Theorem difin2 4291

Description: Represent a class 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 3975	. . . . 5 ⊢ (𝐴 ⊆ 𝐶 → (𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐶))
2	1	pm4.71d 561	. . . 4 ⊢ (𝐴 ⊆ 𝐶 → (𝑥 ∈ 𝐴 ↔ (𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐶)))
3	2	anbi1d 629	. . 3 ⊢ (𝐴 ⊆ 𝐶 → ((𝑥 ∈ 𝐴 ∧ ¬ 𝑥 ∈ 𝐵) ↔ ((𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐶) ∧ ¬ 𝑥 ∈ 𝐵)))
4		eldif 3958	. . 3 ⊢ (𝑥 ∈ (𝐴 ∖ 𝐵) ↔ (𝑥 ∈ 𝐴 ∧ ¬ 𝑥 ∈ 𝐵))
5		elin 3964	. . . 4 ⊢ (𝑥 ∈ ((𝐶 ∖ 𝐵) ∩ 𝐴) ↔ (𝑥 ∈ (𝐶 ∖ 𝐵) ∧ 𝑥 ∈ 𝐴))
6		eldif 3958	. . . . 5 ⊢ (𝑥 ∈ (𝐶 ∖ 𝐵) ↔ (𝑥 ∈ 𝐶 ∧ ¬ 𝑥 ∈ 𝐵))
7	6	anbi1i 623	. . . 4 ⊢ ((𝑥 ∈ (𝐶 ∖ 𝐵) ∧ 𝑥 ∈ 𝐴) ↔ ((𝑥 ∈ 𝐶 ∧ ¬ 𝑥 ∈ 𝐵) ∧ 𝑥 ∈ 𝐴))
8		ancom 460	. . . . 5 ⊢ (((𝑥 ∈ 𝐶 ∧ ¬ 𝑥 ∈ 𝐵) ∧ 𝑥 ∈ 𝐴) ↔ (𝑥 ∈ 𝐴 ∧ (𝑥 ∈ 𝐶 ∧ ¬ 𝑥 ∈ 𝐵)))
9		anass 468	. . . . 5 ⊢ (((𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐶) ∧ ¬ 𝑥 ∈ 𝐵) ↔ (𝑥 ∈ 𝐴 ∧ (𝑥 ∈ 𝐶 ∧ ¬ 𝑥 ∈ 𝐵)))
10	8, 9	bitr4i 278	. . . 4 ⊢ (((𝑥 ∈ 𝐶 ∧ ¬ 𝑥 ∈ 𝐵) ∧ 𝑥 ∈ 𝐴) ↔ ((𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐶) ∧ ¬ 𝑥 ∈ 𝐵))
11	5, 7, 10	3bitri 297	. . 3 ⊢ (𝑥 ∈ ((𝐶 ∖ 𝐵) ∩ 𝐴) ↔ ((𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐶) ∧ ¬ 𝑥 ∈ 𝐵))
12	3, 4, 11	3bitr4g 314	. 2 ⊢ (𝐴 ⊆ 𝐶 → (𝑥 ∈ (𝐴 ∖ 𝐵) ↔ 𝑥 ∈ ((𝐶 ∖ 𝐵) ∩ 𝐴)))
13	12	eqrdv 2729	1 ⊢ (𝐴 ⊆ 𝐶 → (𝐴 ∖ 𝐵) = ((𝐶 ∖ 𝐵) ∩ 𝐴))

Syntax hints:  ¬ wn 3   → wi 4   ∧ wa 395   = wceq 1540   ∈ wcel 2105   ∖ cdif 3945   ∩ cin 3947   ⊆ wss 3948

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1796  ax-4 1810  ax-5 1912  ax-6 1970  ax-7 2010  ax-8 2107  ax-9 2115  ax-ext 2702

This theorem depends on definitions:  df-bi 206  df-an 396  df-tru 1543  df-ex 1781  df-sb 2067  df-clab 2709  df-cleq 2723  df-clel 2809  df-v 3475  df-dif 3951  df-in 3955  df-ss 3965

This theorem is referenced by:  gsumdifsnd  19877  issubdrg  20627  restcld  22996  limcnlp  25727  symgcom2  32682  difelsiga  33596  sigapildsyslem  33624  ldgenpisyslem1  33626  difelcarsg2  33777  ballotlemfp1  33955  asindmre  37037  caragendifcl  45691  gsumdifsndf  47020