Theorem recnprss 15410

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 3692	. 2 |- ( S e. { RR , CC } -> ( S = RR \/ S = CC ) )
2		ax-resscn 8123	. . . 4 |- RR C_ CC
3		sseq1 3250	. . . 4 |- ( S = RR -> ( S C_ CC <-> RR C_ CC ) )
4	2, 3	mpbiri 168	. . 3 |- ( S = RR -> S C_ CC )
5		eqimss 3281	. . 3 |- ( S = CC -> S C_ CC )
6	4, 5	jaoi 723	. 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 715   = wceq 1397   e. wcel 2202   C_ wss 3200   {cpr 3670   CCcc 8029   RRcr 8030

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 716  ax-5 1495  ax-7 1496  ax-gen 1497  ax-ie1 1541  ax-ie2 1542  ax-8 1552  ax-10 1553  ax-11 1554  ax-i12 1555  ax-bndl 1557  ax-4 1558  ax-17 1574  ax-i9 1578  ax-ial 1582  ax-i5r 1583  ax-ext 2213  ax-resscn 8123

This theorem depends on definitions:  df-bi 117  df-tru 1400  df-nf 1509  df-sb 1811  df-clab 2218  df-cleq 2224  df-clel 2227  df-nfc 2363  df-v 2804  df-un 3204  df-in 3206  df-ss 3213  df-sn 3675  df-pr 3676

This theorem is referenced by:  dvfgg  15411  dvidsslem  15416  dvconstss  15421  dvaddxx  15426  dvmulxx  15427  dviaddf  15428  dvimulf  15429  dvmptfsum  15448