Theorem rabss2OLD 4026

Description: Obsolete version of rabss2 4025 as of 1-Feb-2026. (Contributed by NM, 30-May-2006.) (Proof shortened by Andrew Salmon, 26-Jun-2011.) (Proof modification is discouraged.) (New usage is discouraged.)

Assertion

Ref	Expression
rabss2OLD	⊢ (𝐴 ⊆ 𝐵 → {𝑥 ∈ 𝐴 ∣ 𝜑} ⊆ {𝑥 ∈ 𝐵 ∣ 𝜑})

Distinct variable groups:   𝑥,𝐴   𝑥,𝐵

Allowed substitution hint:   𝜑(𝑥)

Proof of Theorem rabss2OLD

Step	Hyp	Ref	Expression
1		pm3.45 634	. . . 4 ⊢ ((𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵) → ((𝑥 ∈ 𝐴 ∧ 𝜑) → (𝑥 ∈ 𝐵 ∧ 𝜑)))
2	1	alimi 1844	. . 3 ⊢ (∀𝑥(𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵) → ∀𝑥((𝑥 ∈ 𝐴 ∧ 𝜑) → (𝑥 ∈ 𝐵 ∧ 𝜑)))
3		df-ss 3916	. . 3 ⊢ (𝐴 ⊆ 𝐵 ↔ ∀𝑥(𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵))
4		ss2ab 4009	. . 3 ⊢ ({𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜑)} ⊆ {𝑥 ∣ (𝑥 ∈ 𝐵 ∧ 𝜑)} ↔ ∀𝑥((𝑥 ∈ 𝐴 ∧ 𝜑) → (𝑥 ∈ 𝐵 ∧ 𝜑)))
5	2, 3, 4	3imtr4i 295	. 2 ⊢ (𝐴 ⊆ 𝐵 → {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜑)} ⊆ {𝑥 ∣ (𝑥 ∈ 𝐵 ∧ 𝜑)})
6		df-rab 3414	. 2 ⊢ {𝑥 ∈ 𝐴 ∣ 𝜑} = {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜑)}
7		df-rab 3414	. 2 ⊢ {𝑥 ∈ 𝐵 ∣ 𝜑} = {𝑥 ∣ (𝑥 ∈ 𝐵 ∧ 𝜑)}
8	5, 6, 7	3sstr4g 3984	1 ⊢ (𝐴 ⊆ 𝐵 → {𝑥 ∈ 𝐴 ∣ 𝜑} ⊆ {𝑥 ∈ 𝐵 ∣ 𝜑})

This proof depends on syntax axioms:   → wi 4   ∧ wa 401  ∀wal 1568   ∈ wcel 2145  {cab 2739  {crab 3413   ⊆ wss 3899

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-10 2178  ax-11 2194  ax-12 2213  ax-ext 2733

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-ex 1813  df-nf 1817  df-sb 2100  df-clab 2740  df-cleq 2753  df-clel 2836  df-nfc 2910  df-rab 3414  df-ss 3916

This theorem is used by: (None)