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Theorem dfrab3ss 3533

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 3259	. . 3 ⊢ (A ⊆ B ↔ (A ∩ B) = A)
2		ineq1 3450	. . . 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 3531	. 2 ⊢ {x ∈ A ∣ φ} = (A ∩ {x ∣ φ})
6		dfrab3 3531	. . . 4 ⊢ {x ∈ B ∣ φ} = (B ∩ {x ∣ φ})
7	6	ineq2i 3454	. . 3 ⊢ (A ∩ {x ∈ B ∣ φ}) = (A ∩ (B ∩ {x ∣ φ}))
8		inass 3465	. . 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}))

Colors of variables: wff setvar class

Syntax hints: → wi 4 = wceq 1642 {cab 2339 {crab 2618 ∩ cin 3208 ⊆ wss 3257

This theorem was proved from axioms: ax-1 5 ax-2 6 ax-3 7 ax-mp 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 theorem 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 2478 df-rab 2623 df-v 2861 df-nin 3211 df-compl 3212 df-in 3213 df-ss 3259

This theorem is referenced by: (None)

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