theorem Th16:
  (p => (q => r)) => ((p => q) => (p => r)) is tautology
  proof
   let M;
   thus (SAT M).((p => (q => r)) => ((p=>q)=>(p=>r)))
   = (SAT M).(p => (q => r)) => (SAT M).((p=>q)=>(p=>r)) by Def11
   .= ((SAT M).p => (SAT M).(q => r)) => (SAT M).((p=>q)=>(p=>r)) by Def11
   .= ((SAT M).p => ((SAT M).q => (SAT M).r))
   => (SAT M).((p=>q)=>(p=>r)) by Def11
   .= ((SAT M).p => ((SAT M).q => (SAT M).r))
   => ((SAT M).(p=>q)=>(SAT M).(p=>r)) by Def11
   .= ((SAT M).p => ((SAT M).q => (SAT M).r))
   => (((SAT M).p=>(SAT M).q)=>(SAT M).(p=>r)) by Def11
   .= ((SAT M).p => ((SAT M).q => (SAT M).r))
   => (((SAT M).p=>(SAT M).q)=>((SAT M).p=>(SAT M).r)) by Def11
   .= 1 by XBOOLEAN:109;
 end;
