Metamath Proof Explorer


Theorem ss2ab

Description: Class abstractions in a subclass relationship. (Contributed by NM, 3-Jul-1994)

Ref Expression
Assertion ss2ab ( { 𝑥𝜑 } ⊆ { 𝑥𝜓 } ↔ ∀ 𝑥 ( 𝜑𝜓 ) )

Proof

Step Hyp Ref Expression
1 nfab1 𝑥 { 𝑥𝜑 }
2 nfab1 𝑥 { 𝑥𝜓 }
3 1 2 dfss2f ( { 𝑥𝜑 } ⊆ { 𝑥𝜓 } ↔ ∀ 𝑥 ( 𝑥 ∈ { 𝑥𝜑 } → 𝑥 ∈ { 𝑥𝜓 } ) )
4 abid ( 𝑥 ∈ { 𝑥𝜑 } ↔ 𝜑 )
5 abid ( 𝑥 ∈ { 𝑥𝜓 } ↔ 𝜓 )
6 4 5 imbi12i ( ( 𝑥 ∈ { 𝑥𝜑 } → 𝑥 ∈ { 𝑥𝜓 } ) ↔ ( 𝜑𝜓 ) )
7 6 albii ( ∀ 𝑥 ( 𝑥 ∈ { 𝑥𝜑 } → 𝑥 ∈ { 𝑥𝜓 } ) ↔ ∀ 𝑥 ( 𝜑𝜓 ) )
8 3 7 bitri ( { 𝑥𝜑 } ⊆ { 𝑥𝜓 } ↔ ∀ 𝑥 ( 𝜑𝜓 ) )