let n be Nat; for K being Field
for A, B being Matrix of K
for nt being Element of n -tuples_on NAT st rng nt c= dom A & dom A = dom B & n > 0 & ( for i being Nat st i in (dom A) \ (rng nt) holds
( Line A,i = (width A) |-> (0. K) & Line B,i = (width B) |-> (0. K) ) ) holds
Solutions_of A,B = Solutions_of (Segm A,nt,(Sgm (Seg (width A)))),(Segm B,nt,(Sgm (Seg (width B))))
let K be Field; for A, B being Matrix of K
for nt being Element of n -tuples_on NAT st rng nt c= dom A & dom A = dom B & n > 0 & ( for i being Nat st i in (dom A) \ (rng nt) holds
( Line A,i = (width A) |-> (0. K) & Line B,i = (width B) |-> (0. K) ) ) holds
Solutions_of A,B = Solutions_of (Segm A,nt,(Sgm (Seg (width A)))),(Segm B,nt,(Sgm (Seg (width B))))
let A, B be Matrix of K; for nt being Element of n -tuples_on NAT st rng nt c= dom A & dom A = dom B & n > 0 & ( for i being Nat st i in (dom A) \ (rng nt) holds
( Line A,i = (width A) |-> (0. K) & Line B,i = (width B) |-> (0. K) ) ) holds
Solutions_of A,B = Solutions_of (Segm A,nt,(Sgm (Seg (width A)))),(Segm B,nt,(Sgm (Seg (width B))))
let nt be Element of n -tuples_on NAT ; ( rng nt c= dom A & dom A = dom B & n > 0 & ( for i being Nat st i in (dom A) \ (rng nt) holds
( Line A,i = (width A) |-> (0. K) & Line B,i = (width B) |-> (0. K) ) ) implies Solutions_of A,B = Solutions_of (Segm A,nt,(Sgm (Seg (width A)))),(Segm B,nt,(Sgm (Seg (width B)))) )
assume that
A1:
rng nt c= dom A
and
A2:
dom A = dom B
and
A3:
n > 0
and
A4:
for i being Nat st i in (dom A) \ (rng nt) holds
( Line A,i = (width A) |-> (0. K) & Line B,i = (width B) |-> (0. K) )
; Solutions_of A,B = Solutions_of (Segm A,nt,(Sgm (Seg (width A)))),(Segm B,nt,(Sgm (Seg (width B))))
set SB = Segm B,nt,(Sgm (Seg (width B)));
set SA = Segm A,nt,(Sgm (Seg (width A)));
A5:
Solutions_of (Segm A,nt,(Sgm (Seg (width A)))),(Segm B,nt,(Sgm (Seg (width B)))) c= Solutions_of A,B
proof
A6:
Seg (len A) = dom B
by A2, FINSEQ_1:def 3;
A7:
width (Segm B,nt,(Sgm (Seg (width B)))) = card (Seg (width B))
by A3, MATRIX_1:24;
then A8:
width (Segm B,nt,(Sgm (Seg (width B)))) = width B
by FINSEQ_1:78;
let x be
set ;
TARSKI:def 3 ( not x in Solutions_of (Segm A,nt,(Sgm (Seg (width A)))),(Segm B,nt,(Sgm (Seg (width B)))) or x in Solutions_of A,B )
assume
x in Solutions_of (Segm A,nt,(Sgm (Seg (width A)))),
(Segm B,nt,(Sgm (Seg (width B))))
;
x in Solutions_of A,B
then consider X being
Matrix of
K such that A9:
x = X
and A10:
len X = width (Segm A,nt,(Sgm (Seg (width A))))
and A11:
width X = width (Segm B,nt,(Sgm (Seg (width B))))
and A12:
(Segm A,nt,(Sgm (Seg (width A)))) * X = Segm B,
nt,
(Sgm (Seg (width B)))
;
set AX =
