Theorem recnprss 15355

Description: Both RR and CC are subsets of CC. (Contributed by Mario Carneiro, 10-Feb-2015.)

Assertion

Ref	Expression
recnprss	|- ( S e. { RR , CC } -> S C_ CC )

Proof of Theorem recnprss

Step	Hyp	Ref	Expression
1		elpri 3689	. 2 |- ( S e. { RR , CC } -> ( S = RR \/ S = CC ) )
2		ax-resscn 8087	. . . 4 |- RR C_ CC
3		sseq1 3247	. . . 4 |- ( S = RR -> ( S C_ CC <-> RR C_ CC ) )
4	2, 3	mpbiri 168	. . 3 |- ( S = RR -> S C_ CC )
5		eqimss 3278	. . 3 |- ( S = CC -> S C_ CC )
6	4, 5	jaoi 721	. 2 |- ( ( S = RR \/ S = CC ) -> S C_ CC )
7	1, 6	syl 14	1 |- ( S e. { RR , CC } -> S C_ CC )

Syntax hints:   -> wi 4   \/ wo 713   = wceq 1395   e. wcel 2200   C_ wss 3197   {cpr 3667   CCcc 7993   RRcr 7994

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-io 714  ax-5 1493  ax-7 1494  ax-gen 1495  ax-ie1 1539  ax-ie2 1540  ax-8 1550  ax-10 1551  ax-11 1552  ax-i12 1553  ax-bndl 1555  ax-4 1556  ax-17 1572  ax-i9 1576  ax-ial 1580  ax-i5r 1581  ax-ext 2211  ax-resscn 8087

This theorem depends on definitions:  df-bi 117  df-tru 1398  df-nf 1507  df-sb 1809  df-clab 2216  df-cleq 2222  df-clel 2225  df-nfc 2361  df-v 2801  df-un 3201  df-in 3203  df-ss 3210  df-sn 3672  df-pr 3673

This theorem is referenced by:  dvfgg  15356  dvidsslem  15361  dvconstss  15366  dvaddxx  15371  dvmulxx  15372  dviaddf  15373  dvimulf  15374  dvmptfsum  15393