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sav08:homework06 [2008/04/08 15:45] vkuncak |
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- | ====== Homework 06 - Due April 9 ====== | ||
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- | ===== Problem 1 ===== | ||
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- | (Recall [[Definition of Resolution for FOL]].) | ||
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- | Let $F_0$ denote formula | ||
- | \[ | ||
- | \forall x. (A_1(x) \rightarrow B_1(x)) \land (A_2(x) \rightarrow B_2(x)) \leftrightarrow | ||
- | (A_1(x) \land B_1(x)) \lor (A_2(x) \land B_2(x)) | ||
- | \] | ||
- | For each of the following formulas, if the formula is valid, use resolution to prove it; if it is invalid, construct at least one Herbrand model for its negation. | ||
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- | **a):** Formula $F_0$ | ||
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- | **b):** Formula | ||
- | \[ | ||
- | \begin{array}{l} | ||
- | (\forall y. \lnot (A_1(y) \land A_2(y))) \rightarrow F_0 | ||
- | \end{array} | ||
- | \] | ||
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- | **c):** Formula | ||
- | \[ | ||
- | \begin{array}{l} | ||
- | (\forall y. A_1(y) \leftrightarrow \lnot A_2(y)) \rightarrow F_0 | ||
- | \end{array} | ||
- | \] | ||
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- | **d):** Formula | ||
- | \[ | ||
- | \begin{array}{l} | ||
- | (\forall y. A_1(y) \leftrightarrow \lnot A_2(y)) \land (\forall z. B_1(z) \leftrightarrow \lnot B_2(z)) \rightarrow F_0 | ||
- | \end{array} | ||
- | \] | ||
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- | **e):** Formula: | ||
- | \[ | ||
- | \begin{array}{l} | ||
- | (\forall x. \lnot R(x,x)) \land (\forall x. R(x,f(x)) \rightarrow (\exists x,y,z.\ R(x,y) \land R(y,z) \land \lnot R(x,z)) | ||
- | \end{array} | ||
- | \] | ||
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- | ===== Problem 2 ===== | ||
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- | (Recall [[Sets and Relations]].) | ||
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- | We say that a binary relation is a partial order iff it is reflexive, antisymmetric, and transitive. Let $D$ be a non-empty set and $r_0 \subseteq D \times D$ a binary relation on $D$. Let $r = r_0^* = \bigcup_{i\ge 0} r_0^n$ be the reflexive transitive closure of $r_0$. | ||
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- | **a)** Give an example $r_0$ for which $r$ is not necessarily a partial order. | ||
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- | **b)** Define $s = r \cap r^{-1}$. Show that $s$ is a congruence with respect to $r$, that is: $s$ is reflexive, symmetric, and transitive and for all $x,x',y,y' \in D$, | ||
- | \[ | ||
- | (x,x') \in s \land (y,y') \in s \rightarrow ((x,y) \in r \leftrightarrow (x',y') \in r) | ||
- | \] | ||
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- | **c)** For each $x \in D$ let $[x] = \{ y \mid (x,y) \in s \}$. Let $[D] = \{ [x] \mid x \in D\}$. Define a new relation, $[r] \subseteq [D] \times [D]$, by | ||
- | \[ | ||
- | [r] = \{ ([x],[y]) \mid (x,y) \in r \} | ||
- | \] | ||
- | Show that $[r]$ is a partial order on $[D]$. | ||
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- | Optional: Explain this constructions using terminology of graphs and strongly connected components. | ||
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- | ===== Problem 3 ===== | ||
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- | (Recall [[Unification]].) | ||
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- | Let $V$ be an infinite set of variables. Let ${\cal L}$ be some first-order language. We will consider terms that contain variables from $V$ and function symbols from ${\cal L}$. | ||
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- | Following Problem 2 above, let $(\sigma_1,\sigma_2) \in r_0$ iff there exists substitution $\tau$ such that $subst(\sigma_2) = subst(\sigma_1) \circ subst(\tau)$ where $\circ$ is the standard relation composition. | ||
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- | **a)** Compute $r = r_0^*$. What is its relationship to $r_0$? | ||
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- | **b)** Compute $s = r \cap r^{-1}$. Show that relation $s$ holds iff $subst(\sigma_2) = subst(\sigma_1) \circ subst(b)$ where $b$ is a relation which is bijection on the set $V$. | ||
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- | Optional: **c)** Let $E$ be a fixed set of syntactic equations. Let $U$ be the set of unifiers for $E$ and $[U] = \{ [\sigma] \mid \sigma \in U \}$. Show that if $U$ is non-empty, then there exists $a \in [U]$ such that for all $b \in [U]$, we have $(a,b) \in [r]$ (that is, $a$ is the least element of $[U]$ with respect to $[r]$ defined as in Problem 2). | ||