theorem :: Modus ponendo ponens
  B\imp((B\impA)\impA) in F
  proof
A1: ((B\impA)\imp(B\impA))\imp(B\imp((B\impA)\impA)) in F by Th41;
    (B\impA)\imp(B\impA) in F by Th34;
    hence thesis by A1,Def38;
  end;
