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10. Translate the following into MPL (not into PL):Some dogs are kelpies, some are beagles, and the rest are corgis.

Question

10. Translate the following into MPL (not into PL):

Some dogs are kelpies, some are beagles, and the rest are corgis.

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Solution

To translate the given statement into Mathematical Predicate Logic (MPL), we will break down the components of the sentence and represent them using predicates.

Step 1: Identify the Components

  1. Terms:

    • Dogs
    • Kelpies
    • Beagles
    • Corgis
  2. Predicates:

    • Let D(x) D(x) represent "x is a dog."
    • Let K(x) K(x) represent "x is a kelpie."
    • Let B(x) B(x) represent "x is a beagle."
    • Let C(x) C(x) represent "x is a corgi."

Step 2: Structure the Statement

The statement "Some dogs are kelpies, some are beagles, and the rest are corgis" can be interpreted in parts as follows:

  • There exists some dogs that are kelpies.
  • There exists some dogs that are beagles.
  • The rest of the dogs must be corgis.

Step 3: Formulate the MPL

Using the existential quantifier, we will express the statement in MPL:

  • There exist dogs that are kelpies: x(D(x)K(x)) \exists x (D(x) \land K(x))
  • There exist dogs that are beagles: x(D(x)B(x)) \exists x (D(x) \land B(x))
  • The rest are corgis:
    • For this part, we can express that if a dog is not a kelpie or a beagle, it is a corgi: x(D(x)¬K(x)¬B(x)C(x)) \forall x (D(x) \land \neg K(x) \land \neg B(x) \rightarrow C(x))

Final MPL Representation

Combining these parts, the full representation in MPL is: x(D(x)K(x))y(D(y)B(y))z(D(z)¬K(z)¬B(z)C(z)) \exists x (D(x) \land K(x)) \land \exists y (D(y) \land B(y)) \land \forall z (D(z) \land \neg K(z) \land \neg B(z) \rightarrow C(z))

This represents the original statement accurately in Mathematical Predicate Logic.

This problem has been solved

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