Theorem recnprss 13450

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 3606	. 2 |- ( S e. { RR , CC } -> ( S = RR \/ S = CC ) )
2		ax-resscn 7866	. . . 4 |- RR C_ CC
3		sseq1 3170	. . . 4 |- ( S = RR -> ( S C_ CC <-> RR C_ CC ) )
4	2, 3	mpbiri 167	. . 3 |- ( S = RR -> S C_ CC )
5		eqimss 3201	. . 3 |- ( S = CC -> S C_ CC )
6	4, 5	jaoi 711	. 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 703   = wceq 1348   e. wcel 2141   C_ wss 3121   {cpr 3584   CCcc 7772   RRcr 7773

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-io 704  ax-5 1440  ax-7 1441  ax-gen 1442  ax-ie1 1486  ax-ie2 1487  ax-8 1497  ax-10 1498  ax-11 1499  ax-i12 1500  ax-bndl 1502  ax-4 1503  ax-17 1519  ax-i9 1523  ax-ial 1527  ax-i5r 1528  ax-ext 2152  ax-resscn 7866

This theorem depends on definitions:  df-bi 116  df-tru 1351  df-nf 1454  df-sb 1756  df-clab 2157  df-cleq 2163  df-clel 2166  df-nfc 2301  df-v 2732  df-un 3125  df-in 3127  df-ss 3134  df-sn 3589  df-pr 3590

This theorem is referenced by:  dvfgg  13451  dvaddxx  13461  dvmulxx  13462  dviaddf  13463  dvimulf  13464