Metamath Proof Explorer


Theorem ididg

Description: A set is identical to itself. (Contributed by NM, 28-May-2008) (Proof shortened by Andrew Salmon, 27-Aug-2011)

Ref Expression
Assertion ididg ⊢ A ∈ V → A I A

Proof

Step Hyp Ref Expression
1 eqid ⊢ A = A
2 ideqg ⊢ A ∈ V → A I A ↔ A = A
3 1 2 mpbiri ⊢ A ∈ V → A I A