Theorem difsnss 3674

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 6411. (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 3225	. 2 ⊢ ((𝐴 ∖ {𝐵}) ∪ {𝐵}) = ({𝐵} ∪ (𝐴 ∖ {𝐵}))
2		snssi 3672	. . 3 ⊢ (𝐵 ∈ 𝐴 → {𝐵} ⊆ 𝐴)
3		undifss 3448	. . 3 ⊢ ({𝐵} ⊆ 𝐴 ↔ ({𝐵} ∪ (𝐴 ∖ {𝐵})) ⊆ 𝐴)
4	2, 3	sylib 121	. 2 ⊢ (𝐵 ∈ 𝐴 → ({𝐵} ∪ (𝐴 ∖ {𝐵})) ⊆ 𝐴)
5	1, 4	eqsstrid 3148	1 ⊢ (𝐵 ∈ 𝐴 → ((𝐴 ∖ {𝐵}) ∪ {𝐵}) ⊆ 𝐴)

Syntax hints:   → wi 4   ∈ wcel 1481   ∖ cdif 3073   ∪ cun 3074   ⊆ wss 3076  {csn 3532

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-in1 604  ax-in2 605  ax-io 699  ax-5 1424  ax-7 1425  ax-gen 1426  ax-ie1 1470  ax-ie2 1471  ax-8 1483  ax-10 1484  ax-11 1485  ax-i12 1486  ax-bndl 1487  ax-4 1488  ax-17 1507  ax-i9 1511  ax-ial 1515  ax-i5r 1516  ax-ext 2122

This theorem depends on definitions:  df-bi 116  df-tru 1335  df-nf 1438  df-sb 1737  df-clab 2127  df-cleq 2133  df-clel 2136  df-nfc 2271  df-v 2691  df-dif 3078  df-un 3080  df-in 3082  df-ss 3089  df-sn 3538

This theorem is referenced by:  fnsnsplitss  5627  dcdifsnid  6408