reserve R for Ring, S for R-monomorphic Ring,
        K for Field, F for K-monomorphic Field,
        T for K-monomorphic comRing;

theorem Th11:
   for f being Monomorphism of K,F st K,F are_disjoint
   holds emb_iso f is additive
   proof
     let f be Monomorphism of K,F;
     assume AS: K,F are_disjoint;
set g = emb_iso f, R = embField f;
    now let a,b be Element of R;
    reconsider x = a, y = b as Element of carr f by defemb;
    per cases;
    suppose A: x in [#]K & y in [#]K; then
    reconsider a1 = a, b1 = b as Element of K;
B:  a + b = (addemb f).(x,y) by defemb  .= addemb(f,x,y) by defadd
         .= a1 + b1 by defaddf;
      a in K & b in K by A; then
C:    g.a = f.a & g.b = f.b by defiso;
      a + b in K by B;
      hence g.(a+b) = f.(a+b) by defiso .= g.a + g.b by C,B,VECTSP_1:def 20;
      end;
    suppose A: x in [#]K & not y in [#]K; then
      reconsider a1 = a as Element of K;
      reconsider b1 = b as Element of F by A,Lm1;
      reconsider fa = f.a1 as Element of F;
B:    a + b = (addemb f).(x,y) by defemb .= addemb(f,x,y) by defadd
           .= fa + b1 by A,defaddf;
      a in K & not b in K by A; then
C:    g.a = f.a & g.b = b by defiso;
      not fa + b1 in K by AS,XBOOLE_0:def 4;
      hence g.(a+b) = g.a + g.b by C,B,defiso;
      end;
    suppose A: y in [#]K & not x in [#]K; then
      reconsider b1 = b as Element of K;
      reconsider a1 = a as Element of F by A,Lm1;
      reconsider fb = f.b1 as Element of F;
B:    a + b = (addemb f).(x,y) by defemb .= addemb(f,x,y) by defadd
           .= a1 + fb by A,defaddf;
      not a in K & b in K by A; then
C:    g.a = a & g.b = f.b by defiso;
      not a1 + fb in K by AS,XBOOLE_0:def 4;
      hence g.(a+b) = g.a + g.b by C,B,defiso;
      end;
    suppose A: not x in [#]K & not y in [#]K &
      (the addF of F).(x,y) in rng f; then
      reconsider a1 = a, b1 = b as Element of F by Lm1;
C:    a + b = (addemb f).(x,y) by defemb .= addemb(f,x,y) by defadd
           .= (f").(a1 + b1) by A,defaddf;
      not a in K & not b in K by A; then
D:    g.a = a & g.b = b by defiso;
      a1 + b1 in dom(f") by A,FUNCT_1:33; then
      (f").(a1 + b1) in rng(f") by FUNCT_1:3; then
      (f").(a1 + b1) in K;
      hence g.(a+b) = f.((f").(a1 + b1)) by C,defiso
                   .= g.a + g.b by D,A,FUNCT_1:35;
      end;
    suppose A: not x in [#]K & not y in [#]K &
       not (the addF of F).(x,y) in rng f; then
       reconsider a1 = a, b1 = b as Element of F by Lm1;
C:     a + b = (addemb f).(x,y) by defemb .= addemb(f,x,y) by defadd
            .= a1 + b1 by A,defaddf;
      not a in K & not b in K by A; then
D:    g.a = a & g.b = b by defiso;
      not a1 + b1 in K by AS,XBOOLE_0:def 4;
      hence g.(a+b) = g.a + g.b by D,C,defiso;
      end;
  end;
hence g is additive;
end;
