- 2070 Bhadra · 8 marks
Resolution: oversmart persons, Ram
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.
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.
Step 1: Statements in FOPL
Predicates: Oversmart(x), Stupid(x), Child(x, y) = x is a child of y, Naughty(x).
- ∀x Oversmart(x) → Stupid(x)
- ∀x ∀y [Oversmart(y) ∧ Child(x, y)] → Naughty(x)
- Child(Ram, Hari)
- 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 (□) |
¬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.)
✓ Answer checked by Claude Opus 5.5 (high effort)
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