theorem
  L is subst-correct vf-qc-correct implies
  \for(x,A\andA)\iff\for(x,A) in G
  proof
    assume A1: L is subst-correct vf-qc-correct;
    A\andA\impA in G & A\impA\andA in G by Def38,Th53;
    then
A2: \for(x,A\andA\impA) in G & \for(x,A\impA\andA) in G by Def39;
    \for(x,A\impA\andA)\imp(\for(x,A)\imp\for(x,A\andA)) in G &
    \for(x,A\andA\impA)\imp(\for(x,A\andA)\imp\for(x,A)) in G by A1,Th109;
    then (\for(x,A)\imp\for(x,A\andA)) in G &
    (\for(x,A\andA)\imp\for(x,A)) in G by A2,Def38;
    hence \for(x,A\andA)\iff\for(x,A) in G by Th43;
  end;
