theorem
  EqR1 /\ (EqR1 "\/" EqR2) = EqR1
proof
  thus EqR1 /\ (EqR1 "\/" EqR2) c= EqR1 by XBOOLE_1:17;
  EqR1 c= EqR1 \/ EqR2 & EqR1 \/ EqR2 c= EqR1 "\/" EqR2 by Def2,XBOOLE_1:7;
  then EqR1 c= EqR1 "\/" EqR2;
  hence thesis by XBOOLE_1:19;
end;
