Metamath Proof Explorer


Theorem rngqiprngghmlem2

Description: Lemma 2 for rngqiprngghm . (Contributed by AV, 25-Feb-2025)

Ref Expression
Hypotheses rng2idlring.r ( 𝜑𝑅 ∈ Rng )
rng2idlring.i ( 𝜑𝐼 ∈ ( 2Ideal ‘ 𝑅 ) )
rng2idlring.j 𝐽 = ( 𝑅s 𝐼 )
rng2idlring.u ( 𝜑𝐽 ∈ Ring )
rng2idlring.b 𝐵 = ( Base ‘ 𝑅 )
rng2idlring.t · = ( .r𝑅 )
rng2idlring.1 1 = ( 1r𝐽 )
Assertion rngqiprngghmlem2 ( ( 𝜑 ∧ ( 𝐴𝐵𝐶𝐵 ) ) → ( ( 1 · 𝐴 ) ( +g𝐽 ) ( 1 · 𝐶 ) ) ∈ ( Base ‘ 𝐽 ) )

Proof

Step Hyp Ref Expression
1 rng2idlring.r ( 𝜑𝑅 ∈ Rng )
2 rng2idlring.i ( 𝜑𝐼 ∈ ( 2Ideal ‘ 𝑅 ) )
3 rng2idlring.j 𝐽 = ( 𝑅s 𝐼 )
4 rng2idlring.u ( 𝜑𝐽 ∈ Ring )
5 rng2idlring.b 𝐵 = ( Base ‘ 𝑅 )
6 rng2idlring.t · = ( .r𝑅 )
7 rng2idlring.1 1 = ( 1r𝐽 )
8 ringrng ( 𝐽 ∈ Ring → 𝐽 ∈ Rng )
9 4 8 syl ( 𝜑𝐽 ∈ Rng )
10 9 adantr ( ( 𝜑 ∧ ( 𝐴𝐵𝐶𝐵 ) ) → 𝐽 ∈ Rng )
11 1 2 3 4 5 6 7 rngqiprngghmlem1 ( ( 𝜑𝐴𝐵 ) → ( 1 · 𝐴 ) ∈ ( Base ‘ 𝐽 ) )
12 11 adantrr ( ( 𝜑 ∧ ( 𝐴𝐵𝐶𝐵 ) ) → ( 1 · 𝐴 ) ∈ ( Base ‘ 𝐽 ) )
13 1 2 3 4 5 6 7 rngqiprngghmlem1 ( ( 𝜑𝐶𝐵 ) → ( 1 · 𝐶 ) ∈ ( Base ‘ 𝐽 ) )
14 13 adantrl ( ( 𝜑 ∧ ( 𝐴𝐵𝐶𝐵 ) ) → ( 1 · 𝐶 ) ∈ ( Base ‘ 𝐽 ) )
15 eqid ( Base ‘ 𝐽 ) = ( Base ‘ 𝐽 )
16 eqid ( +g𝐽 ) = ( +g𝐽 )
17 15 16 rngacl ( ( 𝐽 ∈ Rng ∧ ( 1 · 𝐴 ) ∈ ( Base ‘ 𝐽 ) ∧ ( 1 · 𝐶 ) ∈ ( Base ‘ 𝐽 ) ) → ( ( 1 · 𝐴 ) ( +g𝐽 ) ( 1 · 𝐶 ) ) ∈ ( Base ‘ 𝐽 ) )
18 10 12 14 17 syl3anc ( ( 𝜑 ∧ ( 𝐴𝐵𝐶𝐵 ) ) → ( ( 1 · 𝐴 ) ( +g𝐽 ) ( 1 · 𝐶 ) ) ∈ ( Base ‘ 𝐽 ) )