Theorem dfrab3ss 4273

Description: Restricted class abstraction with a common superset. (Contributed by Stefan O'Rear, 12-Sep-2015.) (Proof shortened by Mario Carneiro, 8-Nov-2015.)

Assertion

Ref	Expression
dfrab3ss	⊢ (𝐴 ⊆ 𝐵 → {𝑥 ∈ 𝐴 ∣ 𝜑} = (𝐴 ∩ {𝑥 ∈ 𝐵 ∣ 𝜑}))

Distinct variable groups:   𝑥,𝐴   𝑥,𝐵

Allowed substitution hint:   𝜑(𝑥)

Proof of Theorem dfrab3ss

Step	Hyp	Ref	Expression
1		dfss2 3920	. . 3 ⊢ (𝐴 ⊆ 𝐵 ↔ (𝐴 ∩ 𝐵) = 𝐴)
2		ineq1 4163	. . . 4 ⊢ ((𝐴 ∩ 𝐵) = 𝐴 → ((𝐴 ∩ 𝐵) ∩ {𝑥 ∣ 𝜑}) = (𝐴 ∩ {𝑥 ∣ 𝜑}))
3	2	eqcomd 2767	. . 3 ⊢ ((𝐴 ∩ 𝐵) = 𝐴 → (𝐴 ∩ {𝑥 ∣ 𝜑}) = ((𝐴 ∩ 𝐵) ∩ {𝑥 ∣ 𝜑}))
4	1, 3	sylbi 219	. 2 ⊢ (𝐴 ⊆ 𝐵 → (𝐴 ∩ {𝑥 ∣ 𝜑}) = ((𝐴 ∩ 𝐵) ∩ {𝑥 ∣ 𝜑}))
5		dfrab3 4269	. 2 ⊢ {𝑥 ∈ 𝐴 ∣ 𝜑} = (𝐴 ∩ {𝑥 ∣ 𝜑})
6		dfrab3 4269	. . . 4 ⊢ {𝑥 ∈ 𝐵 ∣ 𝜑} = (𝐵 ∩ {𝑥 ∣ 𝜑})
7	6	ineq2i 4167	. . 3 ⊢ (𝐴 ∩ {𝑥 ∈ 𝐵 ∣ 𝜑}) = (𝐴 ∩ (𝐵 ∩ {𝑥 ∣ 𝜑}))
8		inass 4177	. . 3 ⊢ ((𝐴 ∩ 𝐵) ∩ {𝑥 ∣ 𝜑}) = (𝐴 ∩ (𝐵 ∩ {𝑥 ∣ 𝜑}))
9	7, 8	eqtr4i 2787	. 2 ⊢ (𝐴 ∩ {𝑥 ∈ 𝐵 ∣ 𝜑}) = ((𝐴 ∩ 𝐵) ∩ {𝑥 ∣ 𝜑})
10	4, 5, 9	3eqtr4g 2821	1 ⊢ (𝐴 ⊆ 𝐵 → {𝑥 ∈ 𝐴 ∣ 𝜑} = (𝐴 ∩ {𝑥 ∈ 𝐵 ∣ 𝜑}))

Syntax hints:   → wi 4   = wceq 1559  {cab 2739  {crab 3413   ∩ cin 3901   ⊆ wss 3902

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1814  ax-4 1828  ax-5 1929  ax-6 1986  ax-7 2027  ax-8 2143  ax-9 2151  ax-ext 2733

This theorem depends on definitions:  df-bi 209  df-an 400  df-3an 1099  df-tru 1562  df-ex 1799  df-sb 2090  df-clab 2740  df-cleq 2753  df-clel 2836  df-rab 3414  df-v 3455  df-in 3909  df-ss 3919

This theorem is referenced by:  mbfposadd  38127  proot1hash  43733