Metamath Proof Explorer
Description: Class substitution into a membership relation. (Contributed by NM, 17-Nov-2006) (Proof shortened by Andrew Salmon, 29-Jun-2011)
|
|
Ref |
Expression |
|
Assertion |
sbcel2gv |
⊢ ( 𝐵 ∈ 𝑉 → ( [ 𝐵 / 𝑥 ] 𝐴 ∈ 𝑥 ↔ 𝐴 ∈ 𝐵 ) ) |
Proof
Step |
Hyp |
Ref |
Expression |
1 |
|
eleq2 |
⊢ ( 𝑥 = 𝑦 → ( 𝐴 ∈ 𝑥 ↔ 𝐴 ∈ 𝑦 ) ) |
2 |
|
eleq2 |
⊢ ( 𝑦 = 𝐵 → ( 𝐴 ∈ 𝑦 ↔ 𝐴 ∈ 𝐵 ) ) |
3 |
1 2
|
sbcie2g |
⊢ ( 𝐵 ∈ 𝑉 → ( [ 𝐵 / 𝑥 ] 𝐴 ∈ 𝑥 ↔ 𝐴 ∈ 𝐵 ) ) |