theorem Th121:
       L is subst-correct vf-qc-correct implies
  \for(x,A\impB)\imp(\ex(x,A)\imp\ex(x,B)) in G
  proof assume
A1: L is subst-correct vf-qc-correct;
A2: \for(x,A\impB)\imp(\for(x,\notB)\imp\for(x,\notA)) in G by A1,Th117;
    \for(x,\notB)\imp\for(x,\notA)\imp(\not\for(x,\notA)\imp\not\for(x,\notB))
    in G by Th57;
    then
A3: \for(x,A\impB)\imp(\not\for(x,\notA)\imp\not\for(x,\notB)) in G by A2,Th45;
    \not\ex(x,A)\iff\for(x,\notA) in G by Def39;
    then \for(x,\notA)\imp\not\ex(x,A) in G by Th43;
    then \not\not\ex(x,A)\imp\not\for(x,\notA) in G &
    \ex(x,A)\imp\not\not\ex(x,A) in G by Th64,Th58;
    then
A4: \ex(x,A)\imp\not\for(x,\notA) in G by Th45;
    \not\ex(x,B)\iff\for(x,\notB) in G by Def39;
    then \not\ex(x,B)\imp\for(x,\notB) in G by Th43;
    then \not\for(x,\notB)\imp\not\not\ex(x,B) in G &
    (\not\not\ex(x,B))\imp\ex(x,B) in G by Th65,Th58;
    then \not\for(x,\notB)\imp\ex(x,B) in G by Th45;
    then \not\for(x,\notA)\imp\not\for(x,\notB)\imp(\ex(x,A)\imp\ex(x,B)) in G
    by A4,Th103;
    hence \for(x,A\impB)\imp(\ex(x,A)\imp\ex(x,B)) in G by A3,Th45;
  end;
