Theorem ssabdv 43367

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 2896	. 2 ⊢ 𝐴 = {𝑥 ∣ 𝑥 ∈ 𝐴}
2		ssabdv.1	. . 3 ⊢ (𝜑 → (𝑥 ∈ 𝐴 → 𝜓))
3	2	ss2abdv 3990	. 2 ⊢ (𝜑 → {𝑥 ∣ 𝑥 ∈ 𝐴} ⊆ {𝑥 ∣ 𝜓})
4	1, 3	eqsstrid 3946	1 ⊢ (𝜑 → 𝐴 ⊆ {𝑥 ∣ 𝜓})

This proof depends on syntax axioms:   → wi 4   ∈ wcel 2144  {cab 2738   ⊆ wss 3895

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1942  ax-6 1999  ax-7 2040  ax-8 2146  ax-9 2154  ax-ext 2732

This proof depends on definitions:  df-bi 210  df-an 402  df-tru 1572  df-ex 1812  df-sb 2099  df-clab 2739  df-cleq 2752  df-clel 2835  df-ss 3896

This theorem is used by: (None)