theorem
  [i,j] in Indices ((n,m)-->a) implies ((n,m)-->a) * (i,j)=a
proof
  reconsider m1=m as Nat;
  set M=(n,m)-->a;
  assume
A1: [i,j] in Indices M;
  then i in dom M by ZFMISC_1:87;
  then i in Seg len M by FINSEQ_1:def 3;
  then
A2: i in Seg n by Def2;
  then
A3: n > 0;
  j in Seg width M by A1,ZFMISC_1:87;
  then j in Seg m by A3,Th23;
  then
A4: (m1|->a).j = a by FUNCOP_1:7;
  M.i=m1|->a by A2,FUNCOP_1:7;
  hence thesis by A1,A4,Def5;
end;
