Theorem recnprss 15159

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 3656	. 2 |- ( S e. { RR , CC } -> ( S = RR \/ S = CC ) )
2		ax-resscn 8017	. . . 4 |- RR C_ CC
3		sseq1 3216	. . . 4 |- ( S = RR -> ( S C_ CC <-> RR C_ CC ) )
4	2, 3	mpbiri 168	. . 3 |- ( S = RR -> S C_ CC )
5		eqimss 3247	. . 3 |- ( S = CC -> S C_ CC )
6	4, 5	jaoi 718	. 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 710   = wceq 1373   e. wcel 2176   C_ wss 3166   {cpr 3634   CCcc 7923   RRcr 7924

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 711  ax-5 1470  ax-7 1471  ax-gen 1472  ax-ie1 1516  ax-ie2 1517  ax-8 1527  ax-10 1528  ax-11 1529  ax-i12 1530  ax-bndl 1532  ax-4 1533  ax-17 1549  ax-i9 1553  ax-ial 1557  ax-i5r 1558  ax-ext 2187  ax-resscn 8017

This theorem depends on definitions:  df-bi 117  df-tru 1376  df-nf 1484  df-sb 1786  df-clab 2192  df-cleq 2198  df-clel 2201  df-nfc 2337  df-v 2774  df-un 3170  df-in 3172  df-ss 3179  df-sn 3639  df-pr 3640

This theorem is referenced by:  dvfgg  15160  dvidsslem  15165  dvconstss  15170  dvaddxx  15175  dvmulxx  15176  dviaddf  15177  dvimulf  15178  dvmptfsum  15197