theorem Th25:
  F|=A implies F|='X' A
 proof
  assume A1: F|=A;
  let M;
  assume A2: M|=F;
  let n be Element of NAT;
  thus(SAT M).[n,'X' A]=(SAT M).[n+1,A] by Th9
   .=1 by A2,Def12,A1;
 end;
