Theorem difdif 3393

Description: Double class difference. Exercise 11 of [TakeutiZaring] p. 22. (Contributed by NM, 17-May-1998.)

Assertion

Ref	Expression
difdif	⊢ (A ∖ (B ∖ A)) = A

Proof of Theorem difdif

Dummy variable x is distinct from all other variables.

Step	Hyp	Ref	Expression
1		pm4.45im 545	. . 3 ⊢ (x ∈ A ↔ (x ∈ A ∧ (x ∈ B → x ∈ A)))
2		iman 413	. . . . 5 ⊢ ((x ∈ B → x ∈ A) ↔ ¬ (x ∈ B ∧ ¬ x ∈ A))
3		eldif 3222	. . . . 5 ⊢ (x ∈ (B ∖ A) ↔ (x ∈ B ∧ ¬ x ∈ A))
4	2, 3	xchbinxr 302	. . . 4 ⊢ ((x ∈ B → x ∈ A) ↔ ¬ x ∈ (B ∖ A))
5	4	anbi2i 675	. . 3 ⊢ ((x ∈ A ∧ (x ∈ B → x ∈ A)) ↔ (x ∈ A ∧ ¬ x ∈ (B ∖ A)))
6	1, 5	bitr2i 241	. 2 ⊢ ((x ∈ A ∧ ¬ x ∈ (B ∖ A)) ↔ x ∈ A)
7	6	difeqri 3388	1 ⊢ (A ∖ (B ∖ A)) = A

Syntax hints:  ¬ wn 3   → wi 4   ∧ wa 358   = wceq 1642   ∈ wcel 1710   ∖ cdif 3207

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 2479  df-v 2862  df-nin 3212  df-compl 3213  df-in 3214  df-dif 3216

This theorem is referenced by:  dif0  3621  undifabs  3628