Metamath Proof Explorer


Theorem submatminr1

Description: If we take a submatrix by removing the row I and column J , then the result is the same on the matrix with row I and column J modified by the minMatR1 operator. (Contributed by Thierry Arnoux, 25-Aug-2020)

Ref Expression
Hypotheses submateq.a A=1NMatR
submateq.b B=BaseA
submateq.n φN
submateq.i φI1N
submateq.j φJ1N
submatminr1.r φRRing
submatminr1.m φMB
submatminr1.e E=I1NminMatR1RMJ
Assertion submatminr1 φIsubMat1MJ=IsubMat1EJ

Proof

Step Hyp Ref Expression
1 submateq.a A=1NMatR
2 submateq.b B=BaseA
3 submateq.n φN
4 submateq.i φI1N
5 submateq.j φJ1N
6 submatminr1.r φRRing
7 submatminr1.m φMB
8 submatminr1.e E=I1NminMatR1RMJ
9 eqid 1R=1R
10 1 2 9 minmar1marrep RRingMB1NminMatR1RM=M1NmatRRepR1R
11 6 7 10 syl2anc φ1NminMatR1RM=M1NmatRRepR1R
12 11 oveqd φI1NminMatR1RMJ=IM1NmatRRepR1RJ
13 8 12 eqtrid φE=IM1NmatRRepR1RJ
14 eqid BaseR=BaseR
15 14 9 ringidcl RRing1RBaseR
16 6 15 syl φ1RBaseR
17 1 2 marrepcl RRingMB1RBaseRI1NJ1NIM1NmatRRepR1RJB
18 6 7 16 4 5 17 syl32anc φIM1NmatRRepR1RJB
19 13 18 eqeltrd φEB
20 13 3ad2ant1 φi1NIj1NJE=IM1NmatRRepR1RJ
21 20 oveqd φi1NIj1NJiEj=iIM1NmatRRepR1RJj
22 7 3ad2ant1 φi1NIj1NJMB
23 16 3ad2ant1 φi1NIj1NJ1RBaseR
24 4 3ad2ant1 φi1NIj1NJI1N
25 5 3ad2ant1 φi1NIj1NJJ1N
26 simp2 φi1NIj1NJi1NI
27 26 eldifad φi1NIj1NJi1N
28 simp3 φi1NIj1NJj1NJ
29 28 eldifad φi1NIj1NJj1N
30 eqid 1NmatRRepR=1NmatRRepR
31 eqid 0R=0R
32 1 2 30 31 marrepeval MB1RBaseRI1NJ1Ni1Nj1NiIM1NmatRRepR1RJj=ifi=Iifj=J1R0RiMj
33 22 23 24 25 27 29 32 syl222anc φi1NIj1NJiIM1NmatRRepR1RJj=ifi=Iifj=J1R0RiMj
34 eldifsn i1NIi1NiI
35 26 34 sylib φi1NIj1NJi1NiI
36 35 simprd φi1NIj1NJiI
37 36 neneqd φi1NIj1NJ¬i=I
38 37 iffalsed φi1NIj1NJifi=Iifj=J1R0RiMj=iMj
39 21 33 38 3eqtrrd φi1NIj1NJiMj=iEj
40 1 2 3 4 5 7 19 39 submateq φIsubMat1MJ=IsubMat1EJ