theorem Th90:
  A\iffB in F implies B\iffA in F
  proof
    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 by A1,Def38; then
A2: (B\impA)\and(A\impB) in F by Th35;
    (B\impA)\and(A\impB)\imp(B\iffA) in F by Def38;
    hence thesis by A2,Def38;
  end;
