Theorem difcom 4460

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 4129	. . 3 ⊢ (𝐵 ∪ 𝐶) = (𝐶 ∪ 𝐵)
2	1	sseq2i 3984	. 2 ⊢ (𝐴 ⊆ (𝐵 ∪ 𝐶) ↔ 𝐴 ⊆ (𝐶 ∪ 𝐵))
3		ssundif 4459	. 2 ⊢ (𝐴 ⊆ (𝐵 ∪ 𝐶) ↔ (𝐴 ∖ 𝐵) ⊆ 𝐶)
4		ssundif 4459	. 2 ⊢ (𝐴 ⊆ (𝐶 ∪ 𝐵) ↔ (𝐴 ∖ 𝐶) ⊆ 𝐵)
5	2, 3, 4	3bitr3i 301	1 ⊢ ((𝐴 ∖ 𝐵) ⊆ 𝐶 ↔ (𝐴 ∖ 𝐶) ⊆ 𝐵)

Syntax hints:   ↔ wb 206   ∖ cdif 3919   ∪ cun 3920   ⊆ wss 3922

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809  ax-5 1910  ax-6 1967  ax-7 2008  ax-8 2111  ax-9 2119  ax-ext 2702

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-tru 1543  df-ex 1780  df-sb 2066  df-clab 2709  df-cleq 2722  df-clel 2804  df-v 3457  df-dif 3925  df-un 3927  df-ss 3939

This theorem is referenced by:  pssdifcom1  4461  pssdifcom2  4462  isreg2  23270  restmetu  24464  conss1  44405  icccncfext  45858