Theorem difcom 4495

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 4168	. . 3 ⊢ (𝐵 ∪ 𝐶) = (𝐶 ∪ 𝐵)
2	1	sseq2i 4025	. 2 ⊢ (𝐴 ⊆ (𝐵 ∪ 𝐶) ↔ 𝐴 ⊆ (𝐶 ∪ 𝐵))
3		ssundif 4494	. 2 ⊢ (𝐴 ⊆ (𝐵 ∪ 𝐶) ↔ (𝐴 ∖ 𝐵) ⊆ 𝐶)
4		ssundif 4494	. 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 3960   ∪ cun 3961   ⊆ wss 3963

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1792  ax-4 1806  ax-5 1908  ax-6 1965  ax-7 2005  ax-8 2108  ax-9 2116  ax-ext 2706

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-tru 1540  df-ex 1777  df-sb 2063  df-clab 2713  df-cleq 2727  df-clel 2814  df-v 3480  df-dif 3966  df-un 3968  df-ss 3980

This theorem is referenced by:  pssdifcom1  4496  pssdifcom2  4497  isreg2  23401  restmetu  24599  conss1  44440  icccncfext  45843