let r be Real; :: thesis: ( (scf r) . 0 > 0 implies for n being Nat holds (c_n r) . n in NAT )
set s1 = c_n r;
set s = scf r;
defpred S2[ Nat] means (c_n r) . $1 in NAT ;
A1: for n being Nat st S2[n] & S2[n + 1] holds
S2[n + 2]
proof
let n be Nat; :: thesis: ( S2[n] & S2[n + 1] implies S2[n + 2] )
assume that
A2: (c_n r) . n in NAT and
A3: (c_n r) . (n + 1) in NAT ; :: thesis: S2[n + 2]
reconsider n2 = (c_n r) . (n + 1) as Element of NAT by A3;
reconsider n1 = (c_n r) . n as Element of NAT by A2;
n + 2 >= 0 + 1 by XREAL_1:7;
then reconsider n3 = (scf r) . (n + 2) as Element of NAT by Th38, INT_1:3;
(n3 * n2) + n1 in NAT ;
hence S2[n + 2] by Def5; :: thesis: verum
end;
assume A4: (scf r) . 0 > 0 ; :: thesis: for n being Nat holds (c_n r) . n in NAT
A5: S2[1]
proof
reconsider n2 = (scf r) . 0 as Element of NAT by A4, INT_1:3;
reconsider n1 = (scf r) . 1 as Element of NAT by Th38, INT_1:3;
(n1 * n2) + 1 in NAT ;
hence S2[1] by Def5; :: thesis: verum
end;
let n be Nat; :: thesis: (c_n r) . n in NAT
(scf r) . 0 in NAT by A4, INT_1:3;
then A6: S2[ 0 ] by Def5;
for n being Nat holds S2[n] from FIB_NUM:sch 1(A6, A5, A1);
 hence (c_n r) . n in NAT ; :: thesis: verum