Theorem difsnid 3855

Description: If we remove a single element from a class then put it back in, we end up with the original class. (Contributed by NM, 2-Oct-2006.)

Assertion

Ref	Expression
difsnid	⊢ (B ∈ A → ((A ∖ {B}) ∪ {B}) = A)

Proof of Theorem difsnid

Step	Hyp	Ref	Expression
1		uncom 3409	. 2 ⊢ ((A ∖ {B}) ∪ {B}) = ({B} ∪ (A ∖ {B}))
2		snssi 3853	. . 3 ⊢ (B ∈ A → {B} ⊆ A)
3		undif 3631	. . 3 ⊢ ({B} ⊆ A ↔ ({B} ∪ (A ∖ {B})) = A)
4	2, 3	sylib 188	. 2 ⊢ (B ∈ A → ({B} ∪ (A ∖ {B})) = A)
5	1, 4	syl5eq 2397	1 ⊢ (B ∈ A → ((A ∖ {B}) ∪ {B}) = A)

Syntax hints:   → wi 4   = wceq 1642   ∈ wcel 1710   ∖ cdif 3207   ∪ cun 3208   ⊆ wss 3258  {csn 3738

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  df-sn 3742

This theorem is referenced by:  pwadjoin  4120  phiall  4619