Kleene Algebra of Weighted Programs with Domain

Abstract

Weighted programs were recently introduced by Batz et al. (Proc. ACM Program. Lang. 2022) as a generalization of probabilistic programs which can also represent optimization problems and, in general, programs whose execution traces carry some sort of weight. Batz et al. show that a weighted version of Dijkstra’s weakest precondition operator can be used to reason about the competitive ratios of weighted programs. In this paper we study a propositional abstraction of weighted programs with three main contributions. First, we formulate a semantics for weighted programs with the weighted weakest precondition operator based on functions from multimonoids to quantales. Second, we show that the weighted weakest precondition operator corresponds to a generalization of the domain operator known from Kleene algebra with domain, and we study the properties of the generalized domain operator. Third, we formulate a weighted version of Kleene algebra with domain as a framework for reasoning about weighted programs with weakest precondition in an abstract setting.

A Technical appendix

1.1 A.1 Identities in multimonoids

Lemma 2

Let M be a multimonoid. For all $x,y \in M$ and $i, j \in I$:

Proof

The proof is based on some of the arguments in [6]. (1.) This holds since $\{ x \} \otimes I = \{ x \} = I \otimes \{ x \}$ by definition. (2.) If $x \in i \otimes j$, then $x = i$ and $x = j$ by the previous item. (3.) By definition, $i \otimes I = \{ i \}$; hence there is $j \in I$ such that $i \in i \otimes j$. By the previous item, $i \in i \otimes i$. If $x \in i \otimes i$, then $x = i$ by the first item. (4.) If the assumption holds, then $\emptyset \ne i \otimes x = i \otimes (j \otimes x) = (i \otimes j) \otimes x$. It follows that $y \in i \otimes j$ for some y, and so $i = j$ by the second item. (5.) is similar.    $\square$

1.2 A.2 Proof of Proposition 1

Before we give the proof, we state a useful lemma of [6] (the following is an excerpt of their Lemmas 3.1 and 3.3). In the lemma, we use the set lifting of $\textsf{s}$ defined in the obvious way: for $X \subseteq M$, $\textsf{s}(X) = \{ \textsf{s}(x) \mid x \in X \}$. We also write xy instead of $x \otimes y$.

Lemma 3

The following holds for each multimonoid M: 1. $\textsf{s}(x) \textsf{s}(x) = \textsf{s}(x)$; 2. $\textsf{s}(\textsf{s}(x) y) = \textsf{s}(x) \textsf{s}(y)$; 3. $\textsf{s}(xy) \subseteq \textsf{s}(x \textsf{s}(y))$. If M is local, then 4. $\textsf{s}(x \textsf{s}(y)) \subseteq \textsf{s}(xy)$.

Now we turn to the proof of Proposition 1.

Proof

(1.) If $h \in Di(Q^{M})$, then $\textsf{d}(h) = h$. Note that $\textsf{d}(h)(x) = \bigvee _{x = \textsf{s}(y)} h(y) = \bigvee _{x = \textsf{s}(y) \& y \in I} h(y)$ (the last equality holds since $h \in Di(Q^{M})$). However, $\textsf{s}(y) = y$ by Lemma 2(1), and so $\bigvee _{x = \textsf{s}(y) \& y \in I} h(y) = h(x)$.

(2.) $\textsf{d}(f \cdot g) \le \textsf{d}(f \cdot \textsf{d}(g))$. We reason as follows:

$$\begin{aligned} \textsf{d}(f \cdot \textsf{d}(g))(x) & = \bigvee _{x = \textsf{s}(y)} \bigvee _{y \in z \otimes u} \left( f(z) \cdot \textsf{d}(g)(u) \right) \quad = \bigvee _{x \in \textsf{s}(z \otimes u)} \left( f(z) \cdot \bigvee _{u = \textsf{s}(v)} g(v) \right) \\ & = \bigvee _{x \in \textsf{s}(z \otimes \textsf{s}(v))} \left( f(z) \cdot g(v) \right) \quad \overset{\text {Lemma 3(3)}}{\ge } \bigvee _{x \in \textsf{s}(z \otimes v)} \left( f(z) \cdot g(v) \right) \quad \\ & = \textsf{d}(f \cdot g)(x) \end{aligned}$$

We note that (5.) is established in a similar fashion using Lemma 3(3).

