Theorem difcom 4489

Description: Swap the arguments of a class difference. (Contributed by NM, 29-Mar-2007.)

Assertion

Ref	Expression
difcom	⊢ ((𝐴 ∖ 𝐵) ⊆ 𝐶 ↔ (𝐴 ∖ 𝐶) ⊆ 𝐵)

Proof of Theorem difcom

Step	Hyp	Ref	Expression
1		uncom 4152	. . 3 ⊢ (𝐵 ∪ 𝐶) = (𝐶 ∪ 𝐵)
2	1	sseq2i 4009	. 2 ⊢ (𝐴 ⊆ (𝐵 ∪ 𝐶) ↔ 𝐴 ⊆ (𝐶 ∪ 𝐵))
3		ssundif 4488	. 2 ⊢ (𝐴 ⊆ (𝐵 ∪ 𝐶) ↔ (𝐴 ∖ 𝐵) ⊆ 𝐶)
4		ssundif 4488	. 2 ⊢ (𝐴 ⊆ (𝐶 ∪ 𝐵) ↔ (𝐴 ∖ 𝐶) ⊆ 𝐵)
5	2, 3, 4	3bitr3i 301	1 ⊢ ((𝐴 ∖ 𝐵) ⊆ 𝐶 ↔ (𝐴 ∖ 𝐶) ⊆ 𝐵)

Syntax hints:   ↔ wb 205   ∖ cdif 3944   ∪ cun 3945   ⊆ wss 3947

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1790  ax-4 1804  ax-5 1906  ax-6 1964  ax-7 2004  ax-8 2101  ax-9 2109  ax-ext 2699

This theorem depends on definitions:  df-bi 206  df-an 396  df-or 847  df-tru 1537  df-ex 1775  df-sb 2061  df-clab 2706  df-cleq 2720  df-clel 2806  df-v 3473  df-dif 3950  df-un 3952  df-in 3954  df-ss 3964

This theorem is referenced by:  pssdifcom1  4490  pssdifcom2  4491  isreg2  23294  restmetu  24492  conss1  43881  icccncfext  45275