If there is one prayer that you should pray/sing every day and every hour, it is the LORD's prayer (Our FATHER in Heaven prayer)
It is the most powerful prayer. A pure heart, a clean mind, and a clear conscience is necessary for it. - Samuel Dominic Chukwuemeka

For in GOD we live, and move, and have our being. - Acts 17:28

The Joy of a Teacher is the Success of his Students. - Samuel Chukwuemeka

Solved Examples on Symbolic Arguments and Syllogistic Arguments

Samuel Dominic Chukwuemeka (SamDom For Peace) Unless stated otherwise:
Determine whether these arguments are valid or invalid. Show all work.

For symbolic arguments, use at least two methods: by Definition, by Formula and/or by the Valid Forms of Arguments and Invalid Forms of Arguments for each argument as applicable.

For syllogistic arguments, use Euler diagrams or Venn diagrams as applicable.

(1.)

$ p \leftrightarrow q \\[2ex] p \underline{\lor} q \\ \rule{1.2in}{0.3pt} \\ \therefore \neg q $


$ Premise\: 1: p \leftrightarrow q \\[2ex] Premise\: 2: p \underline{\lor} q \\ \rule{2.5in}{0.5pt} \\ Conclusion\: \therefore \neg q \\[3ex] $ Second Method: Definition
Let us draw a truth table of the premises and the conclusion.
$p$ $q$ $p \leftrightarrow q$ $p \underline{\lor} q$ $\neg q$
$T$ $T$ $T$ $F$ $F$
$T$ $F$ $F$ $T$ $T$
$F$ $T$ $F$ $T$ $F$
$F$ $F$ $T$ $F$ $T$
Premise 1 Premise 2 Conclusion

There is no case where both premises are true.
Argument is valid by Definition

Third Method: Formula
Let us draw the truth table for the formula: $[(premise\: 1) \land (premise\: 2)] \rightarrow conclusion$
$[(p \leftrightarrow q) \land (p \underline{\lor} q)] \rightarrow \neg q$
$p$ $q$ $p \leftrightarrow q$ $p \underline{\lor} q$ $(p \rightarrow q) \land (p \underline{\lor} q)$ $\neg q$ $[(p \rightarrow q) \land (p \underline{\lor} q)] \rightarrow \neg q$
$T$ $T$ $T$ $F$ $F$ $F$ $T$
$T$ $F$ $F$ $T$ $F$ $T$ $T$
$F$ $T$ $F$ $T$ $F$ $F$ $T$
$F$ $F$ $T$ $F$ $F$ $T$ $T$
The formula is a tautology.
Argument is valid by Formula

(2.)

$ p \lor q \\[2ex] p \\ \rule{1.2in}{0.3pt} \\ \therefore \neg q $


$ Premise\: 1: p \lor q \\[2ex] Premise\: 2: p \\ \rule{2.5in}{0.5pt} \\ Conclusion\: \therefore \neg q \\[3ex] $ First Method: Valid and Invalid Forms of Arguments
Argument is invalid by Misuse of Disjunctive Reasoning

Second Method: Definition
Let us draw a truth table of the premises and the conclusion.
$p$ $q$ $p \lor q$ $\neg q$
$T\checkmark$ $T$ $T\checkmark$ $F\times$
$T\checkmark$ $F$ $T\checkmark$ $T\checkmark$
$F$ $T$ $T$ $F$
$F$ $F$ $F$ $T$
Premise 2 Premise 1 Conclusion
In the $1st$ case, the conclusion is false when both premises are true.
Argument is invalid by Definition

Third Method: Formula
Let us draw the truth table for the formula: $[(premise\: 1) \land (premise\: 2)] \rightarrow conclusion$
$[(p \lor q) \land p] \rightarrow \neg q$
$p$ $q$ $p \lor q$ $(p \lor q) \land p$ $\neg q$ $[(p \lor q) \land p] \rightarrow \neg q$
$T$ $T$ $T$ $T$ $F$ $F$
$T$ $F$ $T$ $T$ $T$ $T$
$F$ $T$ $T$ $F$ $F$ $T$
$F$ $F$ $F$ $F$ $T$ $T$

The formula is not a tautology.
Argument is invalid by Formula

(3.)

$ p \rightarrow \neg q \\[2ex] q \\ \rule{1.2in}{0.3pt} \\ \therefore \neg p $


$ Premise\: 1: p \rightarrow \neg q \\[2ex] Premise\: 2: q \\ \rule{2.5in}{0.5pt} \\ Conclusion\: \therefore \neg p \\[3ex] $ First Method: Valid and Invalid Forms of Arguments
Argument is valid by Modus Tollens

Second Method: Definition
Let us draw a truth table of the premises and the conclusion.
$p$ $q$ $\neg q$ $p \rightarrow \neg q$ $\neg p$
$T$ $T$ $F$ $F$ $F$
$T$ $F$ $T$ $T$ $F$
$F$ $T\checkmark$ $F$ $T\checkmark$ $T\checkmark$
$F$ $F$ $T$ $T$ $T$
Premise 2 Premise 1 Conclusion

In all the cases when both premises are true (3rd case), the conclusion is also true.
Argument is valid by Definition

Third Method: Formula
Let us draw the truth table for the formula: $[(premise\: 1) \land (premise\: 2)] \rightarrow conclusion$
$[(p \rightarrow \neg q) \land q] \rightarrow \neg p$
$p$ $q$ $\neg q$ $p \rightarrow \neg q$ $(p \rightarrow \neg q) \land q$ $\neg p$ $[(p \rightarrow \neg q) \land q] \rightarrow \neg p$
$T$ $T$ $F$ $F$ $F$ $F$ $T$
$T$ $F$ $T$ $T$ $F$ $F$ $T$
$F$ $T$ $F$ $T$ $T$ $T$ $T$
$F$ $F$ $T$ $T$ $F$ $T$ $T$

The formula is a tautology.
Argument is valid by Formula

(4.)

$ p \lor q \\[2ex] \neg p \\ \rule{1.2in}{0.3pt} \\ \therefore q $


$ Premise\: 1: p \lor q \\[2ex] Premise\: 2: \neg p \\ \rule{2.5in}{0.5pt} \\ Conclusion\: \therefore q \\[3ex] $ First Method: Valid and Invalid Forms of Arguments
Argument is valid by Disjunctive Reasoning

Second Method: Definition
Let us draw a truth table of the premises and the conclusion.
$p$ $q$ $p \lor q$ $\neg p$
$T$ $T$ $T$ $F$
$T$ $F$ $T$ $F$
$F$ $T\checkmark$ $T\checkmark$ $T\checkmark$
$F$ $F$ $F$ $T$
Conclusion Premise 1 Premise 2

The conclusion is true when both premises are true.
Yes, it is only one case: Case 3...that is fine.
Argument is valid by Definition

Third Method: Formula
Let us draw the truth table for the formula: $[(premise\: 1) \land (premise\: 2)] \rightarrow conclusion$
$[(p \lor q) \land \neg p] \rightarrow q$
$p$ $q$ $p \lor q$ $\neg p$ $(p \lor q) \land \neg p$ $[(p \lor q) \land \neg p] \rightarrow q$
$T$ $T$ $T$ $F$ $F$ $T$
$T$ $F$ $T$ $F$ $F$ $T$
$F$ $T$ $T$ $T$ $T$ $T$
$F$ $F$ $F$ $T$ $F$ $T$
The formula is a tautology.
Argument is valid by Formula

(5.)

