
theorem mm5a:
for R being Ring
for S being RingExtension of R
for a being Element of R, b being Element of S
for n being Element of NAT st a = b holds n * a = n * b
proof
let F be Ring, E be RingExtension of F, a be Element of F,
    b be Element of E, n be Element of NAT;
assume AS: a = b;
defpred P[Nat] means ($1) * a = ($1) * b;
H: F is Subring of E by FIELD_4:def 1;
0 * a = 0.F by BINOM:12 .= 0.E by H,C0SP1:def 3
     .= 0 * b by BINOM:12; then
A: P[0];
B: now let k be Nat;
   assume C: P[k];
   (k+1) * a = k*a + 1*a by BINOM:15
            .= k*a + a by BINOM:13
            .= k*b + b by H,C,AS,FIELD_6:15
            .= k*b + 1*b by BINOM:13
            .= (k+1) * b by BINOM:15;
   hence P[k+1];
   end;
for k being Nat holds P[k] from NAT_1:sch 2(A,B);
hence thesis;
end;
