theorem
  (i+j)|->x = (i|->x)^(j|->x)
proof
A0: j is Nat by TARSKI:1;
  defpred P[Nat] means (i+$1)|->x = (i|->x)^($1|->x);
A1: for j be Nat st P[j] holds P[j+1]
  proof
    let j be Nat such that
A2: (i+j)|->x = (i|->x)^(j|->x);
    thus (i+(j+1))|->x = (i+j+1)|->x .= ((i+j)|->x) ^ <*x*> by Th58
      .= (i|->x)^((j|->x) ^ <*x*>) by A2,FINSEQ_1:32
      .= (i|->x)^((j+1)|->x) by Th58;
  end;
  (i+0)|->x = (i|->x)^{} by FINSEQ_1:34
    .= (i|->x)^(0|->x);
  then
A3: P[0];
  for j be Nat holds P[j] from NAT_1:sch 2(A3,A1);
  hence thesis by A0;
end;
