
theorem th0e:
for n being Nat holds exp(2,n) = 2|^n
proof
let n be Nat;
defpred P[Nat] means exp(2,$1) = 2|^($1);
IA: P[0]
    proof
    thus exp(2,0) = 1 by CARD_2:25 .= 2|^0 by NEWTON:4;
    end;
IS: now let k be Nat;
    assume IV: P[k];
    exp(2,k+1) = exp(2,k+`1) by th0a
              .= exp(2,k) *` exp(2,1) by CARD_2:28
              .= (2|^k) *` 2 by IV,CARD_2:27
              .= (2|^k) * 2 by th0a
              .= 2|^(k+1) by NEWTON:6;
    hence P[k+1];
    end;
for k being Nat holds P[k] from NAT_1:sch 2(IA,IS);
hence thesis;
end;
