theorem Th11:
  (SAT M).[n,'F' A]=1 iff ex i st(SAT M).[n+i,A]=1
 proof
  hereby assume(SAT M).[n,'F' A]=1;
   then (SAT M).[n,'G'('not' A)]=0 by Th5;
   then consider i such that
A1: not(SAT M).[n+i,'not' A]=1 by Th10;
   not(SAT M).[n+i,A]=0 by A1,Th5;
   hence ex i st(SAT M).[n+i,A]=1 by XBOOLEAN:def 3;
  end;
  assume ex i st(SAT M).[n+i,A]=1;
  then consider i such that
   A2: (SAT M).[n+i,A]=1;
  not(SAT M).[n+i,'not' A]=1 by A2,Th5;
  then not(SAT M).[n,'G'('not' A)]=1 by Th10;
  hence (SAT M).[n,'F' A]=1 by Th5,XBOOLEAN:def 3;
 end;
