Metamath Proof Explorer


Theorem ispridl

Description: The predicate "is a prime ideal". (Contributed by Jeff Madsen, 10-Jun-2010)

Ref Expression
Hypotheses pridlval.1 G = 1 st R
pridlval.2 H = 2 nd R
pridlval.3 X = ran G
Assertion ispridl R RingOps P PrIdl R P Idl R P X a Idl R b Idl R x a y b x H y P a P b P

Proof

Step Hyp Ref Expression
1 pridlval.1 G = 1 st R
2 pridlval.2 H = 2 nd R
3 pridlval.3 X = ran G
4 1 2 3 pridlval R RingOps PrIdl R = i Idl R | i X a Idl R b Idl R x a y b x H y i a i b i
5 4 eleq2d R RingOps P PrIdl R P i Idl R | i X a Idl R b Idl R x a y b x H y i a i b i
6 neeq1 i = P i X P X
7 eleq2 i = P x H y i x H y P
8 7 2ralbidv i = P x a y b x H y i x a y b x H y P
9 sseq2 i = P a i a P
10 sseq2 i = P b i b P
11 9 10 orbi12d i = P a i b i a P b P
12 8 11 imbi12d i = P x a y b x H y i a i b i x a y b x H y P a P b P
13 12 2ralbidv i = P a Idl R b Idl R x a y b x H y i a i b i a Idl R b Idl R x a y b x H y P a P b P
14 6 13 anbi12d i = P i X a Idl R b Idl R x a y b x H y i a i b i P X a Idl R b Idl R x a y b x H y P a P b P
15 14 elrab P i Idl R | i X a Idl R b Idl R x a y b x H y i a i b i P Idl R P X a Idl R b Idl R x a y b x H y P a P b P
16 3anass P Idl R P X a Idl R b Idl R x a y b x H y P a P b P P Idl R P X a Idl R b Idl R x a y b x H y P a P b P
17 15 16 bitr4i P i Idl R | i X a Idl R b Idl R x a y b x H y i a i b i P Idl R P X a Idl R b Idl R x a y b x H y P a P b P
18 5 17 bitrdi R RingOps P PrIdl R P Idl R P X a Idl R b Idl R x a y b x H y P a P b P