Theorem difin2 4306

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 3988	. . . . 5 ⊢ (𝐴 ⊆ 𝐶 → (𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐶))
2	1	pm4.71d 561	. . . 4 ⊢ (𝐴 ⊆ 𝐶 → (𝑥 ∈ 𝐴 ↔ (𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐶)))
3	2	anbi1d 631	. . 3 ⊢ (𝐴 ⊆ 𝐶 → ((𝑥 ∈ 𝐴 ∧ ¬ 𝑥 ∈ 𝐵) ↔ ((𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐶) ∧ ¬ 𝑥 ∈ 𝐵)))
4		eldif 3972	. . 3 ⊢ (𝑥 ∈ (𝐴 ∖ 𝐵) ↔ (𝑥 ∈ 𝐴 ∧ ¬ 𝑥 ∈ 𝐵))
5		ancom 460	. . . 4 ⊢ (((𝑥 ∈ 𝐶 ∧ ¬ 𝑥 ∈ 𝐵) ∧ 𝑥 ∈ 𝐴) ↔ (𝑥 ∈ 𝐴 ∧ (𝑥 ∈ 𝐶 ∧ ¬ 𝑥 ∈ 𝐵)))
6		elin 3978	. . . . 5 ⊢ (𝑥 ∈ ((𝐶 ∖ 𝐵) ∩ 𝐴) ↔ (𝑥 ∈ (𝐶 ∖ 𝐵) ∧ 𝑥 ∈ 𝐴))
7		eldif 3972	. . . . 5 ⊢ (𝑥 ∈ (𝐶 ∖ 𝐵) ↔ (𝑥 ∈ 𝐶 ∧ ¬ 𝑥 ∈ 𝐵))
8	6, 7	bianbi 627	. . . 4 ⊢ (𝑥 ∈ ((𝐶 ∖ 𝐵) ∩ 𝐴) ↔ ((𝑥 ∈ 𝐶 ∧ ¬ 𝑥 ∈ 𝐵) ∧ 𝑥 ∈ 𝐴))
9		anass 468	. . . 4 ⊢ (((𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐶) ∧ ¬ 𝑥 ∈ 𝐵) ↔ (𝑥 ∈ 𝐴 ∧ (𝑥 ∈ 𝐶 ∧ ¬ 𝑥 ∈ 𝐵)))
10	5, 8, 9	3bitr4i 303	. . 3 ⊢ (𝑥 ∈ ((𝐶 ∖ 𝐵) ∩ 𝐴) ↔ ((𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐶) ∧ ¬ 𝑥 ∈ 𝐵))
11	3, 4, 10	3bitr4g 314	. 2 ⊢ (𝐴 ⊆ 𝐶 → (𝑥 ∈ (𝐴 ∖ 𝐵) ↔ 𝑥 ∈ ((𝐶 ∖ 𝐵) ∩ 𝐴)))
12	11	eqrdv 2732	1 ⊢ (𝐴 ⊆ 𝐶 → (𝐴 ∖ 𝐵) = ((𝐶 ∖ 𝐵) ∩ 𝐴))

Syntax hints:  ¬ wn 3   → wi 4   ∧ wa 395   = wceq 1536   ∈ wcel 2105   ∖ cdif 3959   ∩ cin 3961   ⊆ wss 3962

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1791  ax-4 1805  ax-5 1907  ax-6 1964  ax-7 2004  ax-8 2107  ax-9 2115  ax-ext 2705

This theorem depends on definitions:  df-bi 207  df-an 396  df-tru 1539  df-ex 1776  df-sb 2062  df-clab 2712  df-cleq 2726  df-clel 2813  df-v 3479  df-dif 3965  df-in 3969  df-ss 3979

This theorem is referenced by:  gsumdifsnd  19993  issubdrg  20797  restcld  23195  limcnlp  25927  symgcom2  33086  difelsiga  34113  sigapildsyslem  34141  ldgenpisyslem1  34143  difelcarsg2  34294  ballotlemfp1  34472  asindmre  37689  caragendifcl  46469  gsumdifsndf  48024