Theorem difin2 3516

Description: Represent a set difference as an intersection with a larger difference. (Contributed by Jeff Madsen, 2-Sep-2009.)

Assertion

Ref	Expression
difin2	⊢ (A ⊆ C → (A ∖ B) = ((C ∖ B) ∩ A))

Proof of Theorem difin2

Dummy variable x is distinct from all other variables.

Step	Hyp	Ref	Expression
1		ssel 3267	. . . . 5 ⊢ (A ⊆ C → (x ∈ A → x ∈ C))
2	1	pm4.71d 615	. . . 4 ⊢ (A ⊆ C → (x ∈ A ↔ (x ∈ A ∧ x ∈ C)))
3	2	anbi1d 685	. . 3 ⊢ (A ⊆ C → ((x ∈ A ∧ ¬ x ∈ B) ↔ ((x ∈ A ∧ x ∈ C) ∧ ¬ x ∈ B)))
4		eldif 3221	. . 3 ⊢ (x ∈ (A ∖ B) ↔ (x ∈ A ∧ ¬ x ∈ B))
5		elin 3219	. . . 4 ⊢ (x ∈ ((C ∖ B) ∩ A) ↔ (x ∈ (C ∖ B) ∧ x ∈ A))
6		eldif 3221	. . . . 5 ⊢ (x ∈ (C ∖ B) ↔ (x ∈ C ∧ ¬ x ∈ B))
7	6	anbi1i 676	. . . 4 ⊢ ((x ∈ (C ∖ B) ∧ x ∈ A) ↔ ((x ∈ C ∧ ¬ x ∈ B) ∧ x ∈ A))
8		ancom 437	. . . . 5 ⊢ (((x ∈ C ∧ ¬ x ∈ B) ∧ x ∈ A) ↔ (x ∈ A ∧ (x ∈ C ∧ ¬ x ∈ B)))
9		anass 630	. . . . 5 ⊢ (((x ∈ A ∧ x ∈ C) ∧ ¬ x ∈ B) ↔ (x ∈ A ∧ (x ∈ C ∧ ¬ x ∈ B)))
10	8, 9	bitr4i 243	. . . 4 ⊢ (((x ∈ C ∧ ¬ x ∈ B) ∧ x ∈ A) ↔ ((x ∈ A ∧ x ∈ C) ∧ ¬ x ∈ B))
11	5, 7, 10	3bitri 262	. . 3 ⊢ (x ∈ ((C ∖ B) ∩ A) ↔ ((x ∈ A ∧ x ∈ C) ∧ ¬ x ∈ B))
12	3, 4, 11	3bitr4g 279	. 2 ⊢ (A ⊆ C → (x ∈ (A ∖ B) ↔ x ∈ ((C ∖ B) ∩ A)))
13	12	eqrdv 2351	1 ⊢ (A ⊆ C → (A ∖ B) = ((C ∖ B) ∩ A))

Syntax hints:  ¬ wn 3   → wi 4   ∧ wa 358   = wceq 1642   ∈ wcel 1710   ∖ cdif 3206   ∩ cin 3208   ⊆ wss 3257

This theorem was proved from 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 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-v 2861  df-nin 3211  df-compl 3212  df-in 3213  df-dif 3215  df-ss 3259

This theorem is referenced by: (None)