Metamath Proof Explorer


Theorem rngqiprngghmlem3

Description: Lemma 3 for rngqiprngghm . (Contributed by AV, 25-Feb-2025) (Proof shortened by AV, 24-Mar-2025)

Ref Expression
Hypotheses rng2idlring.r φ R Rng
rng2idlring.i φ I 2Ideal R
rng2idlring.j J = R 𝑠 I
rng2idlring.u φ J Ring
rng2idlring.b B = Base R
rng2idlring.t · ˙ = R
rng2idlring.1 1 ˙ = 1 J
Assertion rngqiprngghmlem3 φ A B C B 1 ˙ · ˙ A + R C = 1 ˙ · ˙ A + J 1 ˙ · ˙ C

Proof

Step Hyp Ref Expression
1 rng2idlring.r φ R Rng
2 rng2idlring.i φ I 2Ideal R
3 rng2idlring.j J = R 𝑠 I
4 rng2idlring.u φ J Ring
5 rng2idlring.b B = Base R
6 rng2idlring.t · ˙ = R
7 rng2idlring.1 1 ˙ = 1 J
8 1 2 3 4 5 6 7 rngqiprng1elbas φ 1 ˙ B
9 8 anim1i φ A B C B 1 ˙ B A B C B
10 3anass 1 ˙ B A B C B 1 ˙ B A B C B
11 9 10 sylibr φ A B C B 1 ˙ B A B C B
12 eqid + R = + R
13 5 12 6 rngdi R Rng 1 ˙ B A B C B 1 ˙ · ˙ A + R C = 1 ˙ · ˙ A + R 1 ˙ · ˙ C
14 1 11 13 syl2an2r φ A B C B 1 ˙ · ˙ A + R C = 1 ˙ · ˙ A + R 1 ˙ · ˙ C
15 3 12 ressplusg I 2Ideal R + R = + J
16 2 15 syl φ + R = + J
17 16 oveqd φ 1 ˙ · ˙ A + R 1 ˙ · ˙ C = 1 ˙ · ˙ A + J 1 ˙ · ˙ C
18 17 adantr φ A B C B 1 ˙ · ˙ A + R 1 ˙ · ˙ C = 1 ˙ · ˙ A + J 1 ˙ · ˙ C
19 14 18 eqtrd φ A B C B 1 ˙ · ˙ A + R C = 1 ˙ · ˙ A + J 1 ˙ · ˙ C