- 2075 Bhadra · 8 marks
Rules of inference: Pokhara and Sarankot
Short answer
From ∃x(C(x)∧P(x)), instantiate a with C(a)∧P(a). Then ∀x(P(x)→S(x)) gives P(a)→S(a). Modus ponens yields S(a). So C(a)∧S(a), existentially generalized to ∃x(C(x)∧S(x)). Thus someone in class visited Sarankot.
Show that the premises "There is someone in this class who has been to Pokhara. Everyone who goes to Pokhara visit the Sarankot. Therefore, someone in this class has visited the Sarankot."
The question gives two premises and the conclusion "someone in this class has visited the Sarankot". We prove the conclusion follows from the premises using predicate logic rules.
Symbols
Domain: all people. C(x): x is in this class. P(x): x has been to Pokhara. S(x): x has visited Sarankot.
- Premise 1: ∃x (C(x) ∧ P(x))
- Premise 2: ∀x (P(x) → S(x))
- Conclusion: ∃x (C(x) ∧ S(x))
Proof
| Step | Statement | Reason |
|---|---|---|
| 1 | ∃x (C(x) ∧ P(x)) | Premise 1 |
| 2 | C(a) ∧ P(a) | Existential instantiation from 1 (a is a particular person) |
| 3 | C(a) | Simplification from 2 |
| 4 | P(a) | Simplification from 2 |
| 5 | ∀x (P(x) → S(x)) | Premise 2 |
| 6 | P(a) → S(a) | Universal instantiation from 5 |
| 7 | S(a) | Modus ponens from 4 and 6 |
| 8 | C(a) ∧ S(a) | Conjunction from 3 and 7 |
| 9 | ∃x (C(x) ∧ S(x)) | Existential generalization from 8 |
Step 9 says that someone in this class has visited the Sarankot. Hence the conclusion follows from the premises. ∎
Note: existential instantiation must be done before universal instantiation so that the same name a is used for both.
✓ Answer checked by Claude Opus 5.5 (high effort)
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