Metamath Proof Explorer


Theorem eldioph

Description: Condition for a set to be Diophantine (unpacking existential quantifier). (Contributed by Stefan O'Rear, 5-Oct-2014)

Ref Expression
Assertion eldioph N 0 K N P mzPoly 1 K t | u 0 1 K t = u 1 N P u = 0 Dioph N

Proof

Step Hyp Ref Expression
1 simp1 N 0 K N P mzPoly 1 K N 0
2 simp2 N 0 K N P mzPoly 1 K K N
3 simp3 N 0 K N P mzPoly 1 K P mzPoly 1 K
4 eqidd N 0 K N P mzPoly 1 K t | u 0 1 K t = u 1 N P u = 0 = t | u 0 1 K t = u 1 N P u = 0
5 fveq1 p = P p u = P u
6 5 eqeq1d p = P p u = 0 P u = 0
7 6 anbi2d p = P t = u 1 N p u = 0 t = u 1 N P u = 0
8 7 rexbidv p = P u 0 1 K t = u 1 N p u = 0 u 0 1 K t = u 1 N P u = 0
9 8 abbidv p = P t | u 0 1 K t = u 1 N p u = 0 = t | u 0 1 K t = u 1 N P u = 0
10 9 rspceeqv P mzPoly 1 K t | u 0 1 K t = u 1 N P u = 0 = t | u 0 1 K t = u 1 N P u = 0 p mzPoly 1 K t | u 0 1 K t = u 1 N P u = 0 = t | u 0 1 K t = u 1 N p u = 0
11 3 4 10 syl2anc N 0 K N P mzPoly 1 K p mzPoly 1 K t | u 0 1 K t = u 1 N P u = 0 = t | u 0 1 K t = u 1 N p u = 0
12 oveq2 k = K 1 k = 1 K
13 12 fveq2d k = K mzPoly 1 k = mzPoly 1 K
14 12 oveq2d k = K 0 1 k = 0 1 K
15 14 rexeqdv k = K u 0 1 k t = u 1 N p u = 0 u 0 1 K t = u 1 N p u = 0
16 15 abbidv k = K t | u 0 1 k t = u 1 N p u = 0 = t | u 0 1 K t = u 1 N p u = 0
17 16 eqeq2d k = K t | u 0 1 K t = u 1 N P u = 0 = t | u 0 1 k t = u 1 N p u = 0 t | u 0 1 K t = u 1 N P u = 0 = t | u 0 1 K t = u 1 N p u = 0
18 13 17 rexeqbidv k = K p mzPoly 1 k t | u 0 1 K t = u 1 N P u = 0 = t | u 0 1 k t = u 1 N p u = 0 p mzPoly 1 K t | u 0 1 K t = u 1 N P u = 0 = t | u 0 1 K t = u 1 N p u = 0
19 18 rspcev K N p mzPoly 1 K t | u 0 1 K t = u 1 N P u = 0 = t | u 0 1 K t = u 1 N p u = 0 k N p mzPoly 1 k t | u 0 1 K t = u 1 N P u = 0 = t | u 0 1 k t = u 1 N p u = 0
20 2 11 19 syl2anc N 0 K N P mzPoly 1 K k N p mzPoly 1 k t | u 0 1 K t = u 1 N P u = 0 = t | u 0 1 k t = u 1 N p u = 0
21 eldiophb t | u 0 1 K t = u 1 N P u = 0 Dioph N N 0 k N p mzPoly 1 k t | u 0 1 K t = u 1 N P u = 0 = t | u 0 1 k t = u 1 N p u = 0
22 1 20 21 sylanbrc N 0 K N P mzPoly 1 K t | u 0 1 K t = u 1 N P u = 0 Dioph N