To prove that a consistent heuristic has a value of 0 on a goal state, we can follow these steps: 1. **Definition of Consistent Heuristic**: A heuristic \( h(n) \) is considered consistent (or monotonic) if, for every node \( n \) and every successor \( n' \) of \( n \), the following condition holds: \[ h(n) \leq c(n, n') + h(n') \] where \( c(n, n') \) is the cost of the edge from node \( n \) to its successor \( n' \). 2. **Goal State Definition**: A goal state \( G \) is defined as the state where the search algorithm terminates successfully. 3. **Heuristic Value at Goal State**: By definition of the heuristic, the value at the goal state should represent the estimated cost to reach the goal from that state. Since the goal state is the final destination, the cost to reach the goal from itself is 0. Therefore, we have: \[ h(G) = 0 \] 4. **Applying Consistency**: To further reinforce this, we can apply the consistency condition to the goal state: - Let \( n \) be any node that has \( G \) as a successor. The cost \( c(n, G) \) is the cost to reach the goal state from \( n \). - By the consistency property, we can write: \[ h(n) \leq c(n, G) + h(G) \] Since \( h(G) = 0 \), it simplifies to: \[ h(n) \leq c(n, G) \] This means that the heuristic value at any node \( n \) is less than or equal to the cost to reach the goal, which is consistent with our understanding of heuristics. 5. **Conclusion**: Since the goal state is defined such that reaching it from itself incurs no cost, and since a consistent heuristic must satisfy the above conditions, we conclude that: \[ h(G) = 0 \] Thus, we have proven that a consistent heuristic has a value of 0 on a goal state.

A heuristic is considered consistent if the estimated cost from a given state to the goal is less than or equal to the actual cost of the optimal path leading from that state to the goal.

No, that is the definition of an admissable heuristic. A consistent heuristic is one which satisfies the triangle inequality:

$$ h(N) \leq c(N,P) + h(P) \\ h(G) = 0 $$

To prove that a consistent heuristic has a value of 0 at a goal state, we can follow these steps: 1. **Definition of a Goal State**: A goal state in a search problem is defined as a state where the objective is achieved. In terms of cost, the cost to reach the goal from itself is zero. 2. **Consistent Heuristic**: A heuristic \( h(n) \) is said to be consistent (or monotonic) if, for every node \( n \) and every successor \( n' \) of \( n \), the following condition holds: \[ h(n) \leq c(n, n') + h(n') \] where \( c(n, n') \) is the cost of the edge from \( n \) to \( n' \). 3. **Applying Definition at Goal State**: Let’s denote the goal state as \( g \). By the nature of the goal state, the cost to reach the goal from the goal itself is zero, i.e., \( c(g, g) = 0 \). 4. **Using Consistency**: Now, we can apply the consistency condition at the goal state. Since \( g \) has no successors (or the only successor is itself), we can consider \( n = g \) and \( n' = g \): \[ h(g) \leq c(g, g) + h(g) \] Substituting \( c(g, g) = 0 \) into the inequality gives us: \[ h(g) \leq 0 + h(g) \] This simplifies to: \[ h(g) \leq h(g) \] which is trivially true. 5. **Conclusion**: However, since the heuristic must also be non-negative for all states (as it estimates the cost to reach the goal), we have: \[ h(g) \geq 0 \] Combining these two results, we conclude that: \[ h(g) = 0 \] Thus, a consistent heuristic must have a value of 0 at the goal state.

This is included in the definition of a consistent heuristic as mentioned earlier. While it is also obviously applicable to admissible heuristics, demonstrating this is not required to establish that consistent heuristics are indeed admissible. To prove this, one can apply induction based on the equations provided above (a complete proof is available on the Wikipedia page).