theorem Th111:
       L is subst-correct vf-qc-correct implies
  \ex(x,y,A)\iff\not\for(x,y,\notA) in G
  proof
    assume
A1: L is subst-correct vf-qc-correct;
A2: \ex(x,y,A)\iff\not\for(x,\not\ex(y,A)) in G by Th105;
    \ex(y,A)\iff\not\for(y,\notA) in G &
    \for(y,\notA)\iff\not\not\for(y,\notA) in G by Th66,Th105;
    then \not\ex(y,A)\iff\not\not\for(y,\notA) in G &
    \not\not\for(y,\notA)\iff\for(y,\notA) in G by Th90,Th94;
    then \not\ex(y,A)\iff\for(y,\notA) in G by Th91;
    then \not\ex(y,A)\imp\for(y,\notA) in G &
    \for(y,\notA)\imp\not\ex(y,A) in G by Th43;
    then
A3: \for(x,\not\ex(y,A)\imp\for(y,\notA)) in G &
    \for(x,\for(y,\notA)\imp\not\ex(y,A)) in G by Def39;
    \for(x,\not\ex(y,A)\imp\for(y,\notA))\imp
    (\for(x,\not\ex(y,A))\imp\for(x,\for(y,\notA))) in G &
    \for(x,\for(y,\notA)\imp\not\ex(y,A))\imp
    (\for(x,\for(y,\notA))\imp\for(x,\not\ex(y,A))) in G by A1,Th109;
    then \for(x,\not\ex(y,A))\imp\for(x,\for(y,\notA)) in G &
    \for(x,\for(y,\notA))\imp\for(x,\not\ex(y,A)) in G by A3,Def38;
    then \for(x,\not\ex(y,A))\iff\for(x,y,\notA) in G by Th43;
    then \not\for(x,\not\ex(y,A))\iff\not\for(x,y,\notA) in G by Th94;
    hence thesis by A2,Th91;
  end;
