Theorem difsnss 3740

Description: If we remove a single element from a class then put it back in, we end up with a subset of the original class. If equality is decidable, we can replace subset with equality as seen in nndifsnid 6510. (Contributed by Jim Kingdon, 10-Aug-2018.)

Assertion

Ref	Expression
difsnss	⊢ (𝐵 ∈ 𝐴 → ((𝐴 ∖ {𝐵}) ∪ {𝐵}) ⊆ 𝐴)

Proof of Theorem difsnss

Step	Hyp	Ref	Expression
1		uncom 3281	. 2 ⊢ ((𝐴 ∖ {𝐵}) ∪ {𝐵}) = ({𝐵} ∪ (𝐴 ∖ {𝐵}))
2		snssi 3738	. . 3 ⊢ (𝐵 ∈ 𝐴 → {𝐵} ⊆ 𝐴)
3		undifss 3505	. . 3 ⊢ ({𝐵} ⊆ 𝐴 ↔ ({𝐵} ∪ (𝐴 ∖ {𝐵})) ⊆ 𝐴)
4	2, 3	sylib 122	. 2 ⊢ (𝐵 ∈ 𝐴 → ({𝐵} ∪ (𝐴 ∖ {𝐵})) ⊆ 𝐴)
5	1, 4	eqsstrid 3203	1 ⊢ (𝐵 ∈ 𝐴 → ((𝐴 ∖ {𝐵}) ∪ {𝐵}) ⊆ 𝐴)

Syntax hints:   → wi 4   ∈ wcel 2148   ∖ cdif 3128   ∪ cun 3129   ⊆ wss 3131  {csn 3594

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-in1 614  ax-in2 615  ax-io 709  ax-5 1447  ax-7 1448  ax-gen 1449  ax-ie1 1493  ax-ie2 1494  ax-8 1504  ax-10 1505  ax-11 1506  ax-i12 1507  ax-bndl 1509  ax-4 1510  ax-17 1526  ax-i9 1530  ax-ial 1534  ax-i5r 1535  ax-ext 2159

This theorem depends on definitions:  df-bi 117  df-tru 1356  df-nf 1461  df-sb 1763  df-clab 2164  df-cleq 2170  df-clel 2173  df-nfc 2308  df-v 2741  df-dif 3133  df-un 3135  df-in 3137  df-ss 3144  df-sn 3600

This theorem is referenced by:  fnsnsplitss  5717  dcdifsnid  6507