Metamath Proof Explorer


Theorem epse

Description: The membership relation is set-like on any class. (This is the origin of the term "set-like": a set-like relation "acts like" the membership relation of sets and their elements.) (Contributed by Mario Carneiro, 22-Jun-2015)

Ref Expression
Assertion epse E Se 𝐴

Proof

Step Hyp Ref Expression
1 epel ⊢ ( 𝑦 E 𝑥 ↔ 𝑦 ∈ 𝑥 )
2 1 bicomi ⊢ ( 𝑦 ∈ 𝑥 ↔ 𝑦 E 𝑥 )
3 2 eqabi ⊢ 𝑥 = { 𝑦 ∣ 𝑦 E 𝑥 }
4 vex ⊢ 𝑥 ∈ V
5 3 4 eqeltrri ⊢ { 𝑦 ∣ 𝑦 E 𝑥 } ∈ V
6 rabssab ⊢ { 𝑦 ∈ 𝐴 ∣ 𝑦 E 𝑥 } ⊆ { 𝑦 ∣ 𝑦 E 𝑥 }
7 5 6 ssexi ⊢ { 𝑦 ∈ 𝐴 ∣ 𝑦 E 𝑥 } ∈ V
8 7 rgenw ⊢ ∀ 𝑥 ∈ 𝐴 { 𝑦 ∈ 𝐴 ∣ 𝑦 E 𝑥 } ∈ V
9 df-se ⊢ ( E Se 𝐴 ↔ ∀ 𝑥 ∈ 𝐴 { 𝑦 ∈ 𝐴 ∣ 𝑦 E 𝑥 } ∈ V )
10 8 9 mpbir ⊢ E Se 𝐴