defpred S₁[ Nat] means x |^|^ x is odd ;
A1: S₁[ 0 ] by ORDINAL5:13, POLYFORM:4;

A2: now :: thesis: for k being Nat st S₁[k] holds
S₁[k + 1]

let k be Nat; :: thesis: ( S₁[k] implies S₁[k + 1] )
assume S₁[k] ; :: thesis: S₁[k + 1]
x |^|^ (k + 1) = x |^|^ (Segm (k + 1))
.= x |^|^ (succ (Segm k)) by NAT_1:38
.= exp (x,(x |^|^ k)) by ORDINAL5:14
.= x |^ (x |^|^ k) ;
hence S₁[k + 1] by PEPIN:20; :: thesis: verum

end;

for k being Nat holds S₁[k] from NAT_1:sch 2(A1, A2);
hence x |^|^ i is odd ; :: thesis: verum