theorem
  not a,b '||' a,c & p<>q implies not p,q '||' p,a or not p,q '||' p,b
  or not p,q '||' p,c
proof
  assume that
A1: not a,b '||' a,c and
A2: p<>q;
  assume
A3: not thesis;
  then p,a '||' p,b by A2,Def11;
  then
A4: a,p '||' a,b by Th24;
  p,a '||' p,c by A2,A3,Def11;
  then
A5: a,p '||' a,c by Th24;
  a<>p by A1,A2,A3,Def11;
  hence contradiction by A1,A4,A5,Def11;
end;
