Theorem difin2 3199

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 2939	. . . . 5 ⊢ (𝐴 ⊆ 𝐶 → (𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐶))
2	1	pm4.71d 373	. . . 4 ⊢ (𝐴 ⊆ 𝐶 → (𝑥 ∈ 𝐴 ↔ (𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐶)))
3	2	anbi1d 438	. . 3 ⊢ (𝐴 ⊆ 𝐶 → ((𝑥 ∈ 𝐴 ∧ ¬ 𝑥 ∈ 𝐵) ↔ ((𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐶) ∧ ¬ 𝑥 ∈ 𝐵)))
4		eldif 2927	. . 3 ⊢ (𝑥 ∈ (𝐴 ∖ 𝐵) ↔ (𝑥 ∈ 𝐴 ∧ ¬ 𝑥 ∈ 𝐵))
5		elin 3126	. . . 4 ⊢ (𝑥 ∈ ((𝐶 ∖ 𝐵) ∩ 𝐴) ↔ (𝑥 ∈ (𝐶 ∖ 𝐵) ∧ 𝑥 ∈ 𝐴))
6		eldif 2927	. . . . 5 ⊢ (𝑥 ∈ (𝐶 ∖ 𝐵) ↔ (𝑥 ∈ 𝐶 ∧ ¬ 𝑥 ∈ 𝐵))
7	6	anbi1i 431	. . . 4 ⊢ ((𝑥 ∈ (𝐶 ∖ 𝐵) ∧ 𝑥 ∈ 𝐴) ↔ ((𝑥 ∈ 𝐶 ∧ ¬ 𝑥 ∈ 𝐵) ∧ 𝑥 ∈ 𝐴))
8		ancom 253	. . . . 5 ⊢ (((𝑥 ∈ 𝐶 ∧ ¬ 𝑥 ∈ 𝐵) ∧ 𝑥 ∈ 𝐴) ↔ (𝑥 ∈ 𝐴 ∧ (𝑥 ∈ 𝐶 ∧ ¬ 𝑥 ∈ 𝐵)))
9		anass 381	. . . . 5 ⊢ (((𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐶) ∧ ¬ 𝑥 ∈ 𝐵) ↔ (𝑥 ∈ 𝐴 ∧ (𝑥 ∈ 𝐶 ∧ ¬ 𝑥 ∈ 𝐵)))
10	8, 9	bitr4i 176	. . . 4 ⊢ (((𝑥 ∈ 𝐶 ∧ ¬ 𝑥 ∈ 𝐵) ∧ 𝑥 ∈ 𝐴) ↔ ((𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐶) ∧ ¬ 𝑥 ∈ 𝐵))
11	5, 7, 10	3bitri 195	. . 3 ⊢ (𝑥 ∈ ((𝐶 ∖ 𝐵) ∩ 𝐴) ↔ ((𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐶) ∧ ¬ 𝑥 ∈ 𝐵))
12	3, 4, 11	3bitr4g 212	. 2 ⊢ (𝐴 ⊆ 𝐶 → (𝑥 ∈ (𝐴 ∖ 𝐵) ↔ 𝑥 ∈ ((𝐶 ∖ 𝐵) ∩ 𝐴)))
13	12	eqrdv 2038	1 ⊢ (𝐴 ⊆ 𝐶 → (𝐴 ∖ 𝐵) = ((𝐶 ∖ 𝐵) ∩ 𝐴))

Syntax hints:  ¬ wn 3   → wi 4   ∧ wa 97   = wceq 1243   ∈ wcel 1393   ∖ cdif 2914   ∩ cin 2916   ⊆ wss 2917

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-in1 544  ax-in2 545  ax-io 630  ax-5 1336  ax-7 1337  ax-gen 1338  ax-ie1 1382  ax-ie2 1383  ax-8 1395  ax-10 1396  ax-11 1397  ax-i12 1398  ax-bndl 1399  ax-4 1400  ax-17 1419  ax-i9 1423  ax-ial 1427  ax-i5r 1428  ax-ext 2022

This theorem depends on definitions:  df-bi 110  df-tru 1246  df-nf 1350  df-sb 1646  df-clab 2027  df-cleq 2033  df-clel 2036  df-nfc 2167  df-v 2559  df-dif 2920  df-in 2924  df-ss 2931

This theorem is referenced by: (None)