Metamath Proof Explorer


Theorem rngqiprngfulem1

Description: Lemma 1 for rngqiprngfu (and lemma for rngqiprngu ). (Contributed by AV, 16-Mar-2025)

Ref Expression
Hypotheses rngqiprngfu.r φRRng
rngqiprngfu.i φI2IdealR
rngqiprngfu.j J=R𝑠I
rngqiprngfu.u φJRing
rngqiprngfu.b B=BaseR
rngqiprngfu.t ·˙=R
rngqiprngfu.1 1˙=1J
rngqiprngfu.g ˙=R~QGI
rngqiprngfu.q Q=R/𝑠˙
rngqiprngfu.v φQRing
Assertion rngqiprngfulem1 φxB1Q=x˙

Proof

Step Hyp Ref Expression
1 rngqiprngfu.r φRRng
2 rngqiprngfu.i φI2IdealR
3 rngqiprngfu.j J=R𝑠I
4 rngqiprngfu.u φJRing
5 rngqiprngfu.b B=BaseR
6 rngqiprngfu.t ·˙=R
7 rngqiprngfu.1 1˙=1J
8 rngqiprngfu.g ˙=R~QGI
9 rngqiprngfu.q Q=R/𝑠˙
10 rngqiprngfu.v φQRing
11 eqid BaseQ=BaseQ
12 eqid 1Q=1Q
13 11 12 ringidcl QRing1QBaseQ
14 10 13 syl φ1QBaseQ
15 9 a1i φQ=R/𝑠˙
16 5 a1i φB=BaseR
17 8 ovexi ˙V
18 17 a1i φ˙V
19 15 16 18 1 qusbas φB/˙=BaseQ
20 14 19 eleqtrrd φ1QB/˙
21 fvexd φ1QV
22 elqsg 1QV1QB/˙xB1Q=x˙
23 21 22 syl φ1QB/˙xB1Q=x˙
24 20 23 mpbid φxB1Q=x˙