assume ex k, l, m, n being Nat st
( 0 < k & k < l & l < m & m < n & (Fib m) - (Fib l) = (Fib l) - (Fib k) & (Fib l) - (Fib k) = (Fib n) - (Fib m) & (Fib l) - (Fib k) > 0 ) ; :: thesis: contradiction
then consider k, l, m, n being Nat such that
A1: ( 0 < k & k < l & l < m & m < n & (Fib m) - (Fib l) = (Fib l) - (Fib k) & (Fib l) - (Fib k) = (Fib n) - (Fib m) & (Fib l) - (Fib k) > 0 ) ;

per cases ( ( ( not k = 2 or not l = 3 or not m = 4 ) & ( not k = 1 or not l = 4 or not m = 5 ) ) or ( k = 2 & l = 3 & m = 4 ) or ( k = 1 & l = 4 & m = 5 ) ) ;

suppose ( ( not k = 2 or not l = 3 or not m = 4 ) & ( not k = 1 or not l = 4 or not m = 5 ) ) ; :: thesis: contradiction

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 contradiction by A1, A2, ba, B1, b2, FIB_NUM2:46; :: thesis: verum

end;

suppose ( ( k = 2 & l = 3 & m = 4 ) or ( k = 1 & l = 4 & m = 5 ) ) ; :: thesis: contradiction

per cases then ( ( k = 2 & l = 3 & m = 4 ) or ( k = 1 & l = 4 & m = 5 ) ) ;

suppose SU: ( k = 2 & l = 3 & m = 4 ) ; :: thesis: contradiction

then (Fib l) - (Fib k) = 1 by FIB_NUM2:21, FIB_NUM2:22;
hence contradiction by A1, LemmaGe1, SU; :: thesis: verum

end;

suppose SU: ( k = 1 & l = 4 & m = 5 ) ; :: thesis: contradiction

then (Fib l) - (Fib k) = 2 by FIB_NUM2:23, PRE_FF:1;
hence contradiction by A1, LemmaGe2a, SU; :: thesis: verum

end;

end;

end;

end;