Metamath Proof Explorer


Theorem resmhm2

Description: One direction of resmhm2b . (Contributed by Mario Carneiro, 18-Jun-2015)

Ref Expression
Hypothesis resmhm2.u U = T 𝑠 X
Assertion resmhm2 F S MndHom U X SubMnd T F S MndHom T

Proof

Step Hyp Ref Expression
1 resmhm2.u U = T 𝑠 X
2 mhmrcl1 F S MndHom U S Mnd
3 submrcl X SubMnd T T Mnd
4 2 3 anim12i F S MndHom U X SubMnd T S Mnd T Mnd
5 eqid Base S = Base S
6 eqid Base U = Base U
7 5 6 mhmf F S MndHom U F : Base S Base U
8 1 submbas X SubMnd T X = Base U
9 eqid Base T = Base T
10 9 submss X SubMnd T X Base T
11 8 10 eqsstrrd X SubMnd T Base U Base T
12 fss F : Base S Base U Base U Base T F : Base S Base T
13 7 11 12 syl2an F S MndHom U X SubMnd T F : Base S Base T
14 eqid + S = + S
15 eqid + U = + U
16 5 14 15 mhmlin F S MndHom U x Base S y Base S F x + S y = F x + U F y
17 16 3expb F S MndHom U x Base S y Base S F x + S y = F x + U F y
18 17 adantlr F S MndHom U X SubMnd T x Base S y Base S F x + S y = F x + U F y
19 eqid + T = + T
20 1 19 ressplusg X SubMnd T + T = + U
21 20 ad2antlr F S MndHom U X SubMnd T x Base S y Base S + T = + U
22 21 oveqd F S MndHom U X SubMnd T x Base S y Base S F x + T F y = F x + U F y
23 18 22 eqtr4d F S MndHom U X SubMnd T x Base S y Base S F x + S y = F x + T F y
24 23 ralrimivva F S MndHom U X SubMnd T x Base S y Base S F x + S y = F x + T F y
25 eqid 0 S = 0 S
26 eqid 0 U = 0 U
27 25 26 mhm0 F S MndHom U F 0 S = 0 U
28 27 adantr F S MndHom U X SubMnd T F 0 S = 0 U
29 eqid 0 T = 0 T
30 1 29 subm0 X SubMnd T 0 T = 0 U
31 30 adantl F S MndHom U X SubMnd T 0 T = 0 U
32 28 31 eqtr4d F S MndHom U X SubMnd T F 0 S = 0 T
33 13 24 32 3jca F S MndHom U X SubMnd T F : Base S Base T x Base S y Base S F x + S y = F x + T F y F 0 S = 0 T
34 5 9 14 19 25 29 ismhm F S MndHom T S Mnd T Mnd F : Base S Base T x Base S y Base S F x + S y = F x + T F y F 0 S = 0 T
35 4 33 34 sylanbrc F S MndHom U X SubMnd T F S MndHom T