Editing Frame problem (section)

===Fluent calculus solution===
The [[fluent calculus]] is a variant of the situation calculus. It solves the frame problem by using first-order logic [[First-order logic#Formation rules|terms]], rather than [[Predicate variable|predicates]], to represent the states. Converting predicates into terms in first-order logic is called [[Reification (knowledge representation)|reification]]; the fluent calculus can be seen as a logic in which predicates representing the state of conditions are reified.

The difference between a predicate and a term in first-order logic is that a term is a representation of an object (possibly a complex object composed of other objects), while a predicate represents a condition that can be true or false when evaluated over a given set of terms.

In the fluent calculus, each possible state is represented by a term obtained by composition of other terms, each one representing the conditions that are true in state. For example, the state in which the door is open and the light is on is represented by the term <math>\mathrm{open} \circ \mathrm{on}</math>. It is important to notice that a term is not true or false by itself, as it is an object and not a condition. In other words, the term <math>\mathrm{open} \circ \mathrm{on}</math> represent a possible state, and does not by itself mean that this is the current state. A separate condition can be stated to specify that this is actually the state at a given time, e.g., <math>\mathrm{state}(\mathrm{open} \circ \mathrm{on}, 10)</math> means that this is the state at time <math>10</math>.

The solution to the frame problem given in the fluent calculus is to specify the effects of actions by stating how a term representing the state changes when the action is executed. For example, the action of opening the door at time 0 is represented by the formula:

: <math>\mathrm{state}(s \circ \mathrm{open}, 1) \iff \mathrm{state}(s,0)</math>

The action of closing the door, which makes a condition false instead of true, is represented in a slightly different way:

: <math>\mathrm{state}(s, 1) \iff \mathrm{state}(s \circ \mathrm{open}, 0)</math>

This formula works provided that suitable axioms are given about <math>\mathrm{state}</math> and <math>\circ</math>, e.g., a term containing the same condition twice is not a valid state (for example, <math>\mathrm{state}(\mathrm{open} \circ s \circ \mathrm{open}, t)</math> is always false for every <math>s</math> and <math>t</math>).