theorem
       L is subst-correct vf-qc-correct implies
  \for(x,\notA\imp\notB)\imp(\for(x,B)\imp\for(x,A)) in G
  proof assume
A1: L is subst-correct vf-qc-correct;
    (\notA\imp\notB)\imp(B\impA) in G by Def38;
    then
A2: \for(x,\notA\imp\notB)\imp(\for(x,B\impA)) in G by A1,Th115;
    \for(x,B\impA)\imp(\for(x,B)\imp\for(x,A)) in G by A1,Th109;
    hence \for(x,\notA\imp\notB)\imp(\for(x,B)\imp\for(x,A)) in G by A2,Th45;
  end;
