theorem Th129:
  L is subst-correct vf-qc-correct &
  A\iffB in G implies \ex(x,A)\iff\ex(x,B) in G
  proof assume
A1: L is subst-correct vf-qc-correct;
    assume
A2: A\iffB in G;
    (A\iffB)\imp(A\impB) in G & (A\iffB)\imp(B\impA) in G by Def38;
    then A\impB in G & B\impA in G by A2,Def38;
    then \ex(x,A)\imp\ex(x,B) in G & \ex(x,B)\imp\ex(x,A) in G by A1,Th128;
    then
A3: (\ex(x,A)\imp\ex(x,B))\and(\ex(x,B)\imp\ex(x,A)) in G by Th35;
    (\ex(x,A)\imp\ex(x,B))\and(\ex(x,B)\imp\ex(x,A))\imp(\ex(x,A)\iff\ex(x,B))
    in G by Def38;
    hence thesis by A3,Def38;
  end;
