 𝑃 : ( 0 ... ( ♯ ‘ 𝐹 ) ) –1-1→ ( Vtx ‘ 𝐺 ) ) )
7	2 4 6	3syl	⊢ ( 𝐹 ( Trails ‘ 𝐺 ) 𝑃 → ( Fun ◡ 𝑃 → 𝑃 : ( 0 ... ( ♯ ‘ 𝐹 ) ) –1-1→ ( Vtx ‘ 𝐺 ) ) )
8	7	imp	⊢ ( ( 𝐹 ( Trails ‘ 𝐺 ) 𝑃 ∧ Fun ◡ 𝑃 ) → 𝑃 : ( 0 ... ( ♯ ‘ 𝐹 ) ) –1-1→ ( Vtx ‘ 𝐺 ) )
9	1 8	sylbi	⊢ ( 𝐹 ( SPaths ‘ 𝐺 ) 𝑃 → 𝑃 : ( 0 ... ( ♯ ‘ 𝐹 ) ) –1-1→ ( Vtx ‘ 𝐺 ) )