Q1) Q2) Q3) Use the table below to determine whether this proposed rule of infer

Question

Q1)

Q2)

Q3)

Use the table below to determine whether this proposed rule of inference is valid.

p v ~(q -> r)

------------

~(q <-> r)

Here is a (correct) truth table for the two statements in the rule.

Q4:

Q5:

If we let q be a short name for (i.e., be equivalent to) the expression r ^ s, then which of these is the expression p -> q equivalent to?

Consider the following rule of inference:

a -> b

b -> c

------

a -> c

Which of the following is the result of applying the given rule to the following statements:

p -> (q v r)

q -> s

(If there are multiple correct answers, choose any one.)

p -> q

p -> r

p -> s

p -> (q v r)

Some other statement.

None of these, because the rule does not apply to these statements.

Consider the following rule of inference:

a v b

~a

------

b

Which of the following is the result of applying the given rule to the following statement(s):

(p ^ r) v (q ^ s)

~(p ^ r)

(If there are multiple correct answers, choose any one.)

p ^ r

~(p ^ r)

q ^ s

~(q ^ s)

None of these, but application of the rule to these statements could generate a different statement as a result.

None of these, because the rule does not apply to these statements.

Any help on these questions will be greatly appreciated. Please, if you pick an option to an answer, let me know why

you picked it.

Thank you.

1. p -> s 2. p -> r 3. (p -> r) ^ s 4. p -> (r ^ s) 5. None of these

Use the table below to determine whether this proposed rule of inference is valid. p-> q Here is a (correct) truth table for the three statements in the rule. If the rule is invalid, select any one line of the truth table which proves that the rule is invalid. (There may be more than one. but it only takes one counterexample to its validity to show that a rule is invalid.) Reminder: What makes a rule of inference valid? Anytime all of its preconditions are true, its conclusion must also be true. (What does that say about situations where its preconditions are false?) 1 2 3 4 None of these, because the rule is valid. Use the table below to determine whether this proposed rule of inference is valid. p-> q Here is a (correct) truth table for the three statements in the rule. If the rule is invalid, select any one line of the truth table which proves that the rule is invalid. (There may be more than one. but it only takes one counterexample to its validity to show that a rule is invalid.) 1 2 3 4 None of these, because the rule is valid. If the rule is invalid, select any one line of the truth table which proves that the rule is invalid. (There may be more than one, but it only takes one counterexample to its validity to show that a rule is invalid.) 1 2 3 4 5 6 7 8 None of these, because the rule is valid. Use the table below to determine whether this proposed rule of inference is valid. p^~p Here is a (correct) truth table for the two statements in the rule. If the rule is invalid, select any one line of the truth table which proves that the rule is invalid. (There may be more than one. but it only takes one counterexample to its validity to show that a rule is invalid.) (WARNING: there's a bit of a trick here, but it illustrates just how useful any proof is that starts from flawed premises.. and the reason logicians have struggled for hundreds of years to ensure their logics are built on solid foundations.) (Hint: go back up to the 1st question and reread that reminder about how to determine whether a rule is valid before you answer this tricky question!) 1 2 3 4 None of these, because the rule is valid.

Explanation / Answer

1). Answer = 5

The rule is valid

because, (p -> q) = (~p v q)

and (~p v q) ^ ~q = ~p

2). Answer = 3

because p -> q = T and q = T must imply p = T

but p=F in 3

Hence rule is invalid

3). Answer = 4

because p v ~(q -> r) = T and ~(q <-> r) = F

Hence rule is invalid

4). Answer = 5

The rule is valid

because, (p ^ ~p) = F (always)

5). Answer = 4

since q = r ^ s

p -> q = p -> (r ^ s)

6). Answer = 4 and 6

case 1: p = T and q = F and r = T

then 1 is incorrect

case 2: p = T and q = T and r = F

then 2 is incorrect

case 3: p = T and q = F and r = T and s=F

then 3 is incorrect

4 is always correctin all cases

and

6 is also correct answer because for proving 4 to be correct, we haven't used the given rule.

7).Answer = 3

clearly from the rule given

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