Theorem difsnss 3510

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 6080. (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 3087	. 2 ⊢ ((𝐴 ∖ {𝐵}) ∪ {𝐵}) = ({𝐵} ∪ (𝐴 ∖ {𝐵}))
2		snssi 3508	. . 3 ⊢ (𝐵 ∈ 𝐴 → {𝐵} ⊆ 𝐴)
3		undifss 3303	. . 3 ⊢ ({𝐵} ⊆ 𝐴 ↔ ({𝐵} ∪ (𝐴 ∖ {𝐵})) ⊆ 𝐴)
4	2, 3	sylib 127	. 2 ⊢ (𝐵 ∈ 𝐴 → ({𝐵} ∪ (𝐴 ∖ {𝐵})) ⊆ 𝐴)
5	1, 4	syl5eqss 2989	1 ⊢ (𝐵 ∈ 𝐴 → ((𝐴 ∖ {𝐵}) ∪ {𝐵}) ⊆ 𝐴)

Syntax hints:   → wi 4   ∈ wcel 1393   ∖ cdif 2914   ∪ cun 2915   ⊆ wss 2917  {csn 3375

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-in1 544  ax-in2 545  ax-io 630  ax-5 1336  ax-7 1337  ax-gen 1338  ax-ie1 1382  ax-ie2 1383  ax-8 1395  ax-10 1396  ax-11 1397  ax-i12 1398  ax-bndl 1399  ax-4 1400  ax-17 1419  ax-i9 1423  ax-ial 1427  ax-i5r 1428  ax-ext 2022

This theorem depends on definitions:  df-bi 110  df-tru 1246  df-nf 1350  df-sb 1646  df-clab 2027  df-cleq 2033  df-clel 2036  df-nfc 2167  df-v 2559  df-dif 2920  df-un 2922  df-in 2924  df-ss 2931  df-sn 3381

This theorem is referenced by:  nndifsnid  6080  fidifsnid  6332