A. Directions. Determine whether the following three arguments are valid using t

Question

A. Directions.  Determine whether the following three arguments are valid using the truth table method.  Use the Indirect Truth Table method. Indicate whether each is valid or not. Note that ‘//’ is used as the conclusion indicator and ‘/’ is used to separate the premises.  [Note: Use only the following logical symbols: ‘&’ for conjunctions, ‘v’ for disjunctions, ‘->’ for conditionals, ‘<->’ for biconditionals, ‘~’ for negations.]

1. (E <-> ~W)   /   ~(W v ~G)   //   (E v O)

2. (L -> D)   /   [(L v A) & ~H]   /   ~(H -> ~R)        //    (A -> ~R)

3. (P -> S)   /   (~J -> ~S)   /   (~P -> ~J)   //   (S v J)

B. Directions. Let G be known to be true; let the values of S and R be unknown. Can the truth values of the following two sentences be determined just by using truth tables. If so, say what their truth values are. If not, explain why not. (Please don’t take this one lightly.)

1. [(S v ~G) -> (~S v G)]

2. {~[(S & R) v ~(R v S)] & ~[(~R v ~S) v (~S & ~R)]}

Explanation / Answer

A. 1. (E <-> ~W)   /   ~(W v ~G)   //   (E v O)

ASSUME that the argument is INVALID. That means we will place a T under the main connective of every premise, and an F under the main connective of the conclusion.

(E <-> ~W)   /   ~(W v ~G)   //   (E v O)

T T F

Since, the conclusion (E v O) is F only if both E and O have truth values F. Thus, we have

(E <-> ~W)   /   ~(W v ~G)   //   (E v O)

T T F F F

Since, E has final assignment of F, therefore, in first premse we have following assignments

(E <-> ~W)   /   ~(W v ~G)   //   (E v O)

F T FT T F F F

Now, W has an assignment T, i.e. we have

(E <-> ~W)   /   ~(W v ~G)   //   (E v O)

F T FT T T F F F

This makes a problem or inconsistency in the second premise, which will give T to the expression inside parenthesis no matter what's the assignment of ~G, making outside of the parenthesis F. Which proves that our original assumption (that the argument was invalid) was wrong. So instead of being Invalid, the argument is VALID.

2. (L -> D)   /   [(L v A) & ~H]   /   ~(H -> ~R)        //    (A -> ~R)

ASSUME that the argument is INVALID. That means we will place a T under the main connective of every premise, and an F under the main connective of the conclusion.

(L -> D)   /   [(L v A) & ~H]   /   ~(H -> ~R)        //    (A -> ~R)

T T T F

Now, in the third premise H and R must be T. Therefore, we have

(L -> D)   /   [(L v A) & ~H]   /   ~(H -> ~R)        //    (A -> ~R)

T T FT T T F FT F FT

In concusion premise, A must be T, and thtruth assignments are now,

(L -> D)   /   [(L v A) & ~H]   /   ~(H -> ~R)        //    (A -> ~R)

T T T FT T T F FT T F FT

In second premise L can have T or F values. If L has truth value T then in first premise, to hold D must be T, and therefore, we have

(L -> D)   /   [(L v A) & ~H]   /   ~(H -> ~R)        //    (A -> ~R)

T T T T T T T FT T T F FT T F FT

Thus, we have no inconsistency in the premises. Which proves that our original assumption (that the argument was invalid) was True. So argument is Invalid,

3. (P -> S)   /   (~J -> ~S)   /   (~P -> ~J)   //   (S v J)

ASSUME that the argument is INVALID. That means we will place a T under the main connective of every premise, and an F under the main connective of the conclusion.

(P -> S)   /   (~J -> ~S)   /   (~P -> ~J)   //   (S v J)

T T T F

Since, the conclusion (S v J) is F only if both E and O have truth values F. Thus, we have

(P -> S)   /   (~J -> ~S)   /   (~P -> ~J)   //   (S v J)

T F TF T TF T TF F F F

Now, in the first premise P must be false, i.e.

(P -> S)   /   (~J -> ~S)   /   (~P -> ~J)   //   (S v J)

F T F TF T TF TF T TF F F F

Thus, we have no inconsistency in the premises. Which proves that our original assumption (that the argument was invalid) was True. So argument is Invalid,

B. 1. [(S v ~G) -> (~S v G)]

Given that G is T, therefore we have

  [(S v ~G) -> (~S v G)]

FT T

Now, (~S v G) is T for all truth assignment of S. Therefore, [(S v ~G) -> (~S v G) is T for all possible truth values of (S v ~G). Hence, the given argument is valid.

2. {~[(S & R) v ~(R v S)] & ~[(~R v ~S) v (~S & ~R)]}

For & to have truth value T, both components within brackets must be F, therefore, we have

  {~[(S & R) v ~(R v S)] & ~[(~R v ~S) v (~S & ~R)]}

T F T T F

Now, [(S & R) v ~(R v S)] to be false. Now (S & R) to be false, one of the R or S must be false, and other may be T. Let R has the truth value T, and S has truth value F, then we have

{~[(S & R) v ~(R v S)] & ~[(~R v ~S) v (~S & ~R)]}

T F T F F T T F T T FT TF F TF FT

Now, (~R v ~S) results in T which makes [(~R v ~S) v (~S & ~R)], T, which is a contradiction. Hence the argument is invalid.

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