Theorem ssabdv

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 } )