Metamath Proof Explorer


Theorem elsucg

Description: Membership in a successor. Exercise 5 of TakeutiZaring p. 17. (Contributed by NM, 15-Sep-1995)

Ref Expression
Assertion elsucg ( 𝐴𝑉 → ( 𝐴 ∈ suc 𝐵 ↔ ( 𝐴𝐵𝐴 = 𝐵 ) ) )

Proof

Step Hyp Ref Expression
1 df-suc suc 𝐵 = ( 𝐵 ∪ { 𝐵 } )
2 1 eleq2i ( 𝐴 ∈ suc 𝐵𝐴 ∈ ( 𝐵 ∪ { 𝐵 } ) )
3 elun ( 𝐴 ∈ ( 𝐵 ∪ { 𝐵 } ) ↔ ( 𝐴𝐵𝐴 ∈ { 𝐵 } ) )
4 2 3 bitri ( 𝐴 ∈ suc 𝐵 ↔ ( 𝐴𝐵𝐴 ∈ { 𝐵 } ) )
5 elsng ( 𝐴𝑉 → ( 𝐴 ∈ { 𝐵 } ↔ 𝐴 = 𝐵 ) )
6 5 orbi2d ( 𝐴𝑉 → ( ( 𝐴𝐵𝐴 ∈ { 𝐵 } ) ↔ ( 𝐴𝐵𝐴 = 𝐵 ) ) )
7 4 6 bitrid ( 𝐴𝑉 → ( 𝐴 ∈ suc 𝐵 ↔ ( 𝐴𝐵𝐴 = 𝐵 ) ) )