reserve i,j for Nat;
reserve A,B for Ring;

theorem Th11:
  for n being Element of NAT, A,B be Ring, z be Element of A  st
  A is Subring of B holds
  (power B).(In(z,B),n) = In((power A).(z,n),B)
  proof
    let n be Element of NAT, A,B be Ring, z be Element of A;
    assume
A0: A is Subring of B; then
    z is Element of B by Lm6; then
A2: In(z,B) = z by SUBSET_1:def 8;
A3: 1_A = 1_B by A0,C0SP1:def 3;
    reconsider x = z as Element of B by A0,Lm6;
    (power B).(In(z,B),n) = In((power A).(z,n),B)
    proof
      defpred P[Nat] means (power B).(In(z,B),$1) = In((power A).(z,$1),B);
A5:   P[0]
      proof
A6:     In(1.A,B) = 1.B by A0,Lm5;
        In((power A).(z,0),B) = 1.B by A3,GROUP_1:def 7,A6;
        hence thesis by A3,GROUP_1:def 7;
      end;
A7:   for m be Nat st P[m] holds P[m+1]
      proof
        let m be Nat;
        assume
A9:     P[m];
A10:    (power A).(z,m) is Element of B by A0,Lm6;
A11:    In((power A).(z,m),B) = (power A).(z,m)
        by A10,SUBSET_1:def 8;
A12:    (power A).(z,m+1) is Element of B by A0,Lm6;
        (power B).(In(z,B),m+1)
         = (power B).(In(z,B),m)* In(z,B) by GROUP_1:def 7
        .= (power A).(z,m)*z by A0,A11,A2,Th9,A9
        .= (power A).(z,m+1) by GROUP_1:def 7
        .= In((power A).(z,m+1),B) by A12,SUBSET_1:def 8;
        hence thesis;
      end;
      for m be Nat holds P[m] from NAT_1:sch 2(A5,A7);
      hence thesis;
    end;
    hence thesis;
  end;
