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Both sides previous revision Previous revision Next revision | Previous revision Next revision Both sides next revision | ||
sav08:first-order_logic_syntax [2008/03/18 11:44] vkuncak |
sav08:first-order_logic_syntax [2008/03/18 12:17] vkuncak |
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When in doubt, use parentheses. | When in doubt, use parentheses. | ||
- | Example: Consider language ${\cal L} = \{ P, Q, R, f \}$ with | + | Example: Consider language ${\cal L} = \{ P, Q, R, f \}$ with $ar(P) = 1$, $ar(Q) = 1$, $ar(R) = 2$, $ar(f) = 2$. Then |
- | \[\begin{array}{l} | + | |
- | ar(P) = 1 \\ | + | |
- | ar(Q) = 1 \\ | + | |
- | ar(R) = 2 \\ | + | |
- | ar(f) = 2 | + | |
- | \end{array} | + | |
- | \] | + | |
- | Then | + | |
\[ | \[ | ||
\lnot \forall x.\, \forall y. R(x,y) \land Q(x) \rightarrow Q(f(y,x)) \lor P(x) | \lnot \forall x.\, \forall y. R(x,y) \land Q(x) \rightarrow Q(f(y,x)) \lor P(x) | ||
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$FV$ denotes the set of free variables in the given propositional formula and can be defined recursively as follows: | $FV$ denotes the set of free variables in the given propositional formula and can be defined recursively as follows: | ||
- | \[\begin{array}{l} | + | \[\begin{array}{rcl} |
- | FV(x) = \{ x \}, \mbox{ for } x \in V \\ | + | FV(x) &=& \{ x \}, \mbox{ for } x \in V \\ |
+ | FV(f(t_1,\ldots,t_n) &=& F(t_1) \cup \ldots \cup F(t_n) \\ | ||
+ | FV(R(t_1,\ldots,t_n) &=& F(t_1) \cup \ldots \cup F(t_n) \\ | ||
+ | FV(t_1 = t_2) &=& F(t_1) \cup F(t_2) \\ | ||
FV(\lnot F) = FV(F) \\ | FV(\lnot F) = FV(F) \\ | ||
FV(F_1 \land F_2) = FV(F_1) \cup FV(F_2) \\ | FV(F_1 \land F_2) = FV(F_1) \cup FV(F_2) \\ |