Elementary Ranking Condition (ERC)

Once data is represented in a comparative tableau, ranking information is readily accessible as a set of elementary ranking conditions, or ERCs. The entire grammar (i.e., all necessary ranking information) can and should be given by the right set of ERCs.

An ERC is a single row in a comparative tableau. In logical terms, an ERC is a proposition that must be true in the final grammar for the given analysis. The grammar can the be expressed as an ERC set, where there are no contradictions between any two ERCs. Recall that a CT shows information about whether a particular constraint prefers the winning candidate or some losing candidate. Therefore, for the intended output to be the winner, a ranking must have every winner-preferring constraint dominate all loser-preferring constraint. Observe the following dummy CT:

Notice that in row d, C1 assigns 'e'. This means that it makes no distiinction between the winner and cand3; the violations it assigns to both are equal. This is usually indicated by a blank sell (to avoid clutter), but when discussing the finer points of ERC logic it is better to show this e explicitly.

Each row in this CT is a valid ERC and can thus be "translated" into an English sentence:

The question that remains is: what is the set of constraint orders (or rankings) that satisfies all of these propositions? Note that some are compound propositions, joined with and or or.

There is in fact one total order that satisfies the ERC set in : C1 ≫ C3 ≫ C2. The ERC in row a is satisfied because C1 dominates all other constraints. The ERC in row c is satisfied because C3 dominates C2. And crucially, the ERC in row b is satisfied because C1 dominates C3. The sub-proposition "C2 must dominate C3" is not satisfied, but this is acceptable because it is part of a disjunction. The statement as a whole is still true because the left side of the disjunction is satisfied. (And for reference, or in ERC logic is inclusive.)

Consider the other possible total orders for this set of three constraints. Can you find for each one how the ERC set is not satisfied?

Operations over ERC values

Because ERCs can be translated into propositions over and and or, logical properties of these operators can be straightfowardly applied to individual ERCs as well. Since ERCs are logical propositions, the following operations help find logical entailments between ERCs.

L-retraction

Given an ERC E, replacing any L in this ERC with E results in a new ERC F such that E entails F. What this means in practice is as follows.

In , the ERC WLL is shown, along with its translation. If, for instance, the second L in WLL is replaced with e, the resulting ERC is WLe, which is translated as follows:

• C1 must dominate C2.

From here, the logic is simple:

1. Assume P = "C1 must dominate C2"
2. Assume Q = "C1 must dominate C3"
3. WLL = P & Q (def. ERC)
4. WLe = P (def. ERC)
5. If P & Q, then P (def. '&')
6. Therefore, if WLL, then WLe (substutition)

W-extension

On the disjunctive side, a similar argument can be made: given an ERC E, replacing any e with W will result in an ERC F such that E entails F. Below is the third ERC from the original example CT from this section.

In the ERC eLW, if e is replaced with W, the resulting ERC is WLW, which has the following translation:

• C1 must dominate C2 or C3 must dominate C2.

Again, to show that eLW entails WLW is straightforward:

1. Assume P = "C1 must dominate C2"
2. Assume Q = "C3 must dominate C2"
3. eLW = Q (def. ERC)
4. WLW = P or Q (def. ERC)
5. If Q, then P or Q (def. 'or')
6. Therefore, if eLW, then WLW (substitution)

The pragmatics of 5 are a little more obscure ("if you are a cat, then you are a cat or you are a dog" is logically true), but the logic is sound.

To see if one individual ERC A entails another individual ERC B, look for a chain of L-retractions and W-extensions to get from one to the other. If such a path exists, ERC A entails ERC B.

Fusion

Fusion is an operation over W, L, and e that determins what information is entailed by a set of ERCs. If A and B are ERCs, then the fusion is A◦B. To calculate the fusion of an ERC set, apply the following position-by-position (constraint by constraint) to the ERC set:

To repeat the example from [Prince2002]:9, two ERCs are shown along with their fusion in the CT below.

The fusion operations are applied individually per each column, so taking the definition above, each of the following apply (essentially the above tableau, transposed):

1. C1: W◦e = W
2. C2: L◦W = L
3. C3: e◦L = L

The resulting ERC is WLL, meaning that in an ERC set where both ERCs A and B are true, then it must also be the case that the resulting ERC A◦B is also true. Indivudally, neither A or B says anything jointly about C1 and C3. If we visualize the Hasse diagram for A and B, it's clear to see why A◦B hold, knowing that the domination relation is transitive, but we can also prove this via fusion without leaving our ERCs. Also beware that A◦B is less informative than A and B together; the fustion does not tell us anything about how C2 and C3 relate to each other, which B tells us exactly this. Thus, the fustion of two ERCs A and B is not everything that is entailed by A and B, but the results of fusion will be something entailed by A and B.