theorem Th27:
  [s0,s1] in R implies ex pai being inf_path of R st pai.0 = s0 & pai.1=s1
proof
  consider pai1 be inf_path of R such that
A1: pai1.0 = s1 by Th25;
  deffunc F(object) = PrePath($1,s0,pai1);
A2: for x being object st x in NAT holds F(x) in S;
  consider pai be sequence of S such that
A3: for n being object st n in NAT holds pai.n= F(n) from FUNCT_2:sch 2(A2);
  assume
A4: [s0,s1] in R;
  for n being Nat holds [pai.n,pai.(n+1)] in R
  proof
    let n be Nat;
    set n1=n+1;
    set n0=n-1;
    per cases;
    suppose
A5:   n = 0;
      then
A6:   k_nat(k_nat(n1)-1) = k_nat(1-1) by Def2
        .= 0 by Def2;
A7:   pai.n = F(n) by A3,A5
        .= s0 by A5,Def61;
      pai.n1 = F(n1) by A3
        .= s1 by A1,A6,Def61;
      hence thesis by A4,A7;
    end;
    suppose
A8:   n<>0;
      then reconsider n0 as Element of NAT by NAT_1:20;
A9:   pai.n1 = F(n1) by A3
        .= pai1.(k_nat(k_nat(n1)-1)) by Def61
        .= pai1.(k_nat(n1-1)) by Def2
        .= pai1.(n0+1) by Def2;
A10:    n in NAT by ORDINAL1:def 12;
      then
      pai.n = F(n) by A3
        .= pai1.(k_nat(k_nat(n)-1)) by A8,Def61
        .= pai1.(k_nat(n-1)) by Def2,A10
        .= pai1.n0 by Def2;
      hence thesis by A9,Def39;
    end;
  end;
  then reconsider pai as inf_path of R by Def39;
A11: pai.0= F(0) by A3
    .= s0 by Def61;
  take pai;
  pai.1 = F(1) by A3
    .= pai1.(k_nat(k_nat(1)-1)) by Def61
    .= pai1.(k_nat(1-1)) by Def2
    .= s1 by A1,Def2;
  hence thesis by A11;
end;
