
theorem LM01:
  for m,n be Nat, D be non empty set,
  F1,F2 be Element of m-tuples_on (n-tuples_on D) st for i,j be Nat
  st i in Seg m & j in Seg n
  holds (F1.i).j = (F2.i).j holds F1 = F2
  proof
    let m,n be Nat, D be non empty set,
    F1,F2 be Element of m-tuples_on (n-tuples_on D);
    assume
AS: for i,j be Nat st i in Seg m & j in Seg n
    holds (F1.i).j = (F2.i).j;
    F1 in m-tuples_on (n-tuples_on D);
    then
P1: ex s be Element of (n-tuples_on D)* st F1 = s & len s = m;
    F2 in m-tuples_on (n-tuples_on D);
    then
P2: ex s be Element of (n-tuples_on D)* st F2 = s & len s = m;
    now let i be Nat;
      assume 1 <= i & i <= len F1;
      then
P4:   i in Seg m by P1;
      then i in dom F1 by FINSEQ_1:def 3,P1;
      then F1.i in rng F1 by FUNCT_1:3;
      then F1.i in n-tuples_on D;
      then
P6:   ex s be Element of D* st F1.i = s & len s = n;
      then reconsider F1i = F1.i as Element of D*;
      i in dom F2 by FINSEQ_1:def 3,P2,P4;
      then F2.i in rng F2 by FUNCT_1:3;
      then F2.i in n-tuples_on D;
      then
R6:   ex s be Element of D* st F2.i = s & len s = n;
      then reconsider F2i = F2.i as Element of D*;
      now let j be Nat;
      assume 1 <= j & j <= len F1i;
      then j in Seg n by P6;
      hence F1i.j = F2i.j by AS,P4;
    end;
    hence
    F1.i = F2.i by P6,R6,FINSEQ_1:14;
  end;
  hence F1 = F2 by P1,P2,FINSEQ_1:14;
end;
