Metamath Proof Explorer
Description: Deduction of abstraction subclass from implication. (Contributed by SN, 22-Dec-2024)
|
|
Ref |
Expression |
|
Hypothesis |
ssabdv.1 |
⊢ ( 𝜑 → ( 𝑥 ∈ 𝐴 → 𝜓 ) ) |
|
Assertion |
ssabdv |
⊢ ( 𝜑 → 𝐴 ⊆ { 𝑥 ∣ 𝜓 } ) |
Proof
Step |
Hyp |
Ref |
Expression |
1 |
|
ssabdv.1 |
⊢ ( 𝜑 → ( 𝑥 ∈ 𝐴 → 𝜓 ) ) |
2 |
|
abid1 |
⊢ 𝐴 = { 𝑥 ∣ 𝑥 ∈ 𝐴 } |
3 |
1
|
ss2abdv |
⊢ ( 𝜑 → { 𝑥 ∣ 𝑥 ∈ 𝐴 } ⊆ { 𝑥 ∣ 𝜓 } ) |
4 |
2 3
|
eqsstrid |
⊢ ( 𝜑 → 𝐴 ⊆ { 𝑥 ∣ 𝜓 } ) |