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# Differences

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sav08:homework06 [2008/04/08 15:45]
vkuncak
sav08:homework06 [2015/04/21 17:30] (current)
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Let $F_0$ denote formula Let $F_0$ denote formula
-$+\begin{equation*} ​\forall x. (A_1(x) \rightarrow B_1(x)) \land (A_2(x) \rightarrow B_2(x)) \leftrightarrow ​\forall x. (A_1(x) \rightarrow B_1(x)) \land (A_2(x) \rightarrow B_2(x)) \leftrightarrow (A_1(x) \land B_1(x)) \lor (A_2(x) \land B_2(x)) (A_1(x) \land B_1(x)) \lor (A_2(x) \land B_2(x)) -$+\end{equation*}
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. 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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**b):** Formula **b):** Formula
-$+\begin{equation*} \begin{array}{l} \begin{array}{l} ​(\forall y. \lnot (A_1(y) \land A_2(y))) \rightarrow F_0 ​(\forall y. \lnot (A_1(y) \land A_2(y))) \rightarrow F_0 \end{array} \end{array} -$+\end{equation*}

**c):** Formula **c):** Formula
-$+\begin{equation*} \begin{array}{l} \begin{array}{l} ​(\forall y. A_1(y) \leftrightarrow \lnot A_2(y)) \rightarrow F_0 ​(\forall y. A_1(y) \leftrightarrow \lnot A_2(y)) \rightarrow F_0 \end{array} \end{array} -$+\end{equation*}

**d):** Formula **d):** Formula
-$+\begin{equation*} \begin{array}{l} \begin{array}{l} ​(\forall y. A_1(y) \leftrightarrow \lnot A_2(y)) \land (\forall z. B_1(z) \leftrightarrow \lnot B_2(z)) \rightarrow F_0 ​(\forall y. A_1(y) \leftrightarrow \lnot A_2(y)) \land (\forall z. B_1(z) \leftrightarrow \lnot B_2(z)) \rightarrow F_0 \end{array} \end{array} -$+\end{equation*}

**e):** Formula: **e):** Formula:
-$+\begin{equation*} \begin{array}{l} \begin{array}{l} ​(\forall x. \lnot R(x,x)) \land (\forall x. R(x,f(x)) \rightarrow (\exists x,y,z.\ R(x,y) \land R(y,z) \land \lnot R(x,z)) ​(\forall x. \lnot R(x,x)) \land (\forall x. R(x,f(x)) \rightarrow (\exists x,y,z.\ R(x,y) \land R(y,z) \land \lnot R(x,z)) \end{array} \end{array} -$+\end{equation*}

===== Problem 2 ===== ===== Problem 2 =====
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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 $x,​x',​y,​y'​ \in D$, **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 $x,​x',​y,​y'​ \in D$,
-$+\begin{equation*} (x,x') \in s \land (y,y') \in s \rightarrow ((x,y) \in r \leftrightarrow (x',​y'​) \in r) (x,x') \in s \land (y,y') \in s \rightarrow ((x,y) \in r \leftrightarrow (x',​y'​) \in r) -$+\end{equation*}

**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 **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
-$+\begin{equation*} [r] = \{ ([x],[y]) \mid (x,y) \in r \} [r] = \{ ([x],[y]) \mid (x,y) \in r \} -$+\end{equation*}
Show that $[r]$ is a partial order on $[D]$. Show that $[r]$ is a partial order on $[D]$.

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===== Problem 3 ===== ===== Problem 3 =====

-(Recall [[Unification]].)+(Recall ​[[Substitutions for First-Order Logic]], ​[[Unification]].)

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 ${\cal L}$. 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 ${\cal L}$.