let K be Field; :: thesis: for A, B being Matrix of K
for N being finite without_zero Subset of NAT st N c= dom A & not N is empty & dom A = dom B & ( for i being Nat st i in (dom A) \ N holds
( Line (A,i) = () |-> (0. K) & Line (B,i) = () |-> (0. K) ) ) holds
Solutions_of (A,B) = Solutions_of ((Segm (A,N,(Seg ()))),(Segm (B,N,(Seg ()))))

let A, B be Matrix of K; :: thesis: for N being finite without_zero Subset of NAT st N c= dom A & not N is empty & dom A = dom B & ( for i being Nat st i in (dom A) \ N holds
( Line (A,i) = () |-> (0. K) & Line (B,i) = () |-> (0. K) ) ) holds
Solutions_of (A,B) = Solutions_of ((Segm (A,N,(Seg ()))),(Segm (B,N,(Seg ()))))

let N be finite without_zero Subset of NAT; :: thesis: ( N c= dom A & not N is empty & dom A = dom B & ( for i being Nat st i in (dom A) \ N holds
( Line (A,i) = () |-> (0. K) & Line (B,i) = () |-> (0. K) ) ) implies Solutions_of (A,B) = Solutions_of ((Segm (A,N,(Seg ()))),(Segm (B,N,(Seg ())))) )

assume that
A1: N c= dom A and
A2: ( not N is empty & dom A = dom B & ( for i being Nat st i in (dom A) \ N holds
( Line (A,i) = () |-> (0. K) & Line (B,i) = () |-> (0. K) ) ) ) ; :: thesis: Solutions_of (A,B) = Solutions_of ((Segm (A,N,(Seg ()))),(Segm (B,N,(Seg ()))))
dom A = Seg (len A) by FINSEQ_1:def 3;
then rng (Sgm N) = N by ;
hence Solutions_of (A,B) = Solutions_of ((Segm (A,N,(Seg ()))),(Segm (B,N,(Seg ())))) by A1, A2, Th43; :: thesis: verum