theorem Th29:
  M|=F implies M^\i|=F
 proof
  assume A1: M|=F;
  thus M^\i|=F
  proof
   let p;
   assume A2: p in F;
   thus M^\i|=p
   proof
    let n;
    (SAT M).[i+n,p]=1 by A2,A1,Def12;
    hence (SAT(M^\i)).[n,p]=1 by Th28;
   end;
  end;
 end;
