Theorem rabssrabd 4038

Description: Subclass of a restricted class abstraction. (Contributed by AV, 4-Jun-2022.)

Hypotheses

Ref	Expression
rabssrabd.1	⊢ (𝜑 → 𝐴 ⊆ 𝐵)
rabssrabd.2	⊢ ((𝜑 ∧ 𝜓 ∧ 𝑥 ∈ 𝐴) → 𝜒)

Assertion

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

Distinct variable groups:   𝑥,𝐴   𝑥,𝐵   𝜑,𝑥

Allowed substitution hints:   𝜓(𝑥)   𝜒(𝑥)

Proof of Theorem rabssrabd

Step	Hyp	Ref	Expression
1		3anan32 1113	. . . . 5 ⊢ ((𝜑 ∧ 𝜓 ∧ 𝑥 ∈ 𝐴) ↔ ((𝜑 ∧ 𝑥 ∈ 𝐴) ∧ 𝜓))
2		rabssrabd.2	. . . . 5 ⊢ ((𝜑 ∧ 𝜓 ∧ 𝑥 ∈ 𝐴) → 𝜒)
3	1, 2	sylbir 238	. . . 4 ⊢ (((𝜑 ∧ 𝑥 ∈ 𝐴) ∧ 𝜓) → 𝜒)
4	3	ex 418	. . 3 ⊢ ((𝜑 ∧ 𝑥 ∈ 𝐴) → (𝜓 → 𝜒))
5	4	ss2rabdv 4030	. 2 ⊢ (𝜑 → {𝑥 ∈ 𝐴 ∣ 𝜓} ⊆ {𝑥 ∈ 𝐴 ∣ 𝜒})
6		rabssrabd.1	. . 3 ⊢ (𝜑 → 𝐴 ⊆ 𝐵)
7		rabss2 4032	. . 3 ⊢ (𝐴 ⊆ 𝐵 → {𝑥 ∈ 𝐴 ∣ 𝜒} ⊆ {𝑥 ∈ 𝐵 ∣ 𝜒})
8	6, 7	syl 18	. 2 ⊢ (𝜑 → {𝑥 ∈ 𝐴 ∣ 𝜒} ⊆ {𝑥 ∈ 𝐵 ∣ 𝜒})
9	5, 8	sstrd 3948	1 ⊢ (𝜑 → {𝑥 ∈ 𝐴 ∣ 𝜓} ⊆ {𝑥 ∈ 𝐵 ∣ 𝜒})

This proof depends on syntax axioms:   → wi 4   ∧ wa 401   ∧ w3a 1103   ∈ wcel 2146  {crab 3418   ⊆ wss 3906

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 2148  ax-9 2156  ax-ext 2737

This proof depends on definitions:  df-bi 210  df-an 402  df-3an 1105  df-ex 1813  df-sb 2100  df-clab 2744  df-cleq 2757  df-clel 2840  df-ral 3082  df-rab 3419  df-ss 3923

This theorem is used by:  suppfnss  8191  clwlknon2num  30792  numclwlk1lem2  30794