Prove whether each argument is valid or invalid. First find the form of the argument by defining predicates and expressing the hypotheses and the conclusion using the predicates. If the argument is valid, then use the rules of inference to prove that the form is valid. If the argument is invalid, give values for the predicates you defined for a small domain that demonstrate the argument is invalid. The domain for each problem is the set of students in a class. (e) Every student who missed class or got a detention did not get an A. Penelope is a student in the class. Penelope got an A.

The given argument  ∀x (S(x) ∧ (M(x) V D(x)) --> ¬ A(x)) ∧ S(Penelope) ∧ A(Penelope) -->¬ D(Penelope) is valid.

Explanation:

Solution:

Let us Consider following predicates:

M(x): x missed the class

D(x): x got a detention

S(x): x is a student in the class

A(x): x got an A

Now,

We Express the hypotheses and conclusion as:

The hypotheses: ∀x (S(x) ∧ (M(x) V D(x)) -->  A(x)) and S(Penelope) and A(Penelope)

So,

The Conclusion:  D(Penelope)

Thus,

The Argument:

∀x (S(x) ∧ (M(x) V D(x)) -->  A(x)) ∧ S(Penelope) ∧ A(Penelope) -->D(Penelope)

Then,

The given argument is valid or correct and prove using the inference  rule as follows:

Step                      Premises                    Reason (Rule used)

1.  ∀x (S(x) ∧ (M(x) V D(x)) -->  A(x))                Premise

2. S(Penelope) ∧ (M(Penelope)

  V D(Penelope)) --> ¬ A(Penelope)            Universal instantiation

3.  S(Penelope)                                              Premise

4.  A(Penelope)                                              Premise

5. ¬[S(Penelope) ∧ (M(Penelope)

  V D(Penelope))]                                         2,4, Modus Tollens

6. ¬S(Penelope) V (¬M(Penelope)

  ∧ ¬D(Penelope))                                        De Morgan law

7.¬M(Penelope) ∧ ¬D(Penelope)                 3,6,Disjunctive Syllogism

8¬D(Penelope)                                              7, Simplification

Therefore, the given argument ∀x (S(x) ∧ (M(x) V D(x)) --> ¬ A(x)) ∧ S(Penelope) ∧ A(Penelope) -->¬ D(Penelope) is valid.

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