Theorem abssdv

Description: Deduction of abstraction subclass from implication. (Contributed by NM, 20-Jan-2006) (Proof shortened by SN, 22-Dec-2024)

	Ref	Expression
Hypothesis	abssdv.1	⊢ ( 𝜑 → ( 𝜓 → 𝑥 ∈ 𝐴 ) )
Assertion	abssdv	⊢ ( 𝜑 → { 𝑥 ∣ 𝜓 } ⊆ 𝐴 )

Proof

Step	Hyp	Ref	Expression
1		abssdv.1	⊢ ( 𝜑 → ( 𝜓 → 𝑥 ∈ 𝐴 ) )
2	1	ss2abdv	⊢ ( 𝜑 → { 𝑥 ∣ 𝜓 } ⊆ { 𝑥 ∣ 𝑥 ∈ 𝐴 } )
3		abid1	⊢ 𝐴 = { 𝑥 ∣ 𝑥 ∈ 𝐴 }
4	2 3	sseqtrrdi	⊢ ( 𝜑 → { 𝑥 ∣ 𝜓 } ⊆ 𝐴 )