Theorem ssabdv 42213

Description: Deduction of abstraction subclass from implication. (Contributed by SN, 22-Dec-2024.)

Hypothesis

Ref	Expression
ssabdv.1	⊢ (𝜑 → (𝑥 ∈ 𝐴 → 𝜓))

Assertion

Ref	Expression
ssabdv	⊢ (𝜑 → 𝐴 ⊆ {𝑥 ∣ 𝜓})

Distinct variable groups:   𝜑,𝑥   𝑥,𝐴

Allowed substitution hint:   𝜓(𝑥)

Proof of Theorem ssabdv

Step	Hyp	Ref	Expression
1		abid1 2881	. 2 ⊢ 𝐴 = {𝑥 ∣ 𝑥 ∈ 𝐴}
2		ssabdv.1	. . 3 ⊢ (𝜑 → (𝑥 ∈ 𝐴 → 𝜓))
3	2	ss2abdv 4089	. 2 ⊢ (𝜑 → {𝑥 ∣ 𝑥 ∈ 𝐴} ⊆ {𝑥 ∣ 𝜓})
4	1, 3	eqsstrid 4057	1 ⊢ (𝜑 → 𝐴 ⊆ {𝑥 ∣ 𝜓})

Syntax hints:   → wi 4   ∈ wcel 2108  {cab 2717   ⊆ wss 3976

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1793  ax-4 1807  ax-5 1909  ax-6 1967  ax-7 2007  ax-8 2110  ax-9 2118  ax-ext 2711

This theorem depends on definitions:  df-bi 207  df-an 396  df-tru 1540  df-ex 1778  df-sb 2065  df-clab 2718  df-cleq 2732  df-clel 2819  df-ss 3993

This theorem is referenced by: (None)