1. Assume that two different sequences of FCs exist within the algebraic space that have the same runtime address, i.e the sequence of trios leading to the two FCs are same.
  2. Denote the two sequences as Seq1 and Seq2, and their runtime addresses as Addr1 and Addr2, respectively.
  3. Since Addr1 and Addr2 are the same, it implies that the last FC in Seq1 and the last FC in Seq2 are the same FC.
  4. According to the design-time constraint, each object can have only one particular intention leading to a specific FC. Therefore, if the last FC in Seq1 and Seq2 is the same, it means that the object and intention pairs leading to that FC must be the same.
  5. Let's denote the object and intention pair leading to the last FC in Seq1 and Seq2 as (Object1, Intention1) and (Object2, Intention2), respectively.
  6. Since (Object1, Intention1) and (Object2, Intention2) are the same, it means that both the object and intention in each pair must be the same.
  7. However, this contradicts the assumption that Seq1 and Seq2 are different sequences, as they have the same object and intention pair leading to the last FC.
  8. Therefore, we can conclude that if two different sequences of FCs exist, they must have different runtime addresses in the algebraic space, determined by the unique combination of object and intention pairs.