$ (p \lor q) \rightarrow r \\[2ex] \rule{1.7in}{0.3pt} \\ \therefore (p \land q) \rightarrow r $


$ Premise\: 1: (p \lor q) \rightarrow r \\[2ex] \rule{3.1in}{0.3pt} \\ Conclusion\: \therefore (p \land q) \rightarrow r \\[3ex] $ Second Method: Definition
Let us draw a truth table of the premise and the conclusion.
$p$ $q$ $r$ $p \lor q$ $(p \lor q) \rightarrow r$ $p \land q$ $(p \land q) \rightarrow r$
$T$ $T$ $T$ $T$ $T\checkmark$ $T$ $T\checkmark$
$T$ $T$ $F$ $T$ $F$ $T$ $F$
$T$ $F$ $T$ $T$ $T\checkmark$ $F$ $T\checkmark$
$T$ $F$ $F$ $T$ $F$ $F$ $T$
$F$ $T$ $T$ $T$ $T\checkmark$ $F$ $T\checkmark$
$F$ $T$ $F$ $T$ $F$ $F$ $T$
$F$ $F$ $T$ $F$ $T\checkmark$ $F$ $T\checkmark$
$F$ $F$ $F$ $F$ $T\checkmark$ $F$ $T\checkmark$
Premise 1 Conclusion
The conclusion is true whenever the premise is true.
Argument is valid by Definition

Third Method: Formula
Let us draw the truth table for the formula: $premise\: 1 \rightarrow conclusion$
$[(p \lor q) \rightarrow r] \rightarrow [(p \land q) \rightarrow r]$
$p$ $q$ $r$ $p \lor q$ $(p \lor q) \rightarrow r$ $p \land q$ $(p \land q) \rightarrow r$ $[(p \lor q) \rightarrow r] \rightarrow [(p \land q) \rightarrow r]$
$T$ $T$ $T$ $T$ $T$ $T$ $T$ $T$
$T$ $T$ $F$ $T$ $F$ $T$ $F$ $T$
$T$ $F$ $T$ $T$ $T$ $F$ $T$ $T$
$T$ $F$ $F$ $T$ $F$ $F$ $T$ $T$
$F$ $T$ $T$ $T$ $T$ $F$ $T$ $T$
$F$ $T$ $F$ $T$ $F$ $F$ $T$ $T$
$F$ $F$ $T$ $F$ $T$ $F$ $T$ $T$
$F$ $F$ $F$ $F$ $T$ $F$ $T$ $T$
The formula is a tautology.
Argument is valid by Formula

(6.)

$ p \land q \\[2ex] \rule{1.2in}{0.3pt} \\ \therefore p $


$ Premise\: 1: p \land q \\[2ex] \rule{2.5in}{0.5pt} \\ Conclusion\: \therefore p \\[3ex] $ First Method: Valid and Invalid Forms of Arguments
Argument is valid by Simplification

Second Method: Definition
Let us draw a truth table of the premise and the conclusion.
$p$ $q$ $p \land q$
$T\checkmark$ $T$ $T\checkmark$
$T$ $F$ $F$
$F$ $T$ $F$
$F$ $F$ $F$
Conclusion Premise 1
The conclusion is true whenever the premise is true.
Yes, it is only one case: Case $1$...that is fine.
Argument is valid by Definition

Third Method: Formula
Let us draw the truth table for the formula: $premise\: 1 \rightarrow conclusion$
$(p \land q) \rightarrow p$
$p$ $q$ $p \land q$ $(p \land q) \rightarrow p$
$T$ $T$ $T$ $T$
$T$ $F$ $F$ $T$
$F$ $T$ $F$ $T$
$F$ $F$ $F$ $T$
The formula is a tautology.
Argument is valid by Formula
(7.) If yesterday was Sunday, then today is not Saturday.
Today is Saturday.
Therefore, yesterday was not Sunday.


Let:
$p$: Yesterday was Sunday.
$q$: Today is Saturday.

$ Premise\: 1: p \rightarrow \neg q \\[2ex] Premise\: 2: q \\ \rule{2.5in}{0.5pt} \\ Conclusion\: \therefore \neg p \\[3ex] $ First Method: Valid and Invalid Forms of Arguments
Argument is valid by Modus Tollens

Second Method: Definition
Let us draw a truth table of the premises and the conclusion.
$p$ $q$ $\neg q$ $p \rightarrow \neg q$ $\neg p$
$T$ $T$ $F$ $F$ $F$
$T$ $F$ $T$ $T$ $F$
$F$ $T\checkmark$ $F$ $T\checkmark$ $T\checkmark$
$F$ $F$ $T$ $T$ $T$
Premise 2 Premise 1 Conclusion

In all the cases when both premises are true (3rd case), the conclusion is also true.
Argument is valid by Definition

Third Method: Formula
Let us draw the truth table for the formula: $[(premise\: 1) \land (premise\: 2)] \rightarrow conclusion$
$[(p \rightarrow \neg q) \land q] \rightarrow \neg p$
$p$ $q$ $\neg q$ $p \rightarrow \neg q$ $(p \rightarrow \neg q) \land q$ $\neg p$ $[(p \rightarrow \neg q) \land q] \rightarrow \neg p$
$T$ $T$ $F$ $F$ $F$ $F$ $T$
$T$ $F$ $T$ $T$ $F$ $F$ $T$
$F$ $T$ $F$ $T$ $T$ $T$ $T$
$F$ $F$ $T$ $T$ $F$ $T$ $T$

The formula is a tautology.
Argument is valid by Formula
(8.) If we permit expanded fracking, then there will be more natural gas available.
There is more natural gas available.
We permitted expanded fracking.


Let:
$p$: We permit expanded fracking.
$q$: There will be more natural gas available.

$ Premise\: 1: p \rightarrow q \\[2ex] Premise\: 2: q \\ \rule{2.5in}{0.5pt} \\ Conclusion\: \therefore p \\[3ex] $ First Method: Valid and Invalid Forms of Arguments
Argument is invalid by Fallacy of the Converse (also known as Affirming the Conclusion)

Second Method: Definition
Let us draw a truth table of the premises and the conclusion.
$p$ $q$ $p \rightarrow q$
$T\checkmark$ $T\checkmark$ $T\checkmark$
$T$ $F$ $F$
$F\times$ $T\checkmark$ $T\checkmark$
$F$ $F$ $T$
Conclusion Premise 2 Premise 1

In the 3rd case, the conclusion is false when both premises are true.
Argument is invalid by Definition

Third Method: Formula
Let us draw the truth table for the formula: $[(premise\: 1) \land (premise\: 2)] \rightarrow conclusion$
$[(p \rightarrow q) \land q] \rightarrow p$
$p$ $q$ $p \rightarrow q$ $(p \rightarrow q) \land q$ $[(p \rightarrow q) \land q] \rightarrow p$
$T$ $T$ $T$ $T$ $T$
$T$ $F$ $F$ $F$ $T$
$F$ $T$ $T$ $T$ $F$
$F$ $F$ $T$ $F$ $T$

The formula is not a tautology.
Argument is invalid by Formula
(9.)

