Theorem dfrab3ss 3534

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 3260	. . 3 ⊢ (A ⊆ B ↔ (A ∩ B) = A)
2		ineq1 3451	. . . 4 ⊢ ((A ∩ B) = A → ((A ∩ B) ∩ {x ∣ φ}) = (A ∩ {x ∣ φ}))
3	2	eqcomd 2358	. . 3 ⊢ ((A ∩ B) = A → (A ∩ {x ∣ φ}) = ((A ∩ B) ∩ {x ∣ φ}))
4	1, 3	sylbi 187	. 2 ⊢ (A ⊆ B → (A ∩ {x ∣ φ}) = ((A ∩ B) ∩ {x ∣ φ}))
5		dfrab3 3532	. 2 ⊢ {x ∈ A ∣ φ} = (A ∩ {x ∣ φ})
6		dfrab3 3532	. . . 4 ⊢ {x ∈ B ∣ φ} = (B ∩ {x ∣ φ})
7	6	ineq2i 3455	. . 3 ⊢ (A ∩ {x ∈ B ∣ φ}) = (A ∩ (B ∩ {x ∣ φ}))
8		inass 3466	. . 3 ⊢ ((A ∩ B) ∩ {x ∣ φ}) = (A ∩ (B ∩ {x ∣ φ}))
9	7, 8	eqtr4i 2376	. 2 ⊢ (A ∩ {x ∈ B ∣ φ}) = ((A ∩ B) ∩ {x ∣ φ})
10	4, 5, 9	3eqtr4g 2410	1 ⊢ (A ⊆ B → {x ∈ A ∣ φ} = (A ∩ {x ∈ B ∣ φ}))

This proof depends on syntax axioms:   → wi 4   = wceq 1642  {cab 2339  {crab 2619   ∩ cin 3209   ⊆ wss 3258

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1546  ax-5 1557  ax-17 1616  ax-9 1654  ax-8 1675  ax-6 1729  ax-7 1734  ax-11 1746  ax-12 1925  ax-ext 2334

This proof depends on definitions:  df-bi 177  df-or 359  df-an 360  df-nan 1288  df-tru 1319  df-ex 1542  df-nf 1545  df-sb 1649  df-clab 2340  df-cleq 2346  df-clel 2349  df-nfc 2479  df-rab 2624  df-v 2862  df-nin 3212  df-compl 3213  df-in 3214  df-ss 3260

This theorem is used by: (None)