defpred S₁[ Nat] means not P ^^ P is empty ;
A1: S₁[ 0 ] ;
A10: for k being Nat st S₁[k] holds
S₁[k + 1]

proof

let k be Nat; :: thesis: ( S₁[k] implies S₁[k + 1] )
assume A11: S₁[k] ; :: thesis: S₁[k + 1]
P ^^ (k + 1) = (P ^^ k) ^ P by Th6;
hence S₁[k + 1] by A11; :: thesis: verum

end;

for n being Nat holds S₁[n] from NAT_1:sch 2(A1, A10);
hence not P ^^ n is empty ; :: thesis: verum