Theorem ssabdv 41708

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 2866	. 2 ⊢ 𝐴 = {𝑥 ∣ 𝑥 ∈ 𝐴}
2		ssabdv.1	. . 3 ⊢ (𝜑 → (𝑥 ∈ 𝐴 → 𝜓))
3	2	ss2abdv 4058	. 2 ⊢ (𝜑 → {𝑥 ∣ 𝑥 ∈ 𝐴} ⊆ {𝑥 ∣ 𝜓})
4	1, 3	eqsstrid 4028	1 ⊢ (𝜑 → 𝐴 ⊆ {𝑥 ∣ 𝜓})

Syntax hints:   → wi 4   ∈ wcel 2099  {cab 2705   ⊆ wss 3947

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1790  ax-4 1804  ax-5 1906  ax-6 1964  ax-7 2004  ax-8 2101  ax-9 2109  ax-ext 2699

This theorem depends on definitions:  df-bi 206  df-an 396  df-tru 1537  df-ex 1775  df-sb 2061  df-clab 2706  df-cleq 2720  df-clel 2806  df-v 3473  df-in 3954  df-ss 3964

This theorem is referenced by: (None)