Theorem recnprss 13707

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 3612	. 2 |- ( S e. { RR , CC } -> ( S = RR \/ S = CC ) )
2		ax-resscn 7878	. . . 4 |- RR C_ CC
3		sseq1 3176	. . . 4 |- ( S = RR -> ( S C_ CC <-> RR C_ CC ) )
4	2, 3	mpbiri 168	. . 3 |- ( S = RR -> S C_ CC )
5		eqimss 3207	. . 3 |- ( S = CC -> S C_ CC )
6	4, 5	jaoi 716	. 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 708   = wceq 1353   e. wcel 2146   C_ wss 3127   {cpr 3590   CCcc 7784   RRcr 7785

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 709  ax-5 1445  ax-7 1446  ax-gen 1447  ax-ie1 1491  ax-ie2 1492  ax-8 1502  ax-10 1503  ax-11 1504  ax-i12 1505  ax-bndl 1507  ax-4 1508  ax-17 1524  ax-i9 1528  ax-ial 1532  ax-i5r 1533  ax-ext 2157  ax-resscn 7878

This theorem depends on definitions:  df-bi 117  df-tru 1356  df-nf 1459  df-sb 1761  df-clab 2162  df-cleq 2168  df-clel 2171  df-nfc 2306  df-v 2737  df-un 3131  df-in 3133  df-ss 3140  df-sn 3595  df-pr 3596

This theorem is referenced by:  dvfgg  13708  dvaddxx  13718  dvmulxx  13719  dviaddf  13720  dvimulf  13721