reserve a,b,d,n,k,i,j,x,s for Nat;

theorem Th9:
  k < n <= 7 implies ex i st Fib(i) mod n = k
proof
  assume
A1: k < n <= 7;
  n=0 or ... or n=7 by A1;
  then per cases by A1;
    suppose
A2:     n=1;
      then
A3:     k=0 by A1,NAT_1:14;
      1*1+0 mod 1 = 0;
      hence thesis by A2,A3,PRE_FF:1;
    end;
    suppose
A4:     n=2;
      then
A5:     k=0 or ... or k=2 by A1;
      2*0+1 mod 2 = 1 & 2*1+0 mod 2 = 0;
      hence thesis by A1,A4,A5,FIB_NUM2:22,PRE_FF:1;
    end;
    suppose
A6:     n=3;
      then
A7:     k=0 or ... or k=3 by A1;
      3*0+2 mod 3 = 2 & 3*0+1 mod 3 = 1 & 3*1+0 mod 3 = 0 by NUMBER02:16;
      hence thesis by A1,A6,A7,FIB_NUM2:22,23,PRE_FF:1;
    end;
    suppose
A8:     n=4;
      then
A9:     k=0 or ... or k=4 by A1;
      4*0+1 mod 4 = 1 & 4*0+2 mod 4 = 2 & 4*0+3 mod 4 = 3 &
      4*2+0 mod 4 = 0 by NUMBER02:16;
      hence thesis by A1,A8,A9,FIB_NUM2:21,22,23,Th7;
    end;
    suppose
A10:    n=5;
      then
A11:    k=0 or ... or k=5 by A1;
      5*1+0 mod 5 = 0 & 5*0+1 mod 5 = 1 & 5*0+2 mod 5 = 2 &
      5*0+3 mod 5 = 3 & 5*6+4 mod 5 = 4 by NUMBER02:16;
      hence thesis by A1,A10,A11,FIB_NUM2:22,23,PRE_FF:1,Th7,NUMBER06:46;
    end;
    suppose
A12:    n=6;
      then
A13:    k=0 or ... or k=6 by A1;
      24*6+0 mod 6 = 0 & 6*0+1 mod 6 = 1 & 6*0+2 mod 6 = 2 &
      6*0+3 mod 6 = 3 & 6*5+4 mod 6 = 4 & 6*0+5 mod 6 = 5 by NUMBER02:16;
      hence thesis by A1,A12,A13,FIB_NUM2:22,23,NUMBER06:46,PRE_FF:1,Th7;
    end;
    suppose
A14:    n=7;
      then
A15:    k=0 or ... or k=7 by A1;
      3*7+0 mod 7 = 0 & 7*0+1 mod 7 = 1 & 7*0+2 mod 7 = 2 &
      7*0+3 mod 7 = 3 & 20*7+4 mod 7 = 4 & 7*0+5 mod 7 = 5 &
      7*1+6 mod 7 = 6 by NUMBER02:16;
      hence thesis by A1,A14,A15,FIB_NUM2:21,22,23,NUMBER06:46,Th7;
    end;
end;
