Theorem dfrab3ss 3209

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 2925	. . 3 ⊢ (A ⊆ B ↔ (A ∩ B) = A)
2		ineq1 3125	. . . 4 ⊢ ((A ∩ B) = A → ((A ∩ B) ∩ {x ∣ φ}) = (A ∩ {x ∣ φ}))
3	2	eqcomd 2042	. . 3 ⊢ ((A ∩ B) = A → (A ∩ {x ∣ φ}) = ((A ∩ B) ∩ {x ∣ φ}))
4	1, 3	sylbi 114	. 2 ⊢ (A ⊆ B → (A ∩ {x ∣ φ}) = ((A ∩ B) ∩ {x ∣ φ}))
5		dfrab3 3207	. 2 ⊢ {x ∈ A ∣ φ} = (A ∩ {x ∣ φ})
6		dfrab3 3207	. . . 4 ⊢ {x ∈ B ∣ φ} = (B ∩ {x ∣ φ})
7	6	ineq2i 3129	. . 3 ⊢ (A ∩ {x ∈ B ∣ φ}) = (A ∩ (B ∩ {x ∣ φ}))
8		inass 3141	. . 3 ⊢ ((A ∩ B) ∩ {x ∣ φ}) = (A ∩ (B ∩ {x ∣ φ}))
9	7, 8	eqtr4i 2060	. 2 ⊢ (A ∩ {x ∈ B ∣ φ}) = ((A ∩ B) ∩ {x ∣ φ})
10	4, 5, 9	3eqtr4g 2094	1 ⊢ (A ⊆ B → {x ∈ A ∣ φ} = (A ∩ {x ∈ B ∣ φ}))

Syntax hints:   → wi 4   = wceq 1242  {cab 2023  {crab 2304   ∩ cin 2910   ⊆ wss 2911

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-io 629  ax-5 1333  ax-7 1334  ax-gen 1335  ax-ie1 1379  ax-ie2 1380  ax-8 1392  ax-10 1393  ax-11 1394  ax-i12 1395  ax-bnd 1396  ax-4 1397  ax-17 1416  ax-i9 1420  ax-ial 1424  ax-i5r 1425  ax-ext 2019

This theorem depends on definitions:  df-bi 110  df-tru 1245  df-nf 1347  df-sb 1643  df-clab 2024  df-cleq 2030  df-clel 2033  df-nfc 2164  df-rab 2309  df-v 2553  df-in 2918  df-ss 2925

This theorem is referenced by: (None)