Metamath Proof Explorer


Theorem sucpre

Description: suc is a right-inverse of pre on Suc . This theorem states the partial inverse relation in the direction we most often need. (Contributed by Peter Mazsa, 27-Jan-2026)

Ref Expression
Assertion sucpre ( 𝑁 ∈ Suc → suc pre 𝑁 = 𝑁 )

Proof

Step Hyp Ref Expression
1 presucmap ( 𝑁 ∈ ran SucMap → pre 𝑁 SucMap 𝑁 )
2 preex pre 𝑁 ∈ V
3 brsucmap ( ( pre 𝑁 ∈ V ∧ 𝑁 ∈ ran SucMap ) → ( pre 𝑁 SucMap 𝑁 ↔ suc pre 𝑁 = 𝑁 ) )
4 2 3 mpan ( 𝑁 ∈ ran SucMap → ( pre 𝑁 SucMap 𝑁 ↔ suc pre 𝑁 = 𝑁 ) )
5 1 4 mpbid ( 𝑁 ∈ ran SucMap → suc pre 𝑁 = 𝑁 )
6 df-succl Suc = ran SucMap
7 5 6 eleq2s ( 𝑁 ∈ Suc → suc pre 𝑁 = 𝑁 )