where the two sequences Ω1 and Ω2 are defined as follows: for Ω1 we have the two **sequences**
(11) hb:D < 0.5i (⊓<) : (9)

$$\begin{array}{r l}{{\overline{{\begin{array}{l l}{\langle x\colon D<0.5\rangle}&{{}(\cap_{<}):(10)}\\ {\mathrm{{\small~clash~}}}&{{}(7),(12)}\end{array}}}\end{array}}}\end{array}$$
$$\begin{array}{l}{{\overline{{(12)}}}}\\ {{(13)}}\end{array}$$
(14) hx:C < 0.5i (⊓<) : (10) (15) clash-free
and for Ω2 we have the two **sequences**
(16) hb:C < 0.5i (⊓<) : (9)

$$\begin{array}{r l}{{\overline{{\langle x\!:\!D<0.5\rangle}}}}&{{(\sqcap_{<}):(10)}}\\ {{\mathrm{clash}}}&{{(7),(17)}}\end{array}$$
$$\frac{(17)}{(18)}$$
$$\begin{array}{l|l l l}{{\overline{{(\Gamma_{<}):(10)}}}}&{{\mathrm{~}}}&{{(19)}}&{{\mathrm{~}\langle x{:}C<0.5\rangle\mathrm{~}\mathrm{~}\left(\Gamma_{<}\right):(10)}}\\ {{(7),(17)}}&{{\mathrm{~}}}&{{(20)}}&{{\mathrm{~}\mathrm{class-free}}}\end{array}$$
✷

Example 7 Consider Example 4 and let us prove **that** Σ
′′
|≈h(∃About.Car)(i1) ≥ 0.6i. We
prove the above relation by verifying that all **completions** of S = Σ
′′ ∪ {hi1:∃About.Car < 0.6i}
contain a clash. In fact, we have the following **sequence.**
(1) hi1:∃About.(Car ⊓ Ferrari
∗
) ≥ 0.6i **Hypothesis:**S
(12.:About.(Car $\cap$ "I") $\geq$ 0.6)  (11:About.Car $<$ 0.6)  ((11,x):About $\geq$ 0.6), $\langle$x:(Car $\cap$ Ferrari") $\geq$ 0.6)  ($\exists_{>}$) : (1)  (x:Car $<$ 0.6)  ($\exists_{<}$) : (3), (4)  (x:Car $\geq$ 0.6), $\langle$x:Ferrari${}^{*}$$\geq$ 0.6), $(\sqcap$) : (4)  (5), (6)
(2) hi2:∃About.(Car ⊓ Porsche ∗ ) ≥ 0.8i
✷

Proposition 3 A finite set of fuzzy constraints S is satisfiable iff there exists a clash **free**
completion of S. ⊣
Proof:
⇒ .) Given the termination property, it is easily verified, by case analysis, that the
above rules are sound, i.e. if S1 is satisfiable then there is a satisfiable **completion** S2
of S1 and, thus, S2 contains no clash. For instance, let us show that the (∀≥) **rule** is
sound. Assume I is an interpretation satisfying hw1:∀R.C ≥ ni and h(w1, w2):R ≥ mi,
where m > 1 − n. Let us show that I satisfies hw2:C ≥ ni. Since I satisfies hw1:∀R.C ≥ ni
it follows that max{1 − RI(w1
I

, w2
I
), C
I
(w2
I
)} ≥ n. **But,** RI(w1
I

, w2
I
) ≥ m and, **thus,**
1 − RI(w1
I

, w2
I
) ≤ 1 − m < n. As a **consequence,** C
I
(w2
I
) ≥ n follows, i.e. I **satisfies**
hw2:C ≥ ni.
⇐ .) Suppose that there exists a clash free **completion** S
′ of S. We build **from** S
′ an

interpretation I **satisfying** S
′ **and,** as S ⊆ S
′

, I satisfies S. I is called canonical **model**.
For any primitive **constraint** α ∈ S
′

, we collect its lower and upper bound **restrictions**
in S
′ as follows: let
N
≥[α] = {n : hα ≥ ni ∈ S
′
}
N
$$\begin{array}{r c l}{{}}&{{}}&{{}}\\ {{N^{>}[\alpha]}}&{{=}}&{{\{n:\langle\alpha>n\rangle\in S^{\prime}\}}}\\ {{}}&{{}}&{{}}\\ {{N^{\leq}[\alpha]}}&{{=}}&{{\{n:\langle\alpha\leq n\rangle\in S^{\prime}\}}}\\ {{}}&{{}}&{{}}\\ {{N^{\leq}[\alpha]}}&{{=}}&{{\{n:\langle\alpha<n\rangle\in S^{\prime}\}.}}\end{array}$$
N
N
: $\langle a\leq n\rangle\in S$ . 