reserve A,B,O for Ordinal,
        o for object,
        x,y,z for Surreal,
        n,m for Nat;

theorem Th14:
  for i,j be Integer holds (uInt.i) + (uInt.j) == uInt.(i+j)
  proof
A1: uInt.0=0_No by Def1;
    defpred P[Nat] means for n,m st n+m=$1
      holds (uInt.n) + (uInt.-m) == uInt.(n-m);
A2: P[0]
    proof
      let n,m;
      assume n+m=0;
      then n=0 & m=0;
      hence thesis by A1;
    end;
A3: for k be Nat st P[k] holds P[k+1]
    proof
      let k be Nat such that
A4:   P[k];
      let n,m such that
A5:   n+m=k+1;
      per cases;
      suppose n=0 or m=0;
        hence thesis by A1;
      end;
      suppose n<>0 & m<>0;
        then reconsider n1=n-1,m1=m-1 as Nat by NAT_1:20;
A6:     uInt.(n1+1) = [{uInt.n1},{}] & uInt.-(m1+1) = [{},{uInt.-m1}]
        by Def1;
        then
A7:     {(uInt.n)} ++ L_(uInt.-m)={} = R_(uInt.n) ++ {(uInt.-m)}
        by SURREALR:27;
        n1+m =k by A5;
        then
A8:     (uInt.n1) + (uInt.-m) == uInt.(n1-m) by A4;
        L_(uInt.n) ++ {(uInt.-m)}
        = {(uInt.n1) + (uInt.-m)} by SURREALR:36,A6;
        then
A9:     L_(uInt.n) ++ {(uInt.-m)} <==> {uInt.(n-m-1)} by A8,SURREALO:32;
        n+m1 = k by A5;
        then
A10:    (uInt.n) + (uInt.-m1) == uInt.(n-m1) by A4;
        {(uInt.n)} ++ R_(uInt.-m)
        = {(uInt.n) + (uInt.-m1)} by A6,SURREALR:36;
        then
A11:    {(uInt.n)} ++ R_(uInt.-m) <==> {uInt.(n-m+1)}
        by A10,SURREALO:32;
        reconsider ss = [{uInt.(n-m-1)},{uInt.(n-m+1)}] as Surreal by Th10;
A12:    ss == uInt.(n-m) by Th10;
        (uInt.n) + (uInt.-m) = [(L_(uInt.n) ++ {(uInt.-m)})\/{},
        {} \/({(uInt.n)} ++ R_(uInt.-m))] by A7,SURREALR:28
        .= [L_(uInt.n) ++ {(uInt.-m)},{(uInt.n)} ++ R_(uInt.-m)];
        then (uInt.n) + (uInt.-m) == ss by A9,A11,SURREALO:29;
        hence thesis by A12,SURREALO:4;
      end;
    end;
A13:for k be Nat holds P[k] from NAT_1:sch 2(A2,A3);
    let i,j be Integer;
    i in INT by INT_1:def 2;
    then consider k be Nat such that
A14:i=k or i=-k by INT_1:def 1;
    j in INT by INT_1:def 2;
    then consider n be Nat such that
A15:j=n or j=-n by INT_1:def 1;
    per cases by A14,A15;
    suppose i=k & j=n;
      hence thesis by Th13;
    end;
    suppose
A16:  i=-k & j=n;
      P[n+k] by A13;
      then (uInt.n) + (uInt.-k) == uInt.(n-k);
      hence thesis by A16;
    end;
    suppose
A17:  i=k & j=-n;
      P[k+n] by A13;
      then (uInt.k) + (uInt.-n) == uInt.(k-n);
      hence thesis by A17;
    end;
    suppose
A18: i=-k & j=-n;
      then (uInt.i) + (uInt.j) = (-uInt.k) + (uInt.-n) by Th12
      .=(-uInt.k) + (-uInt.n) by Th12
      .= -(uInt.k + uInt.n) by SURREALR:40
      .= -(uInt.(k +n)) by Th13
      .= uInt.-(k +n) by Th12;
      hence thesis by A18;
    end;
end;
