theorem
       L is subst-correct vf-qc-correct implies
  \for(x,A)\iff\for(x,\not\notA) in G
  proof assume
A1: L is subst-correct vf-qc-correct;
    A\imp\not\notA in G & \not\notA\impA in G by Th65,Th64;
    then \for(x,A)\imp\for(x,\not\notA) in G &
    \for(x,\not\notA)\imp\for(x,A) in G by A1,Th115;
    then
A2: (\for(x,A)\imp\for(x,\not\notA))\and
    (\for(x,\not\notA)\imp\for(x,A)) in G by Th35;
    (\for(x,A)\imp\for(x,\not\notA))\and
    (\for(x,\not\notA)\imp\for(x,A))\imp(\for(x,A)\iff\for(x,\not\notA)) in G
    by Def38;
    hence thesis by A2,Def38;
  end;
