
theorem Z3:
for R being Ring,
    S being RingExtension of R
for a being Element of R, b being Element of S st b = a
for i being Integer holds i '*' a = i '*' b
proof
let R be Ring, S be RingExtension of R;
let a be Element of R, b be Element of S;
assume AS: b = a;
let i be Integer;
K: R is Subring of S by FIELD_4:def 1;
defpred P[Integer] means
  for k being Integer st k = $1 holds k'*'a = k'*'b;
now let k be Integer;
  assume A1: k = 0;
  hence k'*'a = 0.R by RING_3:59
             .= 0.S by K,C0SP1:def 3 .= k'*'b by A1,RING_3:59;
  end; then
A2: P[0];
A6: 1 '*' a = b by AS,RING_3:60 .= 1 '*' b by RING_3:60; then
A7: -(1 '*' a) = -(1 '*' b) by K,FIELD_6:17;
A3: for u being Integer holds P[u] implies P[u - 1] & P[u + 1]
   proof
   let u be Integer;
   assume P[u]; then
   A4: u'*'a = u'*'b;
   now let k be Integer;
     assume A5: k = u-1;
     hence k'*'a
         = u'*'a - 1'*'a by RING_3:64
        .= u'*'b - 1'*'b by A4,A7,K,FIELD_6:15
        .= k'*'b by A5,RING_3:64;
     end;
   hence P[u-1];
   now let k be Integer;
     assume A5:k = u+1;
     hence k'*'a
         = u'*'a + 1'*'a by RING_3:62
        .= u'*'b + 1'*'b by A4,A6,K,FIELD_6:15
        .= k'*'b by A5,RING_3:62;
     end;
   hence P[u+1];
   end;
 for i being Integer holds P[i] from INT_1:sch 4(A2,A3);
hence thesis;
end;
