Theorem ssabdv 43098

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 2898	. 2 ⊢ 𝐴 = {𝑥 ∣ 𝑥 ∈ 𝐴}
2		ssabdv.1	. . 3 ⊢ (𝜑 → (𝑥 ∈ 𝐴 → 𝜓))
3	2	ss2abdv 4016	. 2 ⊢ (𝜑 → {𝑥 ∣ 𝑥 ∈ 𝐴} ⊆ {𝑥 ∣ 𝜓})
4	1, 3	eqsstrid 3972	1 ⊢ (𝜑 → 𝐴 ⊆ {𝑥 ∣ 𝜓})

This proof depends on syntax axioms:   → wi 4   ∈ wcel 2145  {cab 2740   ⊆ wss 3902

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 1943  ax-6 2000  ax-7 2041  ax-8 2147  ax-9 2155  ax-ext 2734

This proof depends on definitions:  df-bi 210  df-an 402  df-tru 1573  df-ex 1813  df-sb 2100  df-clab 2741  df-cleq 2754  df-clel 2837  df-ss 3919

This theorem is used by: (None)