$ p \leftrightarrow q \\[2ex] q \rightarrow r \\ \rule{1.2in}{0.3pt} \\ \therefore \neg r \rightarrow \neg p $


$ Premise\: 1: p \leftrightarrow q \\[2ex] Premise\: 2: q \rightarrow r \\ \rule{2.5in}{0.5pt} \\ Conclusion\: \therefore \neg r \rightarrow \neg p \\[3ex] $ Second Method: Definition
Let us draw a truth table of the premises and the conclusion.
$p$ $q$ $r$ $p \leftrightarrow q$ $q \rightarrow r$ $\neg r$ $\neg p$ $\neg r \rightarrow \neg p$
$T$ $T$ $T$ $T\checkmark$ $T\checkmark$ $F$ $F$ $T\checkmark$
$T$ $T$ $F$ $T$ $F$ $T$ $F$ $F$
$T$ $F$ $T$ $F$ $T$ $F$ $F$ $T$
$T$ $F$ $F$ $F$ $T$ $T$ $F$ $F$
$F$ $T$ $T$ $F$ $T$ $F$ $T$ $T$
$F$ $T$ $F$ $F$ $F$ $T$ $T$ $T$
$F$ $F$ $T$ $T\checkmark$ $T\checkmark$ $F$ $T$ $T\checkmark$
$F$ $F$ $F$ $T\checkmark$ $T\checkmark$ $T$ $T$ $T\checkmark$
Premise 1 Premise 2 Conclusion

In all the cases when both premises are true, the conclusion is also true.
Argument is valid by Definition

Third Method: Formula
Let us draw the truth table for the formula: $[(premise\: 1) \land (premise\: 2)] \rightarrow conclusion$
$[(p \rightarrow \neg q) \land q] \rightarrow \neg p$
$p$ $q$ $r$ $p \leftrightarrow q$ $q \rightarrow r$ $\neg r$ $\neg p$ $\neg r \rightarrow \neg p$ $(p \leftrightarrow q) \land (q \rightarrow r)$ $[(p \leftrightarrow q) \land (q \rightarrow r)] \rightarrow (\neg r \rightarrow \neg p)$
$T$ $T$ $T$ $T$ $T$ $F$ $F$ $T$ $T$ $T$
$T$ $T$ $F$ $T$ $F$ $T$ $F$ $F$ $F$ $T$
$T$ $F$ $T$ $F$ $T$ $F$ $F$ $T$ $F$ $T$
$T$ $F$ $F$ $F$ $T$ $T$ $F$ $F$ $F$ $T$
$F$ $T$ $T$ $F$ $T$ $F$ $T$ $T$ $F$ $T$
$F$ $T$ $F$ $F$ $F$ $T$ $T$ $T$ $F$ $T$
$F$ $F$ $T$ $T$ $T$ $F$ $T$ $T$ $T$ $T$
$F$ $F$ $F$ $T$ $T$ $T$ $T$ $T$ $T$ $T$

The formula is a tautology.
Argument is valid by Formula

(10.) You must eat well or you will not be healthy.
I eat well.
Therefore, I am healthy.


Let:
$p$: You must eat well.
$q$: You will be healthy.

$ Premise\: 1: p \lor \neg q \\[2ex] Premise\: 2: p \\ \rule{2.5in}{0.5pt} \\ Conclusion\: \therefore q \\[3ex] $ First Method: Valid and Invalid Forms of Arguments
Argument is invalid by Misuse of Disjunctive Reasoning

Second Method: Definition
Let us draw a truth table of the premises and the conclusion.
$p$ $q$ $\neg q$ $p \lor \neg q$
$T\checkmark$ $T\checkmark$ $F$ $T\checkmark$
$T\checkmark$ $F\times$ $T$ $T\checkmark$
$F$ $T$ $F$ $F$
$F$ $F$ $T$ $T$
Premise 2 Conclusion Premise 1
In the 2nd case, the conclusion is false when both premises are true.
Argument is invalid by Definition

Third Method: Formula
Let us draw the truth table for the formula: $[(premise\: 1) \land (premise\: 2)] \rightarrow conclusion$
$[(p \lor \neg q) \land p] \rightarrow q$
$p$ $q$ $\neg q$ $p \lor \neg q$ $(p \lor \neg q) \land p$ $[(p \lor \neg q) \land p] \rightarrow q$
$T$ $T$ $F$ $T$ $T$ $T$
$T$ $F$ $T$ $T$ $T$ $F$
$F$ $T$ $F$ $F$ $F$ $T$
$F$ $F$ $T$ $T$ $F$ $T$
The formula is not a tautology.
Argument is invalid by Formula

(11.) Premise: If you live in Baltimore, then you live in Maryland.
Premise: Amanda does not live in Baltimore.
Conclusion: Amanda does not live in Maryland.


Let:
$p$: Amanda lives in Baltimore.
$q$: Amanda lives in Maryland.

$ Premise\: 1: p \rightarrow q \\[2ex] Premise\: 2: \neg p \\ \rule{2.5in}{0.5pt} \\ Conclusion\: \therefore \neg q \\[3ex] $ First Method: Valid and Invalid Forms of Arguments
Argument is invalid by Fallacy of the Inverse

Second Method: Definition
Let us draw a truth table of the premises and the conclusion.
$p$ $q$ $p \rightarrow q$ $\neg p$ $\neg q$
$T$ $T$ $T$ $F$ $F$
$T$ $F$ $F$ $F$ $T$
$F$ $T$ $T\checkmark$ $T\checkmark$ $F\times$
$F$ $F$ $T\checkmark$ $T\checkmark$ $T\checkmark$
Premise 1 Premise 2 Conclusion
In the 3rd case, the conclusion is false when both premises are true.
Argument is invalid by Definition

Third Method: Formula
Let us draw the truth table for the formula: $[(premise\: 1) \land (premise\: 2)] \rightarrow conclusion$
$[(p \rightarrow q) \land \neg p] \rightarrow \neg q$
$p$ $q$ $p \rightarrow q$ $\neg p$ $\neg q$ $(p \rightarrow q) \land \neg p$ $[(p \rightarrow q) \land \neg p] \rightarrow \neg q$
$T$ $T$ $T$ $F$ $F$ $F$ $T$
$T$ $F$ $F$ $F$ $T$ $F$ $T$
$F$ $T$ $T$ $T$ $F$ $T$ $F$
$F$ $F$ $T$ $T$ $T$ $T$ $T$
The formula is not a tautology.
Argument is invalid by Formula

(12.) We must build a hydroelectric plant or a nuclear plant.
We won't build a nuclear plant, so we must build a hydroelectric plant.


Let:
$p$: We must build a hydroelectric plant.
$q$: We must build a nuclear plant.

$ Premise\: 1: p \lor q \\[2ex] Premise\: 2: \neg q \\ \rule{2.5in}{0.5pt} \\ Conclusion\: \therefore p \\[3ex] $ First Method: Valid and Invalid Forms of Arguments
Argument is valid by Disjunctive Reasoning

Second Method: Definition
Let us draw a truth table of the premises and the conclusion.
$p$ $q$ $p \lor q$ $\neg q$
$T$ $T$ $T$ $F$
$T\checkmark$ $F$ $T\checkmark$ $T\checkmark$
$F$ $T$ $T$ $F$
$F$ $F$ $F$ $T$
Conclusion Premise 1 Premise 2

The conclusion is true when both premises are true.
Yes, it is only one case: Case 2...that is fine.
Argument is valid by Definition

Third Method: Formula
Let us draw the truth table for the formula: $[(premise\: 1) \land (premise\: 2)] \rightarrow conclusion$
$[(p \lor q) \land \neg q] \rightarrow p$
$p$ $q$ $p \lor q$ $\neg q$ $(p \lor q) \land \neg q$ $[(p \lor q) \land \neg q] \rightarrow p$
$T$ $T$ $T$ $F$ $F$ $T$
$T$ $F$ $T$ $T$ $T$ $T$
$F$ $T$ $T$ $F$ $F$ $T$
$F$ $F$ $F$ $T$ $F$ $T$

The formula is a tautology.
Argument is valid by Formula

(13.)

