Theorem ssabdv 43467

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 2900	. 2 ⊢ 𝐴 = {𝑥 ∣ 𝑥 ∈ 𝐴}
2		ssabdv.1	. . 3 ⊢ (𝜑 → (𝑥 ∈ 𝐴 → 𝜓))
3	2	ss2abdv 3993	. 2 ⊢ (𝜑 → {𝑥 ∣ 𝑥 ∈ 𝐴} ⊆ {𝑥 ∣ 𝜓})
4	1, 3	eqsstrid 3951	1 ⊢ (𝜑 → 𝐴 ⊆ {𝑥 ∣ 𝜓})

This proof depends on syntax axioms:   → wi 4   ∈ wcel 2147  {cab 2741   ⊆ wss 3900

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1945  ax-6 2002  ax-7 2043  ax-8 2149  ax-9 2157  ax-ext 2735

This proof depends on definitions:  df-bi 210  df-an 402  df-tru 1573  df-ex 1813  df-sb 2102  df-clab 2742  df-cleq 2755  df-clel 2839  df-ss 3901

This theorem is used by: (None)