Metamath Proof Explorer


Theorem rngqiprngghmlem2

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

Ref Expression
Hypotheses rng2idlring.r φRRng
rng2idlring.i φI2IdealR
rng2idlring.j J=R𝑠I
rng2idlring.u φJRing
rng2idlring.b B=BaseR
rng2idlring.t ·˙=R
rng2idlring.1 1˙=1J
Assertion rngqiprngghmlem2 φABCB1˙·˙A+J1˙·˙CBaseJ

Proof

Step Hyp Ref Expression
1 rng2idlring.r φRRng
2 rng2idlring.i φI2IdealR
3 rng2idlring.j J=R𝑠I
4 rng2idlring.u φJRing
5 rng2idlring.b B=BaseR
6 rng2idlring.t ·˙=R
7 rng2idlring.1 1˙=1J
8 ringrng JRingJRng
9 4 8 syl φJRng
10 9 adantr φABCBJRng
11 1 2 3 4 5 6 7 rngqiprngghmlem1 φAB1˙·˙ABaseJ
12 11 adantrr φABCB1˙·˙ABaseJ
13 1 2 3 4 5 6 7 rngqiprngghmlem1 φCB1˙·˙CBaseJ
14 13 adantrl φABCB1˙·˙CBaseJ
15 eqid BaseJ=BaseJ
16 eqid +J=+J
17 15 16 rngacl JRng1˙·˙ABaseJ1˙·˙CBaseJ1˙·˙A+J1˙·˙CBaseJ
18 10 12 14 17 syl3anc φABCB1˙·˙A+J1˙·˙CBaseJ