$ p \rightarrow q \\[2ex] q \land r \\ \rule{1.2in}{0.3pt} \\ \therefore p \lor r $


$ Premise\: 1: p \rightarrow q \\[2ex] Premise\: 2: q \land r \\ \rule{2.5in}{0.5pt} \\ Conclusion\: \therefore p \lor r \\[3ex] $ Second Method: Definition
Let us draw a truth table of the premises and the conclusion.
$p$ $q$ $r$ $p \rightarrow q$ $q \land r$ $p \lor r$
$T$ $T$ $T$ $T\checkmark$ $T\checkmark$ $T\checkmark$
$T$ $T$ $F$ $T$ $F$ $T$
$T$ $F$ $T$ $F$ $F$ $T$
$T$ $F$ $F$ $F$ $F$ $T$
$F$ $T$ $T$ $T\checkmark$ $T\checkmark$ $T\checkmark$
$F$ $T$ $F$ $T$ $F$ $F$
$F$ $F$ $T$ $T$ $F$ $T$
$F$ $F$ $F$ $T$ $F$ $F$
Premise 1 Premise 2 Conclusion

In all the cases when both premises are true, the conclusion is also true.
Argument is valid by Definition

Third Method: Formula
Let us draw the truth table for the formula: $[(premise\: 1) \land (premise\: 2)] \rightarrow conclusion$
$[(p \rightarrow \neg q) \land q] \rightarrow \neg p$
$p$ $q$ $r$ $p \rightarrow q$ $q \land r$ $p \lor r$ $(p \rightarrow q) \land (q \land r)$ $[(p \rightarrow q) \land (q \land r)] \rightarrow (p \lor r)$
$T$ $T$ $T$ $T$ $T$ $T$ $T$ $T$
$T$ $T$ $F$ $T$ $F$ $T$ $F$ $T$
$T$ $F$ $T$ $F$ $F$ $T$ $F$ $T$
$T$ $F$ $F$ $F$ $F$ $T$ $F$ $T$
$F$ $T$ $T$ $T$ $T$ $T$ $T$ $T$
$F$ $T$ $F$ $T$ $F$ $F$ $F$ $T$
$F$ $F$ $T$ $T$ $F$ $T$ $F$ $T$
$F$ $F$ $F$ $T$ $F$ $F$ $F$ $T$

The formula is a tautology.
Argument is valid by Formula

(14.)

$ p \rightarrow q \\[2ex] \neg p \\ \rule{1.2in}{0.3pt} \\ \therefore q $


$ Premise\: 1: p \rightarrow q \\[2ex] Premise\: 2: \neg p \\ \rule{2.5in}{0.5pt} \\ Conclusion\: \therefore q \\[3ex] $ Second Method: Definition
Let us draw a truth table of the premises and the conclusion.
$p$ $q$ $p \rightarrow q$ $\neg p$
$T$ $T$ $T$ $F$
$T$ $F$ $F$ $F$
$F$ $T\checkmark$ $T\checkmark$ $T\checkmark$
$F$ $F\times$ $T\checkmark$ $T\checkmark$
Conclusion Premise 1 Premise 2

In the 4th case, the conclusion is false when both premises are true.
Argument is invalid by Definition

Third Method: Formula
Let us draw the truth table for the formula: $[(premise\: 1) \land (premise\: 2)] \rightarrow conclusion$
$[(p \rightarrow q) \land \neg p] \rightarrow q$
$p$ $q$ $p \rightarrow q$ $\neg p$ $(p \rightarrow q) \land \neg p$ $[(p \rightarrow q) \land \neg p] \rightarrow q$
$T$ $T$ $T$ $F$ $F$ $T$
$T$ $F$ $F$ $F$ $F$ $T$
$F$ $T$ $T$ $T$ $T$ $T$
$F$ $F$ $T$ $T$ $T$ $F$

The formula is not a tautology.
Argument is invalid by Formula
(15.) Premise: If a corrosive is acid, then it is dangerous.
Premise: Salt is not dangerous.
Conclusion: Salt is not a type of acid.


We already know that salt is corrosive.
Hence, let:
$p$: Salt is a type of acid.
$q$: Salt is dangerous.

$ Premise\: 1: p \rightarrow q \\[2ex] Premise\: 2: \neg q \\ \rule{2.5in}{0.5pt} \\ Conclusion\: \therefore \neg p \\[3ex] $ First Method: Valid and Invalid Forms of Arguments
Argument is valid by Modus Tollens

Second Method: Definition
Let us draw a truth table of the premises and the conclusion.
$p$ $q$ $p \rightarrow q$ $\neg q$ $\neg p$
$T$ $T$ $T$ $F$ $F$
$T$ $F$ $F$ $T$ $F$
$F$ $T$ $T$ $F$ $T$
$F$ $F$ $T\checkmark$ $T\checkmark$ $T\checkmark$
Premise 1 Premise 2 Conclusion

In the case where both premises are true, the conclusion is also true.
Argument is valid by Definition

Third Method: Formula
Let us draw the truth table for the formula: $[(premise\: 1) \land (premise\: 2)] \rightarrow conclusion$
$[(p \rightarrow q) \land \neg q] \rightarrow \neg p$
$p$ $q$ $p \rightarrow q$ $\neg p$ $\neg q$ $(p \rightarrow q) \land \neg q$ $[(p \rightarrow q) \land \neg q] \rightarrow \neg p$
$T$ $T$ $T$ $F$ $F$ $F$ $T$
$T$ $F$ $F$ $T$ $F$ $F$ $T$
$F$ $T$ $T$ $F$ $T$ $F$ $T$
$F$ $F$ $T$ $T$ $T$ $T$ $T$

The formula is a tautology.
Argument is valid by Formula
(16.) Answer these questions giving examples.
(I.) State whether it is possible for a deductive argument to be: valid and sound.
If​ so, make a simple​ three-proposition argument that demonstrates your conclusion.

(II.) State whether it is possible for a deductive argument to be: valid and not sound.
If​ so, make a simple​ three-proposition argument that demonstrates your conclusion.

(III.) State whether it is possible for a deductive argument to be: not valid and not sound.
If​ so, make a simple​ three-proposition argument that demonstrates your conclusion.

(IV.) State whether it is possible for a deductive argument to be: not valid and sound.
If​ so, make a simple​ three-proposition argument that demonstrates your conclusion.

(V.) State whether it is possible for a deductive argument to be valid with true premises and a true conclusion.
If​ so, make a simple​ three-proposition argument that demonstrates your conclusion.

(VI.) State whether it is possible for a deductive argument to be valid with true premises and a false conclusion.
If​ so, make a simple​ three-proposition argument that demonstrates your conclusion.

(VII.) State whether it is possible for a deductive argument to be valid with false premises and a true conclusion.
If​ so, make a simple​ three-proposition argument that demonstrates your conclusion.

(VIII.) State whether it is possible for a deductive argument to be valid with false premises and a false conclusion.
If​ so, make a simple​ three-proposition argument that demonstrates your conclusion.

(IX.) State whether it is possible for a deductive argument to be valid with true premises and a false conclusion.
If​ so, make a simple​ three-proposition argument that demonstrates your conclusion.

(X.) State whether it is possible for a deductive argument to be not valid with true premises and a true conclusion.
If​ so, make a simple​ three-proposition argument that demonstrates your conclusion.

(XI.) State whether it is possible for a deductive argument to be not valid with false premises and a true conclusion.
If​ so, make a simple​ three-proposition argument that demonstrates your conclusion.


