Theorem difsnss 3719

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 6475. (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 3266	. 2 ⊢ ((𝐴 ∖ {𝐵}) ∪ {𝐵}) = ({𝐵} ∪ (𝐴 ∖ {𝐵}))
2		snssi 3717	. . 3 ⊢ (𝐵 ∈ 𝐴 → {𝐵} ⊆ 𝐴)
3		undifss 3489	. . 3 ⊢ ({𝐵} ⊆ 𝐴 ↔ ({𝐵} ∪ (𝐴 ∖ {𝐵})) ⊆ 𝐴)
4	2, 3	sylib 121	. 2 ⊢ (𝐵 ∈ 𝐴 → ({𝐵} ∪ (𝐴 ∖ {𝐵})) ⊆ 𝐴)
5	1, 4	eqsstrid 3188	1 ⊢ (𝐵 ∈ 𝐴 → ((𝐴 ∖ {𝐵}) ∪ {𝐵}) ⊆ 𝐴)

Syntax hints:   → wi 4   ∈ wcel 2136   ∖ cdif 3113   ∪ cun 3114   ⊆ wss 3116  {csn 3576

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 1435  ax-7 1436  ax-gen 1437  ax-ie1 1481  ax-ie2 1482  ax-8 1492  ax-10 1493  ax-11 1494  ax-i12 1495  ax-bndl 1497  ax-4 1498  ax-17 1514  ax-i9 1518  ax-ial 1522  ax-i5r 1523  ax-ext 2147

This theorem depends on definitions:  df-bi 116  df-tru 1346  df-nf 1449  df-sb 1751  df-clab 2152  df-cleq 2158  df-clel 2161  df-nfc 2297  df-v 2728  df-dif 3118  df-un 3120  df-in 3122  df-ss 3129  df-sn 3582

This theorem is referenced by:  fnsnsplitss  5684  dcdifsnid  6472