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 φRRng
rng2idlring.i φI2IdealR
rng2idlring.j J=R𝑠I
rng2idlring.u φJRing
rng2idlring.b B=BaseR
rng2idlring.t ·˙=R
rng2idlring.1 1˙=1J
Assertion rngqiprngghmlem3 φABCB1˙·˙A+RC=1˙·˙A+J1˙·˙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 1 2 3 4 5 6 7 rngqiprng1elbas φ1˙B
9 8 anim1i φABCB1˙BABCB
10 3anass 1˙BABCB1˙BABCB
11 9 10 sylibr φABCB1˙BABCB
12 eqid +R=+R
13 5 12 6 rngdi RRng1˙BABCB1˙·˙A+RC=1˙·˙A+R1˙·˙C
14 1 11 13 syl2an2r φABCB1˙·˙A+RC=1˙·˙A+R1˙·˙C
15 3 12 ressplusg I2IdealR+R=+J
16 2 15 syl φ+R=+J
17 16 oveqd φ1˙·˙A+R1˙·˙C=1˙·˙A+J1˙·˙C
18 17 adantr φABCB1˙·˙A+R1˙·˙C=1˙·˙A+J1˙·˙C
19 14 18 eqtrd φABCB1˙·˙A+RC=1˙·˙A+J1˙·˙C