Metamath Proof Explorer


Theorem sucidg

Description: Part of Proposition 7.23 of TakeutiZaring p. 41 (generalized). (Contributed by NM, 25-Mar-1995) (Proof shortened by Scott Fenton, 20-Feb-2012)

Ref Expression
Assertion sucidg ( 𝐴𝑉𝐴 ∈ suc 𝐴 )

Proof

Step Hyp Ref Expression
1 eqid 𝐴 = 𝐴
2 1 olci ( 𝐴𝐴𝐴 = 𝐴 )
3 elsucg ( 𝐴𝑉 → ( 𝐴 ∈ suc 𝐴 ↔ ( 𝐴𝐴𝐴 = 𝐴 ) ) )
4 2 3 mpbiri ( 𝐴𝑉𝐴 ∈ suc 𝐴 )