Theorem ss2rabd 4020

Description: Subclass of a restricted class abstraction (deduction form). Saves ax-10 2169, ax-11 2185, ax-12 2206 over using ss2rab 4017 and sylibr 236. (Contributed by SN, 4-Feb-2026.)

Hypothesis

Ref	Expression
ss2rabd.1	⊢ (𝜑 → ∀𝑥 ∈ 𝐴 (𝜓 → 𝜒))

Assertion

Ref	Expression
ss2rabd	⊢ (𝜑 → {𝑥 ∈ 𝐴 ∣ 𝜓} ⊆ {𝑥 ∈ 𝐴 ∣ 𝜒})

Proof of Theorem ss2rabd

Step	Hyp	Ref	Expression
1		ss2rabd.1	. . . 4 ⊢ (𝜑 → ∀𝑥 ∈ 𝐴 (𝜓 → 𝜒))
2		df-ral 3071	. . . . 5 ⊢ (∀𝑥 ∈ 𝐴 (𝜓 → 𝜒) ↔ ∀𝑥(𝑥 ∈ 𝐴 → (𝜓 → 𝜒)))
3		imdistan 574	. . . . . 6 ⊢ ((𝑥 ∈ 𝐴 → (𝜓 → 𝜒)) ↔ ((𝑥 ∈ 𝐴 ∧ 𝜓) → (𝑥 ∈ 𝐴 ∧ 𝜒)))
4	3	albii 1833	. . . . 5 ⊢ (∀𝑥(𝑥 ∈ 𝐴 → (𝜓 → 𝜒)) ↔ ∀𝑥((𝑥 ∈ 𝐴 ∧ 𝜓) → (𝑥 ∈ 𝐴 ∧ 𝜒)))
5	2, 4	bitri 277	. . . 4 ⊢ (∀𝑥 ∈ 𝐴 (𝜓 → 𝜒) ↔ ∀𝑥((𝑥 ∈ 𝐴 ∧ 𝜓) → (𝑥 ∈ 𝐴 ∧ 𝜒)))
6	1, 5	sylib 220	. . 3 ⊢ (𝜑 → ∀𝑥((𝑥 ∈ 𝐴 ∧ 𝜓) → (𝑥 ∈ 𝐴 ∧ 𝜒)))
7		ss2abim 4008	. . 3 ⊢ (∀𝑥((𝑥 ∈ 𝐴 ∧ 𝜓) → (𝑥 ∈ 𝐴 ∧ 𝜒)) → {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜓)} ⊆ {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜒)})
8	6, 7	syl 17	. 2 ⊢ (𝜑 → {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜓)} ⊆ {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜒)})
9		df-rab 3409	. 2 ⊢ {𝑥 ∈ 𝐴 ∣ 𝜓} = {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜓)}
10		df-rab 3409	. 2 ⊢ {𝑥 ∈ 𝐴 ∣ 𝜒} = {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜒)}
11	8, 9, 10	3sstr4g 3984	1 ⊢ (𝜑 → {𝑥 ∈ 𝐴 ∣ 𝜓} ⊆ {𝑥 ∈ 𝐴 ∣ 𝜒})

Syntax hints:   → wi 4   ∧ wa 398  ∀wal 1552   ∈ wcel 2136  {cab 2734  ∀wral 3070  {crab 3408   ⊆ wss 3899

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1809  ax-4 1823  ax-5 1924  ax-6 1981  ax-7 2022  ax-9 2146  ax-ext 2728

This theorem depends on definitions:  df-bi 209  df-an 399  df-ex 1794  df-sb 2085  df-clab 2735  df-cleq 2748  df-ral 3071  df-rab 3409  df-ss 3916

This theorem is referenced by:  ss2rabdv  4023  ondomon  10510  xrlimcnp  27003  ss2rabdf  45676  pimiooltgt  47232