Theorem difcom 4447

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 4160	. . 3 ⊢ (𝐵 ∪ 𝐶) = (𝐶 ∪ 𝐵)
2	1	sseq2i 3942	. 2 ⊢ (𝐴 ⊆ (𝐵 ∪ 𝐶) ↔ 𝐴 ⊆ (𝐶 ∪ 𝐵))
3		ssundif 4446	. 2 ⊢ (𝐴 ⊆ (𝐵 ∪ 𝐶) ↔ (𝐴 ∖ 𝐵) ⊆ 𝐶)
4		ssundif 4446	. 2 ⊢ (𝐴 ⊆ (𝐶 ∪ 𝐵) ↔ (𝐴 ∖ 𝐶) ⊆ 𝐵)
5	2, 3, 4	3bitr3i 304	1 ⊢ ((𝐴 ∖ 𝐵) ⊆ 𝐶 ↔ (𝐴 ∖ 𝐶) ⊆ 𝐵)

This proof depends on syntax axioms:   ↔ wb 209   ⊆ wss 3900   ∖ cdif 4048   ∪ cun 4152

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 1945  ax-6 2002  ax-7 2043  ax-8 2149  ax-9 2157  ax-ext 2735

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-tru 1573  df-ex 1813  df-sb 2102  df-clab 2742  df-cleq 2755  df-clel 2839  df-v 3456  df-ss 3901  df-dif 4050  df-un 4154

This theorem is used by:  pssdifcom1  4448  pssdifcom2  4449  isreg2  23750  restmetu  24944  conss1  45687  icccncfext  47170