(I.) Yes, it is possible.
Premise: All living mammals breathe.
Premise: All monkeys are mammals.
Conclusion: All living monkeys breathe.

(II.) Yes, it is possible.
Premise: All mammals fly.
Premise: All monkeys are mammals.
Conclusion: All monkeys fly.

(III.) Yes, it is possible.
Premise: All living mammals breathe.
Premise: All monkeys are mammals.
Conclusion: All mammals are monkeys.

(IV.) No, it is not possible.

(V.) Yes, it is possible.
Premise: All living mammals breathe.
Premise: All monkeys are mammals.
Conclusion: All living monkeys breathe.

(VI.) No, it is not possible.

(VII.) Yes, it is possible.
Premise: All mammals swim.
Premise: All fish are mammals.
Conclusion: All fish swim.

(VIII.) Yes, it is possible.
Premise: All mammals swim.
Premise: All monkeys are mammals.
Conclusion: All monkeys swim.

(IX.) Yes, it is possible.
Premise: All mammals breathe.
Premise: All lizards breathe.
Conclusion: All lizards are mammals.

(X.) Yes, it is possible.
Premise: All mammals breathe.
Premise: All mammals have hair.
Conclusion: All hairy animals breathe.

(XI.) Yes, it is possible.
Premise: All animals are lizards.
Premise: All animals are cold-blooded.
Conclusion: All lizards are cold-blooded.
(17.) For these arguments:
Determine the truth values of the premises.
Assess the strength of the argument.
Discuss the truth of the conclusion.

(A.)
Premise 1: Ducks and Herons are birds, and fly.
Premise 2: Pigeons and Eagles are birds, and fly.
Premise 3: Parrots and Doves are birds, and fly.
Conclusion: Penguins are birds, so they fly.

(B.)
Premise 1: 3 + 6 = 9.
Premise 2: 4 + 5 = 9.
Premise 3: 4 + 9 = 13.
Conclusion: The sum of an odd integer and an even integer is an odd integer.

(C.)
Premise 1: (−3) * (−2) = 6.
Premise 2: (−2) * (−7) = 14.
Premise 3: (−5) * (−6) = 30.
Conclusion: When two negative numbers are​ multiplied, the result is a positive number.


(A.)
Premise 1 is true.
Premise 2 is true.
Premise 3 is true.
The argument seems moderately strong.
The conclusion is false.

(B.)
Premise 1 is true.
Premise 2 is true.
Premise 3 is true.
The argument is moderately strong.
The conclusion is true.

(C.)
Premise 1 is true.
Premise 2 is true.
Premise 3 is true.
The argument is moderately strong.
The conclusion is true.
(18.) Consider an argument in which:
Premise 1 is​ "All knights are​ heroes"
Premise 2 is​ "Paul is a​ hero."
Which of the following conclusions is​ true? Explain your reasoning.

A. Paul may or may not be a knight.
By making a Venn diagram of the​ premises, it can be shown that the​ "heroes" circle is fully inside the​ "knights" circle, and Paul is on the border of the​ "heroes" circle.

B. Paul may or may not be a knight.
By making a Venn diagram of the​ premises, it can be shown that the​ "knights" circle is fully inside the​ "heroes" circle, and Paul is on the border of the​ "knights" circle.

C. Paul is not a knight.
By making a Venn diagram of the​ premises, it can be shown that the​ "knights" circle is fully inside the​ "heroes" circle, and Paul is outside both circles.

D. Paul is a knight.
By making a Venn diagram of the​ premises, it can be shown that the​ "heroes" circle is fully inside the​ "knights" circle, and Paul is inside the​ "heroes" circle.

E. Paul is not a knight.
By making a Venn diagram of the​ premises, it can be shown that the​ "heroes" circle is fully inside the​ "knights" circle, and Paul is outside both circles.

F. Paul is a knight.
By making a Venn diagram of the​ premises, it can be shown that the​ "knights" circle is fully inside the​ "heroes" circle, and Paul is inside the​ "knights" circle.


B. Paul may or may not be a knight.
By making a Venn diagram of the​ premises, it can be shown that the​ "knights" circle is fully inside the​ "heroes" circle, and Paul is on the border of the​ "knights" circle.
(19.) If​ necessary, write the given argument as a chain of conditional propositions that have the form if​ p, then q.
Then determine the validity of the entire argument.

Premise: If taxes are increased, then taxpayers will have less disposable income.
Premise: With less disposable income, spending will decrease and the economy will slow down.
Conclusion: A tax increase will slow down the economy.

(a.) Write the second premise as a conditional proposition that has the form if​ p, then q.

(b.) Write the conclusion as a conditional proposition that has the form if​ p, then q.

(c.) Is the entire argument​ valid?


(a.) Premise: With less disposable income, spending will decrease and the economy will slow down.
Rewording as a conditional statement: Premise 2: If taxpayers have less disposable income, then the economy will slow down.

(b.) Conclusion: A tax increase will slow down the economy.
Rewording as a conditional statement: Conclusion: If taxes are increased, then the economy will slow down.

(c.) Let:
$p$: Taxes are increased.
$q$: Taxpayers will have less disposable income
$r$: The economy will slow down.

$ Premise\: 1: p \rightarrow q \\[2ex] Premise\: 2: q \rightarrow r \\[2ex] \rule{2.5in}{0.5pt} \\ Conclusion\: \therefore p \rightarrow r \\[3ex] $ First Method: Valid and Invalid Forms of Arguments
Argument is valid by Law of Hypothetical Syllogism (also known as Transitive Reasoning)
There is a clear chain from​ "taxes are increased​" to​ "the economy will slow down​."

Second Method: Definition
Let us draw a truth table of the premises and the conclusion.
$p$ $q$ $r$ $p \rightarrow q$ $q \rightarrow r$ $p \rightarrow r$
$T$ $T$ $T$ $T\checkmark$ $T\checkmark$ $T\checkmark$
$T$ $T$ $F$ $T$ $F$ $F$
$T$ $F$ $T$ $F$ $T$ $T$
$T$ $F$ $F$ $F$ $T$ $F$
$F$ $T$ $T$ $T\checkmark$ $T\checkmark$ $T\checkmark$
$F$ $T$ $F$ $T$ $F$ $T$
$F$ $F$ $T$ $T\checkmark$ $T\checkmark$ $T\checkmark$
$F$ $F$ $F$ $T\checkmark$ $T\checkmark$ $T\checkmark$
Premise 1 Premise 2 Conclusion

In all the cases when both premises are true (1st, 5th, 7th, and 8th cases), the conclusion is also true.
Argument is valid by Definition

Third Method: Formula
Let us draw the truth table for the formula: $[(premise\: 1) \land (premise\: 2)] \rightarrow conclusion$
$[(p \rightarrow q) \land (q \rightarrow r)] \rightarrow (p \rightarrow r)$
$p$ $q$ $r$ $p \rightarrow q$ $q \rightarrow r$ $p \rightarrow r$ $(p \rightarrow q) \land (q \rightarrow r)$ $[(p \rightarrow q) \land (q \rightarrow r)] \rightarrow (p \rightarrow r)$
$T$ $T$ $T$ $T$ $T$ $T$ $T$ $T$
$T$ $T$ $F$ $T$ $F$ $F$ $F$ $T$
$T$ $F$ $T$ $F$ $T$ $T$ $F$ $T$
$T$ $F$ $F$ $F$ $T$ $F$ $F$ $T$
$F$ $T$ $T$ $T$ $T$ $T$ $T$ $T$
$F$ $T$ $F$ $T$ $F$ $T$ $F$ $T$
$F$ $F$ $T$ $T$ $T$ $T$ $T$ $T$
$F$ $F$ $F$ $T$ $T$ $T$ $T$ $T$

The formula is a tautology.
Argument is valid by Formula
(20.) Consider an argument in which:
Premise 1 is​ "If p, then​ q"
Premise 2 is​ "q is​ true."
Which can be concluded about p? Explain your reasoning.

