Metamath Proof Explorer


Theorem elimasn

Description: Membership in an image of a singleton. (Contributed by NM, 15-Mar-2004) (Proof shortened by Andrew Salmon, 27-Aug-2011) (Proof shortened by BJ, 16-Oct-2024) TODO: replace existing usages by usages of elimasn1 , remove, and relabel elimasn1 to "elimasn".

Ref Expression
Hypotheses elimasn.1 𝐵 ∈ V
elimasn.2 𝐶 ∈ V
Assertion elimasn ( 𝐶 ∈ ( 𝐴 “ { 𝐵 } ) ↔ ⟨ 𝐵 , 𝐶 ⟩ ∈ 𝐴 )

Proof

Step Hyp Ref Expression
1 elimasn.1 𝐵 ∈ V
2 elimasn.2 𝐶 ∈ V
3 elimasng ( ( 𝐵 ∈ V ∧ 𝐶 ∈ V ) → ( 𝐶 ∈ ( 𝐴 “ { 𝐵 } ) ↔ ⟨ 𝐵 , 𝐶 ⟩ ∈ 𝐴 ) )
4 1 2 3 mp2an ( 𝐶 ∈ ( 𝐴 “ { 𝐵 } ) ↔ ⟨ 𝐵 , 𝐶 ⟩ ∈ 𝐴 )