theorem
       L is subst-correct vf-qc-correct &
  A\iffB in G implies \for(x,A)\iff\for(x,B) in G
  proof assume
A1: L is subst-correct vf-qc-correct;
    set p = A\impB, q = B\impA, pq = A\iffB;
    set a = \for(x,pq), b = \for(x,p), c = \for(x,A)\imp\for(x,B);
    pq\impp in G by Def38;
    then
A2: a\impb in G by A1,Th115;
    b\impc in G by A1,Th109;
    then
A3: (\for(x,pq))\imp(\for(x,A)\imp\for(x,B)) in G by A2,Th45;
    (A\iffB)\imp(B\impA) in G by Def38;
    then
A4: \for(x,A\iffB)\imp\for(x,B\impA) in G by A1,Th115;
    \for(x,B\impA)\imp(\for(x,B)\imp\for(x,A)) in G by A1,Th109;
    then
A5: \for(x,A\iffB)\imp(\for(x,B)\imp\for(x,A)) in G by A4,Th45;
    ((\for(x,A\iffB))\imp(\for(x,A)\imp\for(x,B)))\imp
    ((\for(x,A\iffB)\imp(\for(x,B)\imp\for(x,A)))\imp
    (\for(x,A\iffB)\imp((\for(x,A)\imp\for(x,B))\and(\for(x,B)\imp\for(x,A)))))
    in G by Th49;
    then ((\for(x,A\iffB)\imp(\for(x,B)\imp\for(x,A)))\imp
    (\for(x,A\iffB)\imp((\for(x,A)\imp\for(x,B))\and(\for(x,B)\imp\for(x,A)))))
    in G by A3,Def38;
    then
A6: \for(x,A\iffB)\imp((\for(x,A)\imp\for(x,B))\and(\for(x,B)\imp\for(x,A)))
    in G by A5,Def38;
    (\for(x,A)\imp\for(x,B))\and(\for(x,B)\imp\for(x,A))\imp
    (\for(x,A)\iff\for(x,B)) in G by Def38;
    then
A7: \for(x,A\iffB)\imp(\for(x,A)\iff\for(x,B)) in G by A6,Th45;
    assume A\iffB in G;
    then \for(x,A\iffB) in G by Def39;
    hence thesis by A7,Def38;
  end;
