theorem Th37:
  A is axpl1 or A is axpl2 or A is axpl3 implies F |= A
 proof
  assume A1: A is axpl1 or A is axpl2 or A is axpl3;
  let M;
  assume M|=F;
  per cases by A1;
  suppose A is axpl1;
   then consider p,q be Element of PL-WFF such that
A2: A=p => (q => p);
    A is tautology by Th15,A2;
    hence thesis;
  end;
  suppose A is axpl2;
   then consider p,q,r be Element of PL-WFF such that
    A3: A=(p => (q => r)) => ((p=>q)=>(p=>r));
    A is tautology by Th16,A3;
    hence thesis;
  end;
  suppose A is axpl3;
   then consider p,q be Element of PL-WFF such that
    A4: A=('not' q => 'not' p) => (('not' q => p)=>q);
    A is tautology by Th17,A4;
    hence thesis;
  end;
 end;
