---
title: "Resolution: oversmart persons, Ram"
subject: "Artificial Intelligence"
chapter: 4
key: "semester_6/artificial_intelligence:resolution-oversmart-ram-naughty"
source: https://www.bctengineer.com/q/semester_6/artificial_intelligence/resolution-oversmart-ram-naughty
site: BCT Engineering
---

# Resolution: oversmart persons, Ram

Artificial Intelligence · Chapter 4: Knowledge Representation, Inference & Reasoning

- Asked in: 2070 Bhadra, 8 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/artificial_intelligence/resolution-oversmart-ram-naughty
- Whole chapter: https://www.bctengineer.com/md/semester_6/artificial_intelligence/chapter-4

## Question

All oversmart persons are stupid. Children of oversmart persons are naughty. Ram is child of Hari. Hari is oversmart. Show that Ram is naughty. Use FOPL based resolution method.

## Short answer

Convert statements to FOPL clauses, negate goal ¬Naughty(Ram). Resolve with clauses: ¬Naughty(Ram) and C2 yield ¬Oversmart(y) ∨ ¬Child(Ram,y); with C3 (y/Hari) yield ¬Oversmart(Hari); with C4 yield empty clause. Contradiction proves Ram is naughty.

## Answer

#### Step 1: Statements in FOPL
Predicates: Oversmart(x), Stupid(x), Child(x, y) = x is a child of y, Naughty(x).

1. ∀x Oversmart(x) → Stupid(x)
2. ∀x ∀y [Oversmart(y) ∧ Child(x, y)] → Naughty(x)
3. Child(Ram, Hari)
4. Oversmart(Hari)

**Goal:** Naughty(Ram)

#### Step 2: Clauses
| No. | Clause |
|---|---|
| C1 | ¬Oversmart(x) ∨ Stupid(x) |
| C2 | ¬Oversmart(y) ∨ ¬Child(x, y) ∨ Naughty(x) |
| C3 | Child(Ram, Hari) |
| C4 | Oversmart(Hari) |

#### Step 3: Negated goal
C5: **¬Naughty(Ram)**

#### Step 4: Resolution
| Step | Resolve | Substitution | Resolvent |
|---|---|---|---|
| 1 | C5 and C2 | x/Ram | R1: ¬Oversmart(y) ∨ ¬Child(Ram, y) |
| 2 | R1 and C3 | y/Hari | R2: ¬Oversmart(Hari) |
| 3 | R2 and C4 | | **R3: empty clause (□)** |

```text
¬Naughty(Ram)  +  C2 -> ¬Oversmart(y) ∨ ¬Child(Ram,y)
               +  C3 {y/Hari} -> ¬Oversmart(Hari)
               +  C4 -> □  (contradiction)
```
The empty clause is derived, so the negation of the goal is false. **Hence Ram is naughty.** (Clause C1 is not needed for this proof.)

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Which is the correct FOPL for 'Children of oversmart persons are naughty'?
   A. ∀x ∀y [Oversmart(y) ∧ Child(x,y)] → Naughty(x)
   B. ∀x Naughty(x) → Child(x,y)
   C. ∀x Oversmart(x) → Naughty(x)
   D. Child(Ram, Hari) → Oversmart(Ram)
   Answer: A. If y is oversmart and x is a child of y, then x is naughty. (Key point: FOPL of rule 2)

2. What is the negated goal?
   A. Naughty(Ram)
   B. ¬Naughty(Ram)
   C. ¬Oversmart(Hari)
   D. ¬Child(Ram, Hari)
   Answer: B. We assume Ram is not naughty and derive a contradiction. (Key point: Negated goal)

3. Resolving ¬Naughty(Ram) with C2 gives which resolvent?
   A. ¬Oversmart(y) ∨ ¬Child(Ram, y)
   B. Oversmart(Ram)
   C. ¬Stupid(Ram)
   D. Child(Ram, Hari)
   Answer: A. Substituting x/Ram cancels Naughty(x) and leaves the two negated conditions. (Key point: First resolution)

4. Which clause is not needed in the proof?
   A. C1: Oversmart(x) → Stupid(x)
   B. C3: Child(Ram, Hari)
   C. C4: Oversmart(Hari)
   D. C2
   Answer: A. The proof uses C5, C2, C3 and C4 only; C1 about being stupid is irrelevant. (Key point: Unused clause)

## Related in Artificial Intelligence

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