Theorem difin2 4247

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 3925	. . . . 5 ⊢ (𝐴 ⊆ 𝐶 → (𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐶))
2	1	pm4.71d 571	. . . 4 ⊢ (𝐴 ⊆ 𝐶 → (𝑥 ∈ 𝐴 ↔ (𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐶)))
3	2	anbi1d 643	. . 3 ⊢ (𝐴 ⊆ 𝐶 → ((𝑥 ∈ 𝐴 ∧ ¬ 𝑥 ∈ 𝐵) ↔ ((𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐶) ∧ ¬ 𝑥 ∈ 𝐵)))
4		eldif 3909	. . 3 ⊢ (𝑥 ∈ (𝐴 ∖ 𝐵) ↔ (𝑥 ∈ 𝐴 ∧ ¬ 𝑥 ∈ 𝐵))
5		ancom 466	. . . 4 ⊢ (((𝑥 ∈ 𝐶 ∧ ¬ 𝑥 ∈ 𝐵) ∧ 𝑥 ∈ 𝐴) ↔ (𝑥 ∈ 𝐴 ∧ (𝑥 ∈ 𝐶 ∧ ¬ 𝑥 ∈ 𝐵)))
6		elin 3915	. . . . 5 ⊢ (𝑥 ∈ ((𝐶 ∖ 𝐵) ∩ 𝐴) ↔ (𝑥 ∈ (𝐶 ∖ 𝐵) ∧ 𝑥 ∈ 𝐴))
7		eldif 3909	. . . . 5 ⊢ (𝑥 ∈ (𝐶 ∖ 𝐵) ↔ (𝑥 ∈ 𝐶 ∧ ¬ 𝑥 ∈ 𝐵))
8	6, 7	bianbi 639	. . . 4 ⊢ (𝑥 ∈ ((𝐶 ∖ 𝐵) ∩ 𝐴) ↔ ((𝑥 ∈ 𝐶 ∧ ¬ 𝑥 ∈ 𝐵) ∧ 𝑥 ∈ 𝐴))
9		anass 474	. . . 4 ⊢ (((𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐶) ∧ ¬ 𝑥 ∈ 𝐵) ↔ (𝑥 ∈ 𝐴 ∧ (𝑥 ∈ 𝐶 ∧ ¬ 𝑥 ∈ 𝐵)))
10	5, 8, 9	3bitr4i 306	. . 3 ⊢ (𝑥 ∈ ((𝐶 ∖ 𝐵) ∩ 𝐴) ↔ ((𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐶) ∧ ¬ 𝑥 ∈ 𝐵))
11	3, 4, 10	3bitr4g 317	. 2 ⊢ (𝐴 ⊆ 𝐶 → (𝑥 ∈ (𝐴 ∖ 𝐵) ↔ 𝑥 ∈ ((𝐶 ∖ 𝐵) ∩ 𝐴)))
12	11	eqrdv 2759	1 ⊢ (𝐴 ⊆ 𝐶 → (𝐴 ∖ 𝐵) = ((𝐶 ∖ 𝐵) ∩ 𝐴))

This proof depends on syntax axioms:  ¬ wn 3   → wi 4   ∧ wa 401   = wceq 1570   ∈ wcel 2145   ∖ cdif 3896   ∩ cin 3898   ⊆ wss 3899

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-8 2147  ax-9 2155  ax-ext 2733

This proof depends on definitions:  df-bi 210  df-an 402  df-tru 1573  df-ex 1813  df-sb 2100  df-clab 2740  df-cleq 2753  df-clel 2836  df-v 3453  df-dif 3902  df-in 3906  df-ss 3916

This theorem is used by:  gsumdifsnd  20168  issubdrg  21030  restcld  23483  limcnlp  26191  symgcom2  33638  sigapildsyslem  34787  ldgenpisyslem1  34789  difelcarsg2  34938  ballotlemfp1  35117  asindmre  38601  caragendifcl  47493  gsumdifsndf  49247