Metamath Proof Explorer


Theorem rngqiprnglinlem1

Description: Lemma 1 for rngqiprnglin . (Contributed by AV, 28-Feb-2025) (Proof shortened by AV, 24-Mar-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 rngqiprnglinlem1 φABCB1˙·˙A·˙1˙·˙C=1˙·˙A·˙C

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 2 adantr φABCBI2IdealR
9 3 6 ressmulr I2IdealR·˙=J
10 8 9 syl φABCB·˙=J
11 10 oveqd φABCB1˙·˙A·˙1˙=1˙·˙AJ1˙
12 eqid BaseJ=BaseJ
13 eqid J=J
14 4 adantr φABCBJRing
15 1 2 3 4 5 6 7 rngqiprngghmlem1 φAB1˙·˙ABaseJ
16 15 adantrr φABCB1˙·˙ABaseJ
17 12 13 7 14 16 ringridmd φABCB1˙·˙AJ1˙=1˙·˙A
18 11 17 eqtrd φABCB1˙·˙A·˙1˙=1˙·˙A
19 18 oveq1d φABCB1˙·˙A·˙1˙·˙C=1˙·˙A·˙C
20 1 adantr φABCBRRng
21 1 2 3 4 5 6 7 rngqiprng1elbas φ1˙B
22 21 adantr φABCB1˙B
23 simprl φABCBAB
24 5 6 rngcl RRng1˙BAB1˙·˙AB
25 20 22 23 24 syl3anc φABCB1˙·˙AB
26 simprr φABCBCB
27 5 6 rngass RRng1˙·˙AB1˙BCB1˙·˙A·˙1˙·˙C=1˙·˙A·˙1˙·˙C
28 20 25 22 26 27 syl13anc φABCB1˙·˙A·˙1˙·˙C=1˙·˙A·˙1˙·˙C
29 5 6 rngass RRng1˙BABCB1˙·˙A·˙C=1˙·˙A·˙C
30 20 22 23 26 29 syl13anc φABCB1˙·˙A·˙C=1˙·˙A·˙C
31 19 28 30 3eqtr3d φABCB1˙·˙A·˙1˙·˙C=1˙·˙A·˙C