Theorem nnsucelrlem4 4428

Description: Lemma for nnsucelr 4429. Remove and re-adjoin an element to a set. (Contributed by SF, 15-Jan-2015.)

Assertion

Ref	Expression
nnsucelrlem4	⊢ (A ∈ B → ((B ∖ {A}) ∪ {A}) = B)

Proof of Theorem nnsucelrlem4

Step	Hyp	Ref	Expression
1		undif1 3626	. 2 ⊢ ((B ∖ {A}) ∪ {A}) = (B ∪ {A})
2		snssi 3853	. . 3 ⊢ (A ∈ B → {A} ⊆ B)
3		ssequn2 3437	. . 3 ⊢ ({A} ⊆ B ↔ (B ∪ {A}) = B)
4	2, 3	sylib 188	. 2 ⊢ (A ∈ B → (B ∪ {A}) = B)
5	1, 4	syl5eq 2397	1 ⊢ (A ∈ B → ((B ∖ {A}) ∪ {A}) = B)

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:  nnsucelr  4429  enadj  6061