theory Enigme

  predicate a
  predicate b
  predicate c
   
  predicate da = b /\ (not c)
  predicate db = a -> c
  predicate dc = (not c) /\ ( a \/ b )
  
  goal Q1: not (da /\ db /\ dc)
  (* on demande ensuite un contre exemple *)  
  
  goal Q2c: da /\ db -> dc         (* vrai *)
  goal Q2b: dc /\ da -> db         (* faux *)
  goal Q2a: db /\ dc -> da         (* vrai *)
  
  goal Q3a: (not a) /\ (not b) /\ (not c) -> da   (* faux, a ment *)
  goal Q3b: (not a) /\ (not b) /\ (not c) -> db   (* b dit la vérité *)
  goal Q3c: (not a) /\ (not b) /\ (not c) -> dc   (* faux, c ment *)
  
  goal Q4a: da /\ db /\ dc -> a                   (* faux, a innocent *)
  goal Q4b: da /\ db /\ dc -> b                   (* b coupable *)
  goal Q4c: da /\ db /\ dc -> c                   (* faux, c innocent *)
  
  goal Q5: not ( (a -> (not da)) /\ (b -> (not db)) /\ (c -> (not dc)) )
  (* on demande ensuite un contre exemple *)

end