# Differences

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sav08:instantiation_plus_ground_resolution [2009/05/14 11:57] vkuncak |
sav08:instantiation_plus_ground_resolution [2015/04/21 17:30] (current) |
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=== Ground Instantiation Rule === | === Ground Instantiation Rule === | ||

- | \[ | + | \begin{equation*} |

\frac{C}{subst(\sigma)(C)} | \frac{C}{subst(\sigma)(C)} | ||

- | \] | + | \end{equation*} |

where $C$ is a clause and $\sigma$ is a ground substitution. | where $C$ is a clause and $\sigma$ is a ground substitution. | ||

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If $A$ is a ground atom and $C,D$ are ground claues, then | If $A$ is a ground atom and $C,D$ are ground claues, then | ||

- | \[ | + | \begin{equation*} |

\frac{C \cup \{\lnot A\}\ \ \ D \cup \{A\}} | \frac{C \cup \{\lnot A\}\ \ \ D \cup \{A\}} | ||

{C \cup D} | {C \cup D} | ||

- | \] | + | \end{equation*} |

Note that this is propositional resolution where propositional variables have "long names" (they are ground atoms). | Note that this is propositional resolution where propositional variables have "long names" (they are ground atoms). | ||

==== Example ==== | ==== Example ==== | ||

+ | |||

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The proof is based on [[Herbrand's Expansion Theorem]] (see also the proof of [[Compactness for First-Order Logic]]). | The proof is based on [[Herbrand's Expansion Theorem]] (see also the proof of [[Compactness for First-Order Logic]]). | ||

- | Suppose a set $S$ of clauses is contradictory. By [[Herbrand's Expansion Theorem]] and [[Compactness Theorem|Compactness Theorem for Propositional Formulas]], there is some finite subset $S_0 \subseteq expand(S)$ is contradictory. Then there exists a derivation of empty clause from $S_0$ viewed as set of propositional formulas, using propositional resolution. In other words, there exists a derivation of empty clause from $S_0$ using ground resolution rule. Each element of $S_0$ can be obtained from an element of $S$ using instantiation rule. This means that there exists a proof tree whose leaves are followed by a single application of instantiation rule, and inner nodes contain ground resolution steps. | + | Suppose a set $S$ of clauses is contradictory. By [[Herbrand's Expansion Theorem]] and [[Compactness Theorem|Compactness Theorem for Propositional Formulas]], some finite subset $S_0 \subseteq expand(S)$ is contradictory. Then there exists a derivation of empty clause from $S_0$ viewed as set of propositional formulas, using propositional resolution. In other words, there exists a derivation of empty clause from $S_0$ using ground resolution rule. Each element of $S_0$ can be obtained from an element of $S$ using instantiation rule. This means that there exists a proof tree whose leaves are followed by a single application of instantiation rule, and inner nodes contain ground resolution steps. |