Theorem dfrab3ss 4342

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 3994	. . 3 ⊢ (𝐴 ⊆ 𝐵 ↔ (𝐴 ∩ 𝐵) = 𝐴)
2		ineq1 4234	. . . 4 ⊢ ((𝐴 ∩ 𝐵) = 𝐴 → ((𝐴 ∩ 𝐵) ∩ {𝑥 ∣ 𝜑}) = (𝐴 ∩ {𝑥 ∣ 𝜑}))
3	2	eqcomd 2746	. . 3 ⊢ ((𝐴 ∩ 𝐵) = 𝐴 → (𝐴 ∩ {𝑥 ∣ 𝜑}) = ((𝐴 ∩ 𝐵) ∩ {𝑥 ∣ 𝜑}))
4	1, 3	sylbi 217	. 2 ⊢ (𝐴 ⊆ 𝐵 → (𝐴 ∩ {𝑥 ∣ 𝜑}) = ((𝐴 ∩ 𝐵) ∩ {𝑥 ∣ 𝜑}))
5		dfrab3 4338	. 2 ⊢ {𝑥 ∈ 𝐴 ∣ 𝜑} = (𝐴 ∩ {𝑥 ∣ 𝜑})
6		dfrab3 4338	. . . 4 ⊢ {𝑥 ∈ 𝐵 ∣ 𝜑} = (𝐵 ∩ {𝑥 ∣ 𝜑})
7	6	ineq2i 4238	. . 3 ⊢ (𝐴 ∩ {𝑥 ∈ 𝐵 ∣ 𝜑}) = (𝐴 ∩ (𝐵 ∩ {𝑥 ∣ 𝜑}))
8		inass 4249	. . 3 ⊢ ((𝐴 ∩ 𝐵) ∩ {𝑥 ∣ 𝜑}) = (𝐴 ∩ (𝐵 ∩ {𝑥 ∣ 𝜑}))
9	7, 8	eqtr4i 2771	. 2 ⊢ (𝐴 ∩ {𝑥 ∈ 𝐵 ∣ 𝜑}) = ((𝐴 ∩ 𝐵) ∩ {𝑥 ∣ 𝜑})
10	4, 5, 9	3eqtr4g 2805	1 ⊢ (𝐴 ⊆ 𝐵 → {𝑥 ∈ 𝐴 ∣ 𝜑} = (𝐴 ∩ {𝑥 ∈ 𝐵 ∣ 𝜑}))

Syntax hints:   → wi 4   = wceq 1537  {cab 2717  {crab 3443   ∩ cin 3975   ⊆ wss 3976

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1793  ax-4 1807  ax-5 1909  ax-6 1967  ax-7 2007  ax-8 2110  ax-9 2118  ax-ext 2711

This theorem depends on definitions:  df-bi 207  df-an 396  df-3an 1089  df-tru 1540  df-ex 1778  df-sb 2065  df-clab 2718  df-cleq 2732  df-clel 2819  df-rab 3444  df-v 3490  df-in 3983  df-ss 3993

This theorem is referenced by:  mbfposadd  37627  proot1hash  43156