theorem Th9:
  EqR1 (/\) (EqR1 "\/" EqR2) = EqR1
proof
A1: EqR1 (/\) (EqR1 "\/" EqR2) c= EqR1 by PBOOLE:15;
A2: EqR1 c= EqR1 (\/) EqR2 by PBOOLE:14;
  EqR1 (\/) EqR2 c= EqR1 "\/" EqR2 by Th4;
  then EqR1 c= EqR1 "\/" EqR2 by A2,PBOOLE:13;
  then EqR1 c= EqR1 (/\) (EqR1 "\/" EqR2) by PBOOLE:17;
 hence thesis by A1,PBOOLE:146;
end;
