theorem
  R1 c= P1 & R2 c= P2 implies R1 concur R2 c= P1 concur P2
proof
  assume that
A1: R1 c= P1 and
A2: R2 c= P2;
  let x be object;
  assume x in R1 concur R2;
  then ex C st
  x = C & ex q1,q2 being FinSubsequence st C = q1 \/ q2 & q1 misses q2 &
  Seq q1 in R1 & Seq q2 in R2;
  hence thesis by A1,A2;
end;
