Theorem difsnss 3861

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 6780. (Contributed by Jim Kingdon, 10-Aug-2018.)

Assertion

Ref	Expression
difsnss	|- ( B e. A -> ( ( A \ { B } ) u. { B } ) C_ A )

Proof of Theorem difsnss

Step	Hyp	Ref	Expression
1		uncom 3373	. 2 |- ( ( A \ { B } ) u. { B } ) = ( { B } u. ( A \ { B } ) )
2		snssi 3859	. . 3 |- ( B e. A -> { B } C_ A )
3		undifss 3608	. . 3 |- ( { B } C_ A <-> ( { B } u. ( A \ { B } ) ) C_ A )
4	2, 3	sylib 122	. 2 |- ( B e. A -> ( { B } u. ( A \ { B } ) ) C_ A )
5	1, 4	eqsstrid 3294	1 |- ( B e. A -> ( ( A \ { B } ) u. { B } ) C_ A )

This proof depends on syntax axioms:   -> wi 4   e. wcel 2209   \ cdif 3217   u. cun 3218   C_ wss 3220   {csn 3709

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-in1 623  ax-in2 624  ax-io 721  ax-5 1500  ax-7 1501  ax-gen 1502  ax-ie1 1546  ax-ie2 1547  ax-8 1557  ax-10 1558  ax-11 1559  ax-i12 1560  ax-bndl 1562  ax-4 1563  ax-17 1579  ax-i9 1583  ax-ial 1587  ax-i5r 1588  ax-ext 2220

This proof depends on definitions:  df-bi 117  df-tru 1405  df-nf 1514  df-sb 1816  df-clab 2225  df-cleq 2231  df-clel 2234  df-nfc 2381  df-v 2823  df-dif 3222  df-un 3224  df-in 3226  df-ss 3233  df-sn 3715

This theorem is used by:  fnsnsplitss  5914  dcdifsnid  6777