
theorem teven:
for R being Ring
for a being Element of R
for n being even Nat holds (-a) |^ n = a |^ n
proof
let R be Ring, a be Element of R, n be even Nat;
defpred P[Nat] means $1 is even implies (-a)|^($1) = a|^($1);
A: now let k be Nat;
   assume A2: for n being Nat st n < k holds P[n];
   per cases;
   suppose k is odd;
     hence P[k];
     end;
   suppose A3: k is even;
     now per cases by NAT_1:23;
     case A4: k = 0;
       then (-a)|^k = 1_R by BINOM:8 .= a|^k by A4,BINOM:8;
       hence P[k];
       end;
     case k = 1;
       then k = 2 * 0 + 1;
       hence P[k];
       end;
     case k >= 2;
       then k-2 in NAT by INT_1:5;
       then reconsider k2 = k-2 as Nat;
       A4: k2 + 2 = k; then
       A5: k2 is even by A3;
       (-a)|^k = (-a)|^(k2+2)
              .= ((-a)|^k2) * ((-a)|^2) by BINOM:10
              .= (a|^k2) * (-a)|^(1+1) by A2,A5,A4,NAT_1:16
              .= (a|^k2) * ((-a)|^1 * (-a)|^1) by BINOM:10
              .= (a|^k2) * ((-a) * (-a)|^1) by BINOM:8
              .= (a|^k2) * ((-a) * (-a)) by BINOM:8
              .= (a|^k2) * (a * a) by VECTSP_1:10
              .= (a|^k2) * (a|^2) by RING_5:3
              .= a|^k by A4,BINOM:10;
       hence P[k];
       end;
     end;
     hence P[k];
     end;
   end;
for k being Nat holds P[k] from NAT_1:sch 4(A);
hence thesis;
end;
