reserve r1,r2,r3 for non negative Real;
reserve n,m1 for Nat;
reserve s for Real;
reserve cn,cd,i1,j1 for Integer;
reserve r for irrational Real;
reserve q for Rational;

theorem Th5:
   n > 1 implies
   |.r-c_n(r).n/c_d(r).n.| < 1/(sqrt 5 * c_d(r).n|^2) or
   |.r-c_n(r).(n+1)/c_d(r).(n+1).| < 1/(sqrt 5*c_d(r).(n+1)|^2) or
   |.r-c_n(r).(n+2)/c_d(r).(n+2).| < 1/(sqrt 5*c_d(r).(n+2)|^2)
   proof
     assume that
A2:  n > 1 and
A3:  not (|.r-c_n(r).n/c_d(r).n.| < 1/(sqrt 5*c_d(r).n|^2) or
     |.r-c_n(r).(n+1)/c_d(r).(n+1).| < 1/(sqrt 5*c_d(r).(n+1)|^2) or
     |.r-c_n(r).(n+2)/c_d(r).(n+2).| < 1/(sqrt 5*c_d(r).(n+2)|^2) );
A4:  sqrt 5>c_d(r).(n+1)/c_d(r).n+1/(c_d(r).(n+1)/c_d(r).n) by A2,A3,Th3;
     set m = n+1;
     m > 1 + 1 by A2,XREAL_1:8; then
     m > 1 by XXREAL_0:2; then
A7:  sqrt 5 > c_d(r).(m+1)/c_d(r).m+1/(c_d(r).(m+1)/c_d(r).m) by A3,Th3;
A10: c_d(r).(n+2)/c_d(r).(n+1) =
     (scf(r).(n+2)*c_d(r).(n+1)+c_d(r).n)/c_d(r).m by REAL_3:def 6
     .=scf(r).(n+2)*(c_d(r).(n+1)*(c_d(r).m)")+c_d(r).n/c_d(r).(n+1)
     .=scf(r).(n+2)*1 + c_d(r).n/c_d(r).m by XCMPLX_0:def 7
     .=scf(r).(n+2) + (c_d(r).m/c_d(r).n)" by XCMPLX_1:213;
     c_d(r).(n+2)/c_d(r).m > c_d(r).m/c_d(r).m
       by XREAL_1:74,DIOPHAN1:8; then
A11: c_d(r).(n+2)/c_d(r).(n+1) > 1 by XCMPLX_1:60;
     set m1 = n -1;
     c_d(r).(m1+2)/c_d(r).(m1+1) > c_d(r).(m1+1)/c_d(r).(m1+1)
       by XREAL_1:74,A2,DIOPHAN1:8; then
     c_d(r).(n-1+2)/c_d(r).(n-1+1) > 1 by XCMPLX_1:60; then
A13: 1/(c_d(r).(n+1)/c_d(r).n) > (sqrt 5 -1)/2 by A4,Th1;
     scf(r).(n+1+1) > 0 by DIOPHAN1:5; then
     scf(r).(n+2) >= 0+1 by INT_1:7; then
     scf(r).(n+2)+1/(c_d(r).(n+1)/c_d(r).n) > 1+(sqrt 5 -1)/2
       by A13,XREAL_1:8;
     hence contradiction by A7,A10,A11,Th1;
   end;