A. This argument is affirming the​ hypothesis, which is a valid argument.
So, it can be concluded that p is true.

B. This argument is denying the​ hypothesis, which is an invalid argument.​
So, we cannot conclude anything about p.

C. This argument is denying the​ conclusion, which is a valid argument.​
So, it can be concluded that p is true.

D. This argument is denying the​ hypothesis, which is an invalid argument.​
So, it can be concluded that p is not true.

E. This argument is affirming the​ conclusion, which is an invalid argument.
So, we cannot conclude anything about p.

F. This argument is affirming the​ conclusion, which is an invalid argument.​
So, it can be concluded that p is not true.


Premise 1 is​ "If p, then​ q"
Premise 2 is​ "q is​ true."
Symbolic Logic:
Premise 1: $p \rightarrow q$
Premise 2: $q$
If we review the Valid and Invalid Forms of Arguments, we see:

$ p \rightarrow q \\[2ex] q \\ \rule{0.9in}{0.3pt} \\ \therefore p \\[3ex] $ This is: Fallacy of the Converse or Affirming the Conclusion
E. This argument is affirming the​ conclusion, which is an invalid argument.
So, we cannot conclude anything about p.




Top




(21.) Mary sings soprano or alto or both.
Mary sings alto or tenor or both.
Therefore, Mary sings soprano or tenor.


Let:
$p$: Mary sings soprano.
$q$: Mary sings alto.
$r$: Mary sings tenor.

$ Premise\: 1: p \lor q \\[2ex] Premise\: 2: q \lor r \\[2ex] \rule{2.5in}{0.5pt} \\ Conclusion \therefore p \lor r \\[3ex] $ Second Method: Definition
Let us draw a truth table of the premise and the conclusion.
$p$ $q$ $r$ $p \lor q$ $q \lor r$ $p \lor r$
$T$ $T$ $T$ $T\checkmark$ $T\checkmark$ $T\checkmark$
$T$ $T$ $F$ $T\checkmark$ $T\checkmark$ $T\checkmark$
$T$ $F$ $T$ $T\checkmark$ $T\checkmark$ $T\checkmark$
$T$ $F$ $F$ $T$ $F$ $T$
$F$ $T$ $T$ $T\checkmark$ $T\checkmark$ $T\checkmark$
$F$ $T$ $F$ $T\checkmark$ $T\checkmark$ $F\times$
$F$ $F$ $T$ $F$ $T$ $T$
$F$ $F$ $F$ $F$ $F$ $F$
Premise 1 Premise 2 Conclusion

In the 6th case, the conclusion is false when both premises are true.
Argument is invalid by Definition

Third Method: Formula
Let us draw the truth table for the formula: $[(premise\: 1) \land (premise\: 2)] \rightarrow conclusion$
$[(p \lor q) \land (q \lor r)] \rightarrow (p \lor r)$
$p$ $q$ $r$ $p \lor q$ $q \lor r$ $(p \lor q) \land (q \lor r)$ $p \lor r$ $[(p \lor q) \land (q \lor r)] \rightarrow (p \lor r)$
$T$ $T$ $T$ $T$ $T$ $T$ $T$ $T$
$T$ $T$ $F$ $T$ $T$ $T$ $T$ $T$
$T$ $F$ $T$ $T$ $T$ $T$ $T$ $T$
$T$ $F$ $F$ $T$ $F$ $F$ $T$ $T$
$F$ $T$ $T$ $T$ $T$ $T$ $T$ $T$
$F$ $T$ $F$ $T$ $T$ $T$ $F$ $F$
$F$ $F$ $T$ $F$ $T$ $F$ $T$ $T$
$F$ $F$ $F$ $F$ $F$ $F$ $F$ $T$
The formula is not a tautology.
Argument is not valid by Formula
(22.)


(23.) (I.) Premise: If a body orbits the earth, then it is a satellite.
Premise: The moon orbits the earth.
Conclusion: The moon is a satellite.

(II.) Let X represent the moon in the Venn diagram.
Which Venn diagram below shows the correct placement of the variable​ X?

Number 23

(III.) Is the argument sound?


We already know that the moon is a body.
Hence, let:
$p$: The moon orbits the earth.
$q$: The moon is a satellite.

$ Premise\: 1: p \rightarrow q \\[2ex] Premise\: 2: p \\ \rule{2.5in}{0.5pt} \\ Conclusion\: \therefore q \\[3ex] $ First Method: Valid and Invalid Forms of Arguments
Argument is valid by Modus Ponens (also known as Law of Detachment or Direct Reasoning or Affirming the Hypothesis)

Second Method: Definition
Let us draw a truth table of the premises and the conclusion.
$p$ $q$ $p \rightarrow q$
$T\checkmark$ $T\checkmark$ $T\checkmark$
$T$ $F$ $F$
$F$ $T$ $T$
$F$ $F$ $T$
Premise 2 Conclusion Premise 1

The conclusion is true when both premises are true.
Argument is valid by Definition

Third Method: Formula
Let us draw the truth table for the formula: $[(premise\: 1) \land (premise\: 2)] \rightarrow conclusion$
$[(p \rightarrow q) \land p] \rightarrow q$
$p$ $q$ $p \rightarrow q$ $(p \rightarrow q) \land p$ $[(p \rightarrow q) \land p] \rightarrow q$
$T$ $T$ $T$ $T$ $T$
$T$ $F$ $F$ $F$ $T$
$F$ $T$ $T$ $F$ $T$
$F$ $F$ $T$ $F$ $T$

The formula is a tautology.
Argument is valid by Formula

(II.) The diagram is:
Number 23

(III.) Because the argument is valid and all it's premises are true, the argument is sound.
(24.) Consider the following​ proposition:

"If such procedures are not​ followed[, then] apes ... act as if they​ don't understand the problem at​ hand." — Frans de​ Waal, Are We Smart Enough to Know How Smart Animals​ Are?

(I.) What is the logical conclusion if the usual procedures are not​ followed?
A. The usual procedures were followed.
B. Apes act as if they understand the problem at hand.
C. Apes act as if they​ don't understand the problem at hand.
D. None of the above is a logical conclusion.

(II.) What is the logical conclusion if apes act as if they understand the problem at​ hand?
A. Apes act as if they​ don't understand the problem at hand.
B. Apes act as if they understand the problem at hand.
C. The usual procedures were followed.
D. None of the above is a logical conclusion.

(III.) What is the logical conclusion if the usual procedures are​ followed?
A. The usual procedures are not followed.
B. Apes act as if they​ don't understand the problem at hand.
C. Apes act as if they understand the problem at hand.
D. None of the above is a logical conclusion.


(I.) C. Apes act as if they​ don't understand the problem at hand.

(II.) C. The usual procedures were followed.

(III.) D. None of the above is a logical conclusion.
(25.) No doctor is poor.
Some teachers are poor.
Therefore, some teachers are not doctors.


Let:
set of doctors = D
set of poor people = P
set of teachers = T

Number 25

Premise 1: No doctor is poor means D and P are disjoint sets
Premise 2: Some teachers are poor means T and P have some elements in common...intersection of sets
Conclusion: Therefore, some teachers are not doctors.

