theorem
  L is vf-qc-correct subst-correct implies
  \ex(x,y,A)\imp\ex(y,x,A) in G
  proof
    assume A1: L is vf-qc-correct subst-correct;
    then \for(y,x,\notA)\imp\for(x,y,\notA) in G by Th138;
    then
A2: \not\for(x,y,\notA)\imp\not\for(y,x,\notA) in G by Th58;
    \ex(x,y,A)\iff\not\for(x,y,\notA) in G &
    \ex(y,x,A)\iff\not\for(y,x,\notA) in G by A1,Th111;
    then \ex(x,y,A)\imp\not\for(y,x,\notA) in G &
    \not\for(y,x,\notA)\iff\ex(y,x,A) in G by A2,Th90,Th92;
    hence \ex(x,y,A)\imp\ex(y,x,A) in G by Th93;
  end;
