Theorem difcom 3635

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 3409	. . 3 ⊢ (B ∪ C) = (C ∪ B)
2	1	sseq2i 3297	. 2 ⊢ (A ⊆ (B ∪ C) ↔ A ⊆ (C ∪ B))
3		ssundif 3634	. 2 ⊢ (A ⊆ (B ∪ C) ↔ (A ∖ B) ⊆ C)
4		ssundif 3634	. 2 ⊢ (A ⊆ (C ∪ B) ↔ (A ∖ C) ⊆ B)
5	2, 3, 4	3bitr3i 266	1 ⊢ ((A ∖ B) ⊆ C ↔ (A ∖ C) ⊆ B)

This proof depends on syntax axioms:   ↔ wb 176   ∖ cdif 3207   ∪ cun 3208   ⊆ wss 3258

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1546  ax-5 1557  ax-17 1616  ax-9 1654  ax-8 1675  ax-6 1729  ax-7 1734  ax-11 1746  ax-12 1925  ax-ext 2334

This proof depends on definitions:  df-bi 177  df-or 359  df-an 360  df-nan 1288  df-tru 1319  df-ex 1542  df-nf 1545  df-sb 1649  df-clab 2340  df-cleq 2346  df-clel 2349  df-nfc 2479  df-v 2862  df-nin 3212  df-compl 3213  df-in 3214  df-un 3215  df-dif 3216  df-ss 3260

This theorem is used by:  pssdifcom1  3636  pssdifcom2  3637