theorem Th128:
  L is subst-correct vf-qc-correct &
  A\impB in G implies \ex(x,A)\imp\ex(x,B) in G
  proof assume
A1: L is subst-correct vf-qc-correct;
    assume
A2: A\impB in G;
    A\impB\imp(\notB\imp\notA) in G by Th57;
    then \notB\imp\notA in G by A2,Def38;
    then
A3: \for(x,\notB)\imp\for(x,\notA) in G by A1,Th115;
    \for(x,\notB)\imp\for(x,\notA)\imp(\not\for(x,\notA)\imp\not\for(x,\notB))
    in G by Th57;
    then
A4: \not\for(x,\notA)\imp\not\for(x,\notB) in G by A3,Def38;
    \ex(x,A)\iff\not\for(x,\notA) in G by Th105;
    then
A5: \ex(x,A)\imp\not\for(x,\notB) in G by A4,Th92;
    \ex(x,B)\iff\not\for(x,\notB) in G by Th105;
    then \not\for(x,\notB)\iff\ex(x,B) in G by Th90;
    hence thesis by A5,Th93;
  end;
