Theorem difcom 4424

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 4091	. . 3 ⊢ (𝐵 ∪ 𝐶) = (𝐶 ∪ 𝐵)
2	1	sseq2i 3954	. 2 ⊢ (𝐴 ⊆ (𝐵 ∪ 𝐶) ↔ 𝐴 ⊆ (𝐶 ∪ 𝐵))
3		ssundif 4423	. 2 ⊢ (𝐴 ⊆ (𝐵 ∪ 𝐶) ↔ (𝐴 ∖ 𝐵) ⊆ 𝐶)
4		ssundif 4423	. 2 ⊢ (𝐴 ⊆ (𝐶 ∪ 𝐵) ↔ (𝐴 ∖ 𝐶) ⊆ 𝐵)
5	2, 3, 4	3bitr3i 300	1 ⊢ ((𝐴 ∖ 𝐵) ⊆ 𝐶 ↔ (𝐴 ∖ 𝐶) ⊆ 𝐵)

Syntax hints:   ↔ wb 205   ∖ cdif 3888   ∪ cun 3889   ⊆ wss 3891

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1801  ax-4 1815  ax-5 1916  ax-6 1974  ax-7 2014  ax-8 2111  ax-9 2119  ax-ext 2710

This theorem depends on definitions:  df-bi 206  df-an 396  df-or 844  df-tru 1544  df-ex 1786  df-sb 2071  df-clab 2717  df-cleq 2731  df-clel 2817  df-v 3432  df-dif 3894  df-un 3896  df-in 3898  df-ss 3908

This theorem is referenced by:  pssdifcom1  4425  pssdifcom2  4426  isreg2  22509  restmetu  23707  conss1  42015  icccncfext  43382