Metamath Proof Explorer


Theorem minmar1val

Description: Third substitution for the definition of a matrix for a minor. (Contributed by AV, 31-Dec-2018)

Ref Expression
Hypotheses minmar1fval.a A=NMatR
minmar1fval.b B=BaseA
minmar1fval.q Q=NminMatR1R
minmar1fval.o 1˙=1R
minmar1fval.z 0˙=0R
Assertion minmar1val MBKNLNKQML=iN,jNifi=Kifj=L1˙0˙iMj

Proof

Step Hyp Ref Expression
1 minmar1fval.a A=NMatR
2 minmar1fval.b B=BaseA
3 minmar1fval.q Q=NminMatR1R
4 minmar1fval.o 1˙=1R
5 minmar1fval.z 0˙=0R
6 1 2 3 4 5 minmar1val0 MBQM=kN,lNiN,jNifi=kifj=l1˙0˙iMj
7 6 3ad2ant1 MBKNLNQM=kN,lNiN,jNifi=kifj=l1˙0˙iMj
8 simp2 MBKNLNKN
9 simpl3 MBKNLNk=KLN
10 1 2 matrcl MBNFinRV
11 10 simpld MBNFin
12 11 11 jca MBNFinNFin
13 12 3ad2ant1 MBKNLNNFinNFin
14 13 adantr MBKNLNk=Kl=LNFinNFin
15 mpoexga NFinNFiniN,jNifi=kifj=l1˙0˙iMjV
16 14 15 syl MBKNLNk=Kl=LiN,jNifi=kifj=l1˙0˙iMjV
17 eqeq2 k=Ki=ki=K
18 17 adantr k=Kl=Li=ki=K
19 eqeq2 l=Lj=lj=L
20 19 ifbid l=Lifj=l1˙0˙=ifj=L1˙0˙
21 20 adantl k=Kl=Lifj=l1˙0˙=ifj=L1˙0˙
22 18 21 ifbieq1d k=Kl=Lifi=kifj=l1˙0˙iMj=ifi=Kifj=L1˙0˙iMj
23 22 mpoeq3dv k=Kl=LiN,jNifi=kifj=l1˙0˙iMj=iN,jNifi=Kifj=L1˙0˙iMj
24 23 adantl MBKNLNk=Kl=LiN,jNifi=kifj=l1˙0˙iMj=iN,jNifi=Kifj=L1˙0˙iMj
25 8 9 16 24 ovmpodv2 MBKNLNQM=kN,lNiN,jNifi=kifj=l1˙0˙iMjKQML=iN,jNifi=Kifj=L1˙0˙iMj
26 7 25 mpd MBKNLNKQML=iN,jNifi=Kifj=L1˙0˙iMj