Theorem dfrab3ss 3405

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		df-ss 3134	. . 3 ⊢ (𝐴 ⊆ 𝐵 ↔ (𝐴 ∩ 𝐵) = 𝐴)
2		ineq1 3321	. . . 4 ⊢ ((𝐴 ∩ 𝐵) = 𝐴 → ((𝐴 ∩ 𝐵) ∩ {𝑥 ∣ 𝜑}) = (𝐴 ∩ {𝑥 ∣ 𝜑}))
3	2	eqcomd 2176	. . 3 ⊢ ((𝐴 ∩ 𝐵) = 𝐴 → (𝐴 ∩ {𝑥 ∣ 𝜑}) = ((𝐴 ∩ 𝐵) ∩ {𝑥 ∣ 𝜑}))
4	1, 3	sylbi 120	. 2 ⊢ (𝐴 ⊆ 𝐵 → (𝐴 ∩ {𝑥 ∣ 𝜑}) = ((𝐴 ∩ 𝐵) ∩ {𝑥 ∣ 𝜑}))
5		dfrab3 3403	. 2 ⊢ {𝑥 ∈ 𝐴 ∣ 𝜑} = (𝐴 ∩ {𝑥 ∣ 𝜑})
6		dfrab3 3403	. . . 4 ⊢ {𝑥 ∈ 𝐵 ∣ 𝜑} = (𝐵 ∩ {𝑥 ∣ 𝜑})
7	6	ineq2i 3325	. . 3 ⊢ (𝐴 ∩ {𝑥 ∈ 𝐵 ∣ 𝜑}) = (𝐴 ∩ (𝐵 ∩ {𝑥 ∣ 𝜑}))
8		inass 3337	. . 3 ⊢ ((𝐴 ∩ 𝐵) ∩ {𝑥 ∣ 𝜑}) = (𝐴 ∩ (𝐵 ∩ {𝑥 ∣ 𝜑}))
9	7, 8	eqtr4i 2194	. 2 ⊢ (𝐴 ∩ {𝑥 ∈ 𝐵 ∣ 𝜑}) = ((𝐴 ∩ 𝐵) ∩ {𝑥 ∣ 𝜑})
10	4, 5, 9	3eqtr4g 2228	1 ⊢ (𝐴 ⊆ 𝐵 → {𝑥 ∈ 𝐴 ∣ 𝜑} = (𝐴 ∩ {𝑥 ∈ 𝐵 ∣ 𝜑}))

Syntax hints:   → wi 4   = wceq 1348  {cab 2156  {crab 2452   ∩ cin 3120   ⊆ wss 3121

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-io 704  ax-5 1440  ax-7 1441  ax-gen 1442  ax-ie1 1486  ax-ie2 1487  ax-8 1497  ax-10 1498  ax-11 1499  ax-i12 1500  ax-bndl 1502  ax-4 1503  ax-17 1519  ax-i9 1523  ax-ial 1527  ax-i5r 1528  ax-ext 2152

This theorem depends on definitions:  df-bi 116  df-tru 1351  df-nf 1454  df-sb 1756  df-clab 2157  df-cleq 2163  df-clel 2166  df-nfc 2301  df-rab 2457  df-v 2732  df-in 3127  df-ss 3134

This theorem is referenced by: (None)