A * X;
width (Segm A,nt,(Sgm (Seg (width A)))) = card (Seg (width A))
by A3, MATRIX_1:24;
then A13:
width (Segm A,nt,(Sgm (Seg (width A)))) = width A
by FINSEQ_1:78;
then A14:
width (A * X) = width X
by A10, MATRIX_3:def 4;
A15:
len (A * X) = len A
by A10, A13, MATRIX_3:def 4;
A16:
now A17:
dom (A * X) = Seg (len A)
by A15, FINSEQ_1:def 3;
let j,
k be
Nat;
( [j,k] in Indices (A * X) implies (A * X) * j,k = B * j,k )assume A18:
[j,k] in Indices (A * X)
;
(A * X) * j,k = B * j,kA19:
k in Seg (width (A * X))
by A18, ZFMISC_1:106;
reconsider j9 =
j,
k9 =
k as
Element of
NAT by ORDINAL1:def 13;
A20:
j in dom (A * X)
by A18, ZFMISC_1:106;
now per cases
( j9 in rng nt or not j9 in rng nt )
;
suppose A21:
j9 in rng nt
;
(A * X) * j,k = B * j,kA22:
dom nt = Seg n
by FINSEQ_2:144;
Sgm (Seg (width B)) = idseq (width B)
by FINSEQ_3:54;
then A23:
(Sgm (Seg (width B))) . k9 = k
by A11, A8, A14, A19, FINSEQ_2:57;
consider p being
set such that A24:
p in dom nt
and A25:
nt . p = j9
by A21, FUNCT_1:def 5;
reconsider p =
p as
Element of
NAT by A24;
Indices (Segm B,nt,(Sgm (Seg (width B)))) = [:(Seg n),(Seg (card (Seg (width B)))):]
by A3, MATRIX_1:24;
then A26:
[p,k] in Indices (Segm B,nt,(Sgm (Seg (width B))))
by A11, A7, A14, A19, A24, A22, ZFMISC_1:106;
Line (Segm A,nt,(Sgm (Seg (width A)))),
p = Line A,
j9
by A24, A25, A22, Lm6;
hence (A * X) * j,
k =
(Line (Segm A,nt,(Sgm (Seg (width A)))),p) "*" (Col X,k)
by A10, A13, A18, MATRIX_3:def 4
.=
(Segm B,nt,(Sgm (Seg (width B)))) * p,
k9
by A10, A12, A26, MATRIX_3:def 4
.=
B * j,
k
by A25, A26, A23, MATRIX13:def 1
;
verum end; suppose
not
j9 in rng nt
;
(A * X) * j,k = B * j,kthen A27:
j9 in (dom A) \ (rng nt)
by A2, A6, A20, A17, XBOOLE_0:def 5;
then A28:
Line B,
j = (width B) |-> (0. K)
by A4;
Line A,
j = (width A) |-> (0. K)
by A4, A27;
hence (A * X) * j,
k =
((width A) |-> (0. K)) "*" (Col X,k)
by A10, A13, A18, MATRIX_3:def 4
.=
Sum ((0. K) * (Col X,k))
by A10, A13, FVSUM_1:80
.=
(0. K) * (Sum (Col X,k))
by FVSUM_1:92
.=
0. K
by VECTSP_1:36
.=
(Line B,j) . k
by A11, A8, A14, A19, A28, FINSEQ_2:71
.=
B * j,
k
by A11, A8, A14, A19, MATRIX_1:def 8
;
verum end; end; end; hence
(A * X) * j,
k = B * j,
k
;
verum end;
len (A * X) = len B
by A15, A6, FINSEQ_1:def 3;
then
A * X = B
by A11, A7, A14, A16, FINSEQ_1:78, MATRIX_1:21;
hence
x in Solutions_of A,
B
by A9, A10, A11, A13, A8;
verum
end;
Solutions_of A,B c= Solutions_of (Segm A,nt,(Sgm (Seg (width A)))),(Segm B,nt,(Sgm (Seg (width B))))
by A1, A3, Th42;
hence
Solutions_of A,B = Solutions_of (Segm A,nt,(Sgm (Seg (width A)))),(Segm B,nt,(Sgm (Seg (width B))))
by A5, XBOOLE_0:def 10; verum