Because some teachers are poor, this means that those teachers that are poor, are not doctors. This is because no doctor is poor. This is represented by the shaded area in the diagram (one of the possible Euler diagrams).
Other possible Euler diagrams must include that shaded area.
Hence, the conclusion follows from the premises.
The argument is valid.

Would you like further explanations?
Let us draw all the possibilities (cases) for Premise 1 and Premise 2

Let:
set of doctors = D
set of poor people = P
set of teachers = T

Number 25-1

These are the two cases for Premise 1 and Premise 2
In the 1st case, some teachers are doctors and some teachers are not doctors.
In the 2nd case, no teacher is a doctor.
However, this does not contradict the conclusion that some teachers are not doctors.
The argument is valid.
(26.)


(27.) (I.) Premise: If a gas is a noble gas, then it is inert.
Premise: Argon is inert.
Conclusion: Argon is a noble gas.

(II.) Let X represent Argon in the Venn diagram.
Which Venn diagram below shows the correct placement of the variable​ X?

Number 27

(III.) Is the argument sound?


We already know that Argon is a gas.
Hence, let:
$p$: Argon is a noble gas.
$q$: Argon is inert.

$ Premise\: 1: p \rightarrow q \\[2ex] Premise\: 2: q \\ \rule{2.5in}{0.5pt} \\ Conclusion\: \therefore p \\[3ex] $ First Method: Valid and Invalid Forms of Arguments
Argument is invalid by Fallacy of the Converse (also known as Affirming the Conclusion)

Second Method: Definition
Let us draw a truth table of the premises and the conclusion.
$p$ $q$ $p \rightarrow q$
$T\checkmark$ $T\checkmark$ $T\checkmark$
$T$ $F$ $F$
$F\times$ $T\checkmark$ $T\checkmark$
$F$ $F$ $T$
Conclusion Premise 2 Premise 1

In the 3rd case, the conclusion is false when both premises are true.
Argument is invalid by Definition

Third Method: Formula
Let us draw the truth table for the formula: $[(premise\: 1) \land (premise\: 2)] \rightarrow conclusion$
$[(p \rightarrow q) \land q] \rightarrow p$
$p$ $q$ $p \rightarrow q$ $(p \rightarrow q) \land q$ $[(p \rightarrow q) \land q] \rightarrow p$
$T$ $T$ $T$ $T$ $T$
$T$ $F$ $F$ $F$ $T$
$F$ $T$ $T$ $T$ $F$
$F$ $F$ $T$ $F$ $T$

The formula is not a tautology.
Argument is invalid by Formula

(II.) The diagram is:
Number 27

(III.) Because the argument is invalid, the argument is not sound.
(28.) Consider the following​ proposition:

"If we insist too adamantly on protecting our​ privacy, [then] we will sacrifice both free enterprise and​ security" — Ted ​Koppel, Lights Out?

(I.) What is the logical conclusion if we insist too adamantly on protecting our​ privacy?
A. We did not insist too adamantly on protecting our privacy.
B. We will sacrifice both free enterprise and security.
C. We will not sacrifice both free enterprise and security.
D. None of the above is a logical conclusion.

(II.) What is the logical conclusion if we​ don't sacrifice free enterprise and​ security?
A. We will not sacrifice both free enterprise and security.
B. We did not insist too adamantly on protecting our privacy.
C. We will sacrifice both free enterprise and security.
D. None of the above is a logical conclusion.

(III.) What is the logical conclusion if we​ don't insist too adamantly on protecting our​ privacy?
A. We will sacrifice both free enterprise and security.
B. We insisted too adamantly on protecting our privacy.
C. We will not sacrifice both free enterprise and security.
D. None of the above is a logical conclusion.


(I.) B. We will sacrifice both free enterprise and security.

(II.) B. We did not insist too adamantly on protecting our privacy.

(III.) D. None of the above is a logical conclusion.

(29.) All policemen are rude.
Nobody who is meticulous is a policeman.
Therefore, rude people are not meticulous.


Let:
set of policemen = P
set of meticulous people = M
set of rude people = R

Number 29

Premise 1: All policemen are rude means P is a subset of R
Premise 2: Nobody who is meticulous is a policeman means M and P are disjoint sets
Conclusion: Therefore, rude people are not meticulous.

Based on the diagram (one of the possible Euler diagrams), it is possible that some rude people are meticulous because one of the options is the shaded area: which represents the set that there is at least one rude person that is meticulous.
Hence, the conclusion does not follow from the premises.
The argument is invalid.

Would you like further explanations?
Let us draw all the possibilities (cases) for Premise 1 and Premise 2

Number 29-1

These are the three cases for Premise 1 and Premise 2
In the 1st case, rude people are meticulous.
In the 2nd case, some rude people are meticulous.
In the 3rd case, no rude person is meticulous.
Because of the 1st Case and the 2nd Case, the argument is invalid.

(30.)


(31.) All politicians are rich.
Rich people are never happy.
Therefore, no politician is happy.


Let:
set of politicians = P
set of rich people = R
set of happy people = H

Number 31

Premise 1: All politicians are rich means P is a subset of R
Premise 2: Rich people are never happy means R and H are disjoint sets
Conclusion: Therefore, no politician is happy.

Based on the diagram, it is evident that no politician is happy because P and H are also disjoint.
Hence, the conclusion follows from the premises.
The argument is valid.
(32.)


(33.) It is wrong to smoke in public if secondary cigarette smoke is a health risk.
If secondary cigarette smoke were not a health risk, the American Lung Association would not say that it is.
The American Lung Association says that secondary cigarette smoke is a health risk.
Therefore, it is wrong to smoke in public.

Use p to represent "Secondary cigarette smoke is a health risk," use q to represent "It is wrong to smoke in public," and use r to represent "The American Lung Association says that secondary cigarette smoke is a health risk."


Let:
$p$: Secondary cigarette smoke is a health risk
$q$: It is wrong to smoke in public
$r$: The American Lung Association says that secondary cigarette smoke is a health risk

$ Premise\: 1: p \rightarrow q \\[2ex] Premise\: 2: \neg p \rightarrow \neg r \\[2ex] Premise\: 3: r \\[2ex] \rule{2.5in}{0.5pt} \\ Conclusion \therefore q \\[3ex] $ Second Method: Definition
Let us draw a truth table of the premises and the conclusion.
$p$ $q$ $r$ $p \rightarrow q$ $\neg p$ $\neg r$ $\neg p \rightarrow \neg r$
$T$ $T$✓ $T$✓ $T$✓ $F$ $F$ $T$✓
$T$ $T$ $F$ $T$ $F$ $T$ $T$
$T$ $F$ $T$ $F$ $F$ $F$ $T$
$T$ $F$ $F$ $F$ $F$ $T$ $T$
$F$ $T$ $T$ $T$ $T$ $F$ $F$
$F$ $T$ $F$ $T$ $T$ $T$ $T$
$F$ $F$ $T$ $T$ $T$ $F$ $F$
$F$ $F$ $F$ $T$ $T$ $T$ $T$
Conclusion Premise 3 Premise 1 Premise 2

In the case where all premises are true (only one case), the conclusion is also true.
Argument is valid by Definition

