
theorem :: Problem 64 for four terms
  not ex k,l,m,n being Nat st 0 < k < l < m < n &
    Fib (m) - Fib (l) = Fib (l) - Fib (k) = Fib n - Fib m &
      Fib l - Fib k > 0
  proof
    assume ex k,l,m,n being Nat st 0 < k < l < m < n &
    Fib (m) - Fib (l) = Fib (l) - Fib (k) = Fib n - Fib m &
      Fib l - Fib k > 0; then
    consider k,l,m,n being Nat such that
A1: 0 < k < l < m < n &
    Fib (m) - Fib (l) = Fib (l) - Fib (k) = Fib n - Fib m &
      Fib l - Fib k > 0;
    per cases;
    suppose
      not (k = 2 & l = 3 & m = 4) &
      not (k = 1 & l = 4 & m = 5); then
A2:   l > 2 & k = l - 2 & m = l + 1 by A1,Problem64Part1; then
      n >= l + 1 + 1 by A1,NAT_1:13; then
P1:   Fib n >= Fib (l+2) by FIB_NUM2:45;
      Fib (l+1+1) = Fib (l+1) + Fib l by PRE_FF:1; then
S1:   Fib (l+2) - Fib (l+1) = Fib l;
      reconsider l1 = l - 1 as Element of NAT by A2,INT_1:5,XXREAL_0:2;
b2:   l - 1 > 2 - 1 by A2,XREAL_1:14;
B1:   Fib n - Fib (l+1) >= Fib l by S1,P1,XREAL_1:13;
ba:   l > l1 by XREAL_1:44;
      Fib (l1+1+1) = Fib (l1+1) + Fib l1 by PRE_FF:1;
      hence thesis by A1,A2,ba,B1,b2,FIB_NUM2:46;
    end;
    suppose not
      (not (k = 2 & l = 3 & m = 4) &
      not (k = 1 & l = 4 & m = 5)); then
      per cases;
      suppose
SU:     k = 2 & l = 3 & m = 4; then
        Fib l - Fib k = 1 by FIB_NUM2:21,22;
        hence thesis by A1,LemmaGe1,SU;
      end;
      suppose
SU:     k = 1 & l = 4 & m = 5; then
        Fib l - Fib k = 2 by FIB_NUM2:23,PRE_FF:1;
        hence thesis by A1,LemmaGe2a,SU;
      end;
    end;
  end;
