theorem Th94:
  A\iffB in F iff \notA\iff\notB in F
  proof
    hereby
      assume
A1:   A\iffB in F;
      A\iffB\imp(A\impB) in F & A\iffB\imp(B\impA) in F by Def38; then
      A\impB in F & B\impA in F & A\impB\imp(\notB\imp\notA) in F &
      B\impA\imp(\notA\imp\notB) in F by A1,Def38,Th57; then
      \notA\imp\notB in F & \notB\imp\notA in F by Def38; then
      (\notA\imp\notB)\and(\notB\imp\notA) in F &
      (\notA\imp\notB)\and(\notB\imp\notA)\imp(\notA\iff\notB) in F
      by Def38,Th35;
      hence \notA\iff\notB in F by Def38;
    end;
    assume
A2: \notA\iff\notB in F;
    \notA\iff\notB\imp(\notA\imp\notB) in F &
    \notA\iff\notB\imp(\notB\imp\notA) in F by Def38; then
    \notA\imp\notB in F & \notB\imp\notA in F & \notA\imp\notB\imp(B\impA) in F
    & \notB\imp\notA\imp(A\impB) in F by A2,Def38; then
    A\impB in F & B\impA in F by Def38; then
    (A\impB)\and(B\impA) in F & (A\impB)\and(B\impA)\imp(A\iffB) in F
    by Def38,Th35;
    hence A\iffB in F by Def38;
  end;
