Theorem recnprss 15788

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 3732	. 2 |- ( S e. { RR , CC } -> ( S = RR \/ S = CC ) )
2		ax-resscn 8271	. . . 4 |- RR C_ CC
3		sseq1 3271	. . . 4 |- ( S = RR -> ( S C_ CC <-> RR C_ CC ) )
4	2, 3	mpbiri 168	. . 3 |- ( S = RR -> S C_ CC )
5		eqimss 3302	. . 3 |- ( S = CC -> S C_ CC )
6	4, 5	jaoi 728	. 2 |- ( ( S = RR \/ S = CC ) -> S C_ CC )
7	1, 6	syl 14	1 |- ( S e. { RR , CC } -> S C_ CC )

This proof depends on syntax axioms:   -> wi 4   \/ wo 720   = wceq 1402   e. wcel 2209   C_ wss 3220   {cpr 3710   CCcc 8177   RRcr 8178

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  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  ax-resscn 8271

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-un 3224  df-in 3226  df-ss 3233  df-sn 3715  df-pr 3716

This theorem is used by:  dvfgg  15789  dvidsslem  15794  dvconstss  15799  dvaddxx  15804  dvmulxx  15805  dviaddf  15806  dvimulf  15807  dvmptfsum  15826