Theorem ss2rabd 4000

Description: Subclass of a restricted class abstraction (deduction form). Saves ax-10 2180, ax-11 2197, ax-12 2216 over using ss2rab 3997 and sylibr 237. (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 3081	. . . . 5 ⊢ (∀𝑥 ∈ 𝐴 (𝜓 → 𝜒) ↔ ∀𝑥(𝑥 ∈ 𝐴 → (𝜓 → 𝜒)))
3		imdistan 578	. . . . . 6 ⊢ ((𝑥 ∈ 𝐴 → (𝜓 → 𝜒)) ↔ ((𝑥 ∈ 𝐴 ∧ 𝜓) → (𝑥 ∈ 𝐴 ∧ 𝜒)))
4	3	albii 1852	. . . . 5 ⊢ (∀𝑥(𝑥 ∈ 𝐴 → (𝜓 → 𝜒)) ↔ ∀𝑥((𝑥 ∈ 𝐴 ∧ 𝜓) → (𝑥 ∈ 𝐴 ∧ 𝜒)))
5	2, 4	bitri 278	. . . 4 ⊢ (∀𝑥 ∈ 𝐴 (𝜓 → 𝜒) ↔ ∀𝑥((𝑥 ∈ 𝐴 ∧ 𝜓) → (𝑥 ∈ 𝐴 ∧ 𝜒)))
6	1, 5	sylib 221	. . 3 ⊢ (𝜑 → ∀𝑥((𝑥 ∈ 𝐴 ∧ 𝜓) → (𝑥 ∈ 𝐴 ∧ 𝜒)))
7		ss2abim 3988	. . 3 ⊢ (∀𝑥((𝑥 ∈ 𝐴 ∧ 𝜓) → (𝑥 ∈ 𝐴 ∧ 𝜒)) → {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜓)} ⊆ {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜒)})
8	6, 7	syl 18	. 2 ⊢ (𝜑 → {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜓)} ⊆ {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜒)})
9		df-rab 3417	. 2 ⊢ {𝑥 ∈ 𝐴 ∣ 𝜓} = {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜓)}
10		df-rab 3417	. 2 ⊢ {𝑥 ∈ 𝐴 ∣ 𝜒} = {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜒)}
11	8, 9, 10	3sstr4g 3966	1 ⊢ (𝜑 → {𝑥 ∈ 𝐴 ∣ 𝜓} ⊆ {𝑥 ∈ 𝐴 ∣ 𝜒})

This proof depends on syntax axioms:   → wi 4   ∧ wa 401  ∀wal 1568   ∈ wcel 2147  {cab 2741  ∀wral 3080  {crab 3416   ⊆ 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-9 2157  ax-ext 2735

This proof depends on definitions:  df-bi 210  df-an 402  df-ex 1813  df-sb 2102  df-clab 2742  df-cleq 2755  df-ral 3081  df-rab 3417  df-ss 3901

This theorem is used by:  ss2rabdv  4003  ondomon  10674  xrlimcnp  27326  ss2rabdf  46370  pimiooltgt  47925