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 = N Mat R
minmar1fval.b B = Base A
minmar1fval.q Q = N minMatR1 R
minmar1fval.o 1 ˙ = 1 R
minmar1fval.z 0 ˙ = 0 R
Assertion minmar1val M B K N L N K Q M L = i N , j N if i = K if j = L 1 ˙ 0 ˙ i M j

Proof

Step Hyp Ref Expression
1 minmar1fval.a A = N Mat R
2 minmar1fval.b B = Base A
3 minmar1fval.q Q = N minMatR1 R
4 minmar1fval.o 1 ˙ = 1 R
5 minmar1fval.z 0 ˙ = 0 R
6 1 2 3 4 5 minmar1val0 M B Q M = k N , l N i N , j N if i = k if j = l 1 ˙ 0 ˙ i M j
7 6 3ad2ant1 M B K N L N Q M = k N , l N i N , j N if i = k if j = l 1 ˙ 0 ˙ i M j
8 simp2 M B K N L N K N
9 simpl3 M B K N L N k = K L N
10 1 2 matrcl M B N Fin R V
11 10 simpld M B N Fin
12 11 11 jca M B N Fin N Fin
13 12 3ad2ant1 M B K N L N N Fin N Fin
14 13 adantr M B K N L N k = K l = L N Fin N Fin
15 mpoexga N Fin N Fin i N , j N if i = k if j = l 1 ˙ 0 ˙ i M j V
16 14 15 syl M B K N L N k = K l = L i N , j N if i = k if j = l 1 ˙ 0 ˙ i M j V
17 eqeq2 k = K i = k i = K
18 17 adantr k = K l = L i = k i = K
19 eqeq2 l = L j = l j = L
20 19 ifbid l = L if j = l 1 ˙ 0 ˙ = if j = L 1 ˙ 0 ˙
21 20 adantl k = K l = L if j = l 1 ˙ 0 ˙ = if j = L 1 ˙ 0 ˙
22 18 21 ifbieq1d k = K l = L if i = k if j = l 1 ˙ 0 ˙ i M j = if i = K if j = L 1 ˙ 0 ˙ i M j
23 22 mpoeq3dv k = K l = L i N , j N if i = k if j = l 1 ˙ 0 ˙ i M j = i N , j N if i = K if j = L 1 ˙ 0 ˙ i M j
24 23 adantl M B K N L N k = K l = L i N , j N if i = k if j = l 1 ˙ 0 ˙ i M j = i N , j N if i = K if j = L 1 ˙ 0 ˙ i M j
25 8 9 16 24 ovmpodv2 M B K N L N Q M = k N , l N i N , j N if i = k if j = l 1 ˙ 0 ˙ i M j K Q M L = i N , j N if i = K if j = L 1 ˙ 0 ˙ i M j
26 7 25 mpd M B K N L N K Q M L = i N , j N if i = K if j = L 1 ˙ 0 ˙ i M j