theorem Th24:
  f a.e.cpfunc= g,M & g a.e.cpfunc= h,M implies f a.e.cpfunc= h,M
proof
  assume that
A1: f a.e.cpfunc= g,M and
A2: g a.e.cpfunc= h,M;
  consider EQ1 being Element of S such that
A3: M.EQ1=0 and
A4: f|EQ1` = g|EQ1` by A1;
  consider EQ2 being Element of S such that
A5: M.EQ2=0 and
A6: g|EQ2` = h|EQ2` by A2;
A7: M.(EQ1 \/ EQ2) = 0 by A3,A5,Lm4;
A8: (EQ1\/EQ2)` c= EQ2` by XBOOLE_1:7,34;
A9: (EQ1\/EQ2)` c= EQ1` by XBOOLE_1:7,34;
  then f|(EQ1\/EQ2)` =g|EQ1`|(EQ1\/EQ2)` by A4,FUNCT_1:51
    .=g|(EQ1\/EQ2)` by A9,FUNCT_1:51
    .=h|EQ2`|(EQ1\/EQ2)` by A6,A8,FUNCT_1:51
    .=h|(EQ1\/EQ2)` by A8,FUNCT_1:51;
  hence thesis by A7;
end;
