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Note on Paramodulation

Here's a better description of the paramodulation rule we talked about in class. Let

Then, given a clause

\begin{displaymath}\alpha \vee (s = t)\end{displaymath}

and another clause

\begin{displaymath}\beta \vee \gamma[r]\;\;,\end{displaymath}

you may conclude

\begin{displaymath}(\alpha \vee \beta \vee \gamma[t])\theta\;\;.\end{displaymath}

Here's an example. Given

\begin{displaymath}n(g(C)) \vee (f(g(B)) = A)\end{displaymath}

and

\begin{displaymath}q(x) \vee p(h(f(x)))\end{displaymath}

we can conclude

\begin{displaymath}n(g(C)) \vee q(g(B)) \vee p(h(A))\end{displaymath}

by setting $s = f(g(B))$, $t = A$, $r = f(x)$, $\gamma[\cdot] =
p(h(\cdot))$, and $\theta = \{g(B)/x\}$.





Michael G. Ross 2001-03-09