Theorem undif 3631

Description: Union of complementary parts into whole. (Contributed by NM, 22-Mar-1998.)

Assertion

Ref	Expression
undif	⊢ (A ⊆ B ↔ (A ∪ (B ∖ A)) = B)

Proof of Theorem undif

Step	Hyp	Ref	Expression
1		ssequn1 3434	. 2 ⊢ (A ⊆ B ↔ (A ∪ B) = B)
2		undif2 3627	. . 3 ⊢ (A ∪ (B ∖ A)) = (A ∪ B)
3	2	eqeq1i 2360	. 2 ⊢ ((A ∪ (B ∖ A)) = B ↔ (A ∪ B) = B)
4	1, 3	bitr4i 243	1 ⊢ (A ⊆ B ↔ (A ∪ (B ∖ A)) = B)

Syntax hints:   ↔ wb 176   = wceq 1642   ∖ cdif 3207   ∪ cun 3208   ⊆ wss 3258

This theorem was proved from 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 theorem 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-ne 2519  df-v 2862  df-nin 3212  df-compl 3213  df-in 3214  df-un 3215  df-dif 3216  df-ss 3260  df-nul 3552

This theorem is referenced by:  difsnid  3855