Theorem ssdifeq0 4449

Description: A class is a subclass of itself subtracted from another iff it is the empty set. (Contributed by Steve Rodriguez, 20-Nov-2015.)

Assertion

Ref	Expression
ssdifeq0	⊢ (𝐴 ⊆ (𝐵 ∖ 𝐴) ↔ 𝐴 = ∅)

Proof of Theorem ssdifeq0

Step	Hyp	Ref	Expression
1		inidm 4179	. . 3 ⊢ (𝐴 ∩ 𝐴) = 𝐴
2		ssdifin0 4448	. . 3 ⊢ (𝐴 ⊆ (𝐵 ∖ 𝐴) → (𝐴 ∩ 𝐴) = ∅)
3	1, 2	eqtr3id 2814	. 2 ⊢ (𝐴 ⊆ (𝐵 ∖ 𝐴) → 𝐴 = ∅)
4		0ss 4357	. . 3 ⊢ ∅ ⊆ (𝐵 ∖ ∅)
5		id 23	. . . 4 ⊢ (𝐴 = ∅ → 𝐴 = ∅)
6		difeq2 4075	. . . 4 ⊢ (𝐴 = ∅ → (𝐵 ∖ 𝐴) = (𝐵 ∖ ∅))
7	5, 6	sseq12d 3971	. . 3 ⊢ (𝐴 = ∅ → (𝐴 ⊆ (𝐵 ∖ 𝐴) ↔ ∅ ⊆ (𝐵 ∖ ∅)))
8	4, 7	mpbiri 261	. 2 ⊢ (𝐴 = ∅ → 𝐴 ⊆ (𝐵 ∖ 𝐴))
9	3, 8	impbii 212	1 ⊢ (𝐴 ⊆ (𝐵 ∖ 𝐴) ↔ 𝐴 = ∅)

This proof depends on syntax axioms:   ↔ wb 209   = wceq 1570   ∖ cdif 3903   ∩ cin 3905   ⊆ wss 3906  ∅c0 4286

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 2148  ax-9 2156  ax-ext 2737

This proof depends on definitions:  df-bi 210  df-an 402  df-tru 1573  df-fal 1583  df-ex 1813  df-sb 2100  df-clab 2744  df-cleq 2757  df-clel 2840  df-rab 3419  df-v 3459  df-dif 3909  df-in 3913  df-ss 3923  df-nul 4287

This theorem is used by:  disjdifprg  32995  measxun2  34669  measssd  34674  pmeasmono  34783