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 φ R Rng
rngqiprngfu.i φ I 2Ideal R
rngqiprngfu.j J = R 𝑠 I
rngqiprngfu.u φ J Ring
rngqiprngfu.b B = Base R
rngqiprngfu.t · ˙ = R
rngqiprngfu.1 1 ˙ = 1 J
rngqiprngfu.g ˙ = R ~ QG I
rngqiprngfu.q Q = R / 𝑠 ˙
rngqiprngfu.v φ Q Ring
Assertion rngqiprngfulem1 φ x B 1 Q = x ˙

Proof

Step Hyp Ref Expression
1 rngqiprngfu.r φ R Rng
2 rngqiprngfu.i φ I 2Ideal R
3 rngqiprngfu.j J = R 𝑠 I
4 rngqiprngfu.u φ J Ring
5 rngqiprngfu.b B = Base R
6 rngqiprngfu.t · ˙ = R
7 rngqiprngfu.1 1 ˙ = 1 J
8 rngqiprngfu.g ˙ = R ~ QG I
9 rngqiprngfu.q Q = R / 𝑠 ˙
10 rngqiprngfu.v φ Q Ring
11 eqid Base Q = Base Q
12 eqid 1 Q = 1 Q
13 11 12 ringidcl Q Ring 1 Q Base Q
14 10 13 syl φ 1 Q Base Q
15 9 a1i φ Q = R / 𝑠 ˙
16 5 a1i φ B = Base R
17 8 ovexi ˙ V
18 17 a1i φ ˙ V
19 15 16 18 1 qusbas φ B / ˙ = Base Q
20 14 19 eleqtrrd φ 1 Q B / ˙
21 fvexd φ 1 Q V
22 elqsg 1 Q V 1 Q B / ˙ x B 1 Q = x ˙
23 21 22 syl φ 1 Q B / ˙ x B 1 Q = x ˙
24 20 23 mpbid φ x B 1 Q = x ˙