(3.) $\textsf{d}(\textsf{d}(f) \cdot g) = \textsf{d}(f) \cdot \textsf{d}(g)$. We reason as follows:

$$ \begin{aligned} \textsf{d}(\textsf{d}(f) \cdot g)(x) & = \bigvee _{x \in \textsf{s}(y \otimes z)} \left( \textsf{d}(f)(y) \cdot g(z) \right) \quad = \bigvee _{x \in \textsf{s}(y \otimes z) \& y \in I} \left( \textsf{d}(f)(y) \cdot g(z) \right) \\ & \overset{\text {Lemma 2(1)}}{=} \bigvee _{x \in \textsf{s}(\textsf{s}(x) \otimes z)} \left( \textsf{d}(f)(x) \cdot g(z) \right) \\ & \overset{\text {Lemma 3(1, 2)}}{=} \left( \textsf{d}(f)(x) \cdot \bigvee _{x = \textsf{s}(z)} g(z) \right) \quad = \, \textsf{d}(f)(x) \cdot \textsf{d}(g)(x)\\ & \overset{\text {Lemma 2(1--3)}}{=} \left( \textsf{d}(f) \cdot \textsf{d}(g)\right) (x) \end{aligned}$$

(4.) $\textsf{d}\left( \bigvee _{i \in I} f_i \right) = \left( \bigvee _{i \in I} \textsf{d}(f_i) \right)$. This is established as follows:

$$\begin{aligned} \textsf{d}\left( \bigvee _{i \in I} f_i \right) (x) & = \bigvee _{x = \textsf{s}(y)} \left( \bigvee _{i \in I} f_i \right) (y) \quad = \quad \bigvee _{x = \textsf{s}(y)} \left( \bigvee _{i \in I} f_i(y) \right) \\ & \quad \bigvee _{i \in I} \left( \bigvee _{x = \textsf{s}(y)} f_i(y) \right) \quad = \quad \bigvee _{i \in I} \left( \textsf{d}(f_i)(x) \right) \quad = \quad \left( \bigvee _{i \in I}\textsf{d}(f_i) \right) (x) \end{aligned}$$

   $\square$

1.3 A.3 Proof of Proposition 2

Proof

(1.) $\textsf{d}(f) \le 1$ is not valid: it is sufficient to consider a non-integral Q (1 is not the top element). (2.) $\textsf{d}(f) \cdot f = f$ is not valid: consider the pair multimonoid $\{ w \} \times \{ w \}$ and the non-idempotent product quantale $\varPi$ where $f(w,w) = 0.5$.    $\square$

1.4 A.4 Proof of Proposition 3

Proof

We reason as follows:

$$ \begin{aligned} \textsf{d}(f \cdot g)(x) & = \bigvee _{x = \textsf{s}(y)} \left( (f \cdot g)(y) \right) \quad = \quad \bigvee _{x = \textsf{s}(y)} \left( \bigvee _{y \in u \otimes v} \left( f(u) \cdot g(v) \right) \right) \\ & = \bigvee _{x = \textsf{s}(y)} \left( \bigvee _{y \in u \otimes v \& v \in I} \left( f(u) \cdot g(v)\right) \right) \quad = \quad \bigvee _{x = \textsf{s}(y)} \left( f(y) \cdot g(\textsf{t}(y)) \right) \end{aligned}$$

The first two equalities hold by definition. The third equality holds since g is diagonal. The fourth equality is established as follows: $y \in y \otimes \textsf{t}(y)$, and so the right hand side is less or equal than the left hand side; conversely, if $y \in u \otimes v \& v \in I$, then $y = u$ by Lemma 2(1) and $v = \textsf{t}(y)$ since $\textsf{t}(y)$ is the unique right identity of y by definition. Hence, the left hand side is less or equal than the right hand side.    $\square$