Metamath Proof Explorer


Theorem rngqiprngghmlem2

Description: Lemma 2 for rngqiprngghm . (Contributed by AV, 25-Feb-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 rngqiprngghmlem2 φ A B C B 1 ˙ · ˙ A + J 1 ˙ · ˙ C Base J

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 ringrng J Ring J Rng
9 4 8 syl φ J Rng
10 9 adantr φ A B C B J Rng
11 1 2 3 4 5 6 7 rngqiprngghmlem1 φ A B 1 ˙ · ˙ A Base J
12 11 adantrr φ A B C B 1 ˙ · ˙ A Base J
13 1 2 3 4 5 6 7 rngqiprngghmlem1 φ C B 1 ˙ · ˙ C Base J
14 13 adantrl φ A B C B 1 ˙ · ˙ C Base J
15 eqid Base J = Base J
16 eqid + J = + J
17 15 16 rngacl J Rng 1 ˙ · ˙ A Base J 1 ˙ · ˙ C Base J 1 ˙ · ˙ A + J 1 ˙ · ˙ C Base J
18 10 12 14 17 syl3anc φ A B C B 1 ˙ · ˙ A + J 1 ˙ · ˙ C Base J