Metamath Proof Explorer


Theorem lmod1lem2

Description: Lemma 2 for lmod1 . (Contributed by AV, 28-Apr-2019)

Ref Expression
Hypothesis lmod1.m M = Base ndx I + ndx I I I Scalar ndx R ndx x Base R , y I y
Assertion lmod1lem2 I V R Ring r Base R r M I + M I = r M I + M r M I

Proof

Step Hyp Ref Expression
1 lmod1.m M = Base ndx I + ndx I I I Scalar ndx R ndx x Base R , y I y
2 fvex Base R V
3 snex I V
4 2 3 pm3.2i Base R V I V
5 mpoexga Base R V I V x Base R , y I y V
6 4 5 mp1i I V R Ring r Base R x Base R , y I y V
7 1 lmodvsca x Base R , y I y V x Base R , y I y = M
8 6 7 syl I V R Ring r Base R x Base R , y I y = M
9 8 eqcomd I V R Ring r Base R M = x Base R , y I y
10 simprr I V R Ring r Base R x = r y = I y = I
11 simp3 I V R Ring r Base R r Base R
12 snidg I V I I
13 12 3ad2ant1 I V R Ring r Base R I I
14 9 10 11 13 13 ovmpod I V R Ring r Base R r M I = I
15 snex I I I V
16 1 lmodplusg I I I V I I I = + M
17 15 16 mp1i I V R Ring r Base R I I I = + M
18 17 eqcomd I V R Ring r Base R + M = I I I
19 18 oveqd I V R Ring r Base R I + M I = I I I I I
20 df-ov I I I I I = I I I I I
21 opex I I V
22 simp1 I V R Ring r Base R I V
23 fvsng I I V I V I I I I I = I
24 21 22 23 sylancr I V R Ring r Base R I I I I I = I
25 20 24 eqtrid I V R Ring r Base R I I I I I = I
26 19 25 eqtrd I V R Ring r Base R I + M I = I
27 26 oveq2d I V R Ring r Base R r M I + M I = r M I
28 3 a1i I V R Ring r Base R I V
29 2 28 5 sylancr I V R Ring r Base R x Base R , y I y V
30 29 7 syl I V R Ring r Base R x Base R , y I y = M
31 30 eqcomd I V R Ring r Base R M = x Base R , y I y
32 31 10 11 13 13 ovmpod I V R Ring r Base R r M I = I
33 32 32 oveq12d I V R Ring r Base R r M I + M r M I = I + M I
34 33 26 eqtrd I V R Ring r Base R r M I + M r M I = I
35 14 27 34 3eqtr4d I V R Ring r Base R r M I + M I = r M I + M r M I