Third Method: Formula
Let us draw the truth table for the formula: $[(premise\: 1) \land (premise\: 2) \land (premise\: 3)] \rightarrow conclusion$
$[(p \rightarrow q) \land (\neg p \rightarrow \neg r) \land r] \rightarrow q$
$p$ $q$ $r$ $p \rightarrow q$ $\neg p$ $\neg r$ $\neg p \rightarrow \neg r$ $(p \rightarrow q) \land (\neg p \rightarrow \neg r)$ $(p \rightarrow q) \land (\neg p \rightarrow \neg r) \land r$ $[(p \rightarrow q) \land (\neg p \rightarrow \neg r) \land r] \rightarrow q$
$T$ $T$ $T$ $T$ $F$ $F$ $T$ $T$ $T$ $T$
$T$ $T$ $F$ $T$ $F$ $T$ $T$ $T$ $F$ $T$
$T$ $F$ $T$ $F$ $F$ $F$ $T$ $F$ $F$ $T$
$T$ $F$ $F$ $F$ $F$ $T$ $T$ $F$ $F$ $T$
$F$ $T$ $T$ $T$ $T$ $F$ $F$ $F$ $F$ $T$
$F$ $T$ $F$ $T$ $T$ $T$ $T$ $T$ $F$ $T$
$F$ $F$ $T$ $T$ $T$ $F$ $F$ $F$ $F$ $T$
$F$ $F$ $F$ $T$ $T$ $T$ $T$ $T$ $F$ $T$
Tautology

The formula is a tautology.
Argument is valid by Formula
(34.)


(35.) If​ necessary, write the given argument as a chain of conditional propositions that have the form if​ p, then q.
Then determine the validity of the entire argument.

Premise: If prices are increased, then consumers will have less disposable income.
Premise: With less disposable income, spending will decrease and the economy will slow down.
Conclusion: A price increase will slow down the economy.

(a.) Write the second premise as a conditional proposition that has the form if​ p, then q.

(b.) Write the conclusion as a conditional proposition that has the form if​ p, then q.

(c.) Is the entire argument​ valid?


(a.) Premise: With less disposable income, spending will decrease and the economy will slow down.
Rewording as a conditional statement: Premise 2: If consumers have less disposable income, then the economy will slow down.

(b.) Conclusion: A price increase will slow down the economy.
Rewording as a conditional statement: Conclusion: If prices are increased, then the economy will slow down.

(c.) Let:
$p$: Prices are increased.
$q$: Consumers will have less disposable income
$r$: The economy will slow down.

$ Premise\: 1: p \rightarrow q \\[2ex] Premise\: 2: q \rightarrow r \\[2ex] \rule{2.5in}{0.5pt} \\ Conclusion\: \therefore p \rightarrow r \\[3ex] $ First Method: Valid and Invalid Forms of Arguments
Argument is valid by Law of Hypothetical Syllogism (also known as Transitive Reasoning)
There is a clear chain from​ "taxes are increased​" to​ "the economy will slow down​."

Second Method: Definition
Let us draw a truth table of the premises and the conclusion.
$p$ $q$ $r$ $p \rightarrow q$ $q \rightarrow r$ $p \rightarrow r$
$T$ $T$ $T$ $T\checkmark$ $T\checkmark$ $T\checkmark$
$T$ $T$ $F$ $T$ $F$ $F$
$T$ $F$ $T$ $F$ $T$ $T$
$T$ $F$ $F$ $F$ $T$ $F$
$F$ $T$ $T$ $T\checkmark$ $T\checkmark$ $T\checkmark$
$F$ $T$ $F$ $T$ $F$ $T$
$F$ $F$ $T$ $T\checkmark$ $T\checkmark$ $T\checkmark$
$F$ $F$ $F$ $T\checkmark$ $T\checkmark$ $T\checkmark$
Premise 1 Premise 2 Conclusion

In all the cases when both premises are true (1st, 5th, 7th, and 8th cases), the conclusion is also true.
Argument is valid by Definition

Third Method: Formula
Let us draw the truth table for the formula: $[(premise\: 1) \land (premise\: 2)] \rightarrow conclusion$
$[(p \rightarrow q) \land (q \rightarrow r)] \rightarrow (p \rightarrow r)$
$p$ $q$ $r$ $p \rightarrow q$ $q \rightarrow r$ $p \rightarrow r$ $(p \rightarrow q) \land (q \rightarrow r)$ $[(p \rightarrow q) \land (q \rightarrow r)] \rightarrow (p \rightarrow r)$
$T$ $T$ $T$ $T$ $T$ $T$ $T$ $T$
$T$ $T$ $F$ $T$ $F$ $F$ $F$ $T$
$T$ $F$ $T$ $F$ $T$ $T$ $F$ $T$
$T$ $F$ $F$ $F$ $T$ $F$ $F$ $T$
$F$ $T$ $T$ $T$ $T$ $T$ $T$ $T$
$F$ $T$ $F$ $T$ $F$ $T$ $F$ $T$
$F$ $F$ $T$ $T$ $T$ $T$ $T$ $T$
$F$ $F$ $F$ $T$ $T$ $T$ $T$ $T$

The formula is a tautology.
Argument is valid by Formula
(36.)


(37.) If a natural number is divisible by 75​, then it is divisible by 15.
If a natural number is divisible by 15​, then it is divisible by 5.
If a natural number is divisible by 75​, then it is divisible by 5.


Let:
$p$: A natural number is divisible by 75
$q$: A natural number is divisible by 15
$r$: A natural number is divisible by 5

$ Premise\: 1: p \rightarrow q \\[2ex] Premise\: 2: q \rightarrow r \\[2ex] \rule{2.5in}{0.5pt} \\ Conclusion\: \therefore p \rightarrow r \\[3ex] $ First Method: Valid and Invalid Forms of Arguments
Argument is valid by Law of Hypothetical Syllogism (also known as Transitive Reasoning)
There is a clear chain from​ "a natural number is divisible by 75​" to​ "a natural number is divisible by 5​."

Second Method: Definition
Let us draw a truth table of the premises and the conclusion.
$p$ $q$ $r$ $p \rightarrow q$ $q \rightarrow r$ $p \rightarrow r$
$T$ $T$ $T$ $T\checkmark$ $T\checkmark$ $T\checkmark$
$T$ $T$ $F$ $T$ $F$ $F$
$T$ $F$ $T$ $F$ $T$ $T$
$T$ $F$ $F$ $F$ $T$ $F$
$F$ $T$ $T$ $T\checkmark$ $T\checkmark$ $T\checkmark$
$F$ $T$ $F$ $T$ $F$ $T$
$F$ $F$ $T$ $T\checkmark$ $T\checkmark$ $T\checkmark$
$F$ $F$ $F$ $T\checkmark$ $T\checkmark$ $T\checkmark$
Premise 1 Premise 2 Conclusion

In all the cases when both premises are true (1st, 5th, 7th, and 8th cases), the conclusion is also true.
Argument is valid by Definition

Third Method: Formula
Let us draw the truth table for the formula: $[(premise\: 1) \land (premise\: 2)] \rightarrow conclusion$
$[(p \rightarrow q) \land (q \rightarrow r)] \rightarrow (p \rightarrow r)$
$p$ $q$ $r$ $p \rightarrow q$ $q \rightarrow r$ $p \rightarrow r$ $(p \rightarrow q) \land (q \rightarrow r)$ $[(p \rightarrow q) \land (q \rightarrow r)] \rightarrow (p \rightarrow r)$
$T$ $T$ $T$ $T$ $T$ $T$ $T$ $T$
$T$ $T$ $F$ $T$ $F$ $F$ $F$ $T$
$T$ $F$ $T$ $F$ $T$ $T$ $F$ $T$
$T$ $F$ $F$ $F$ $T$ $F$ $F$ $T$
$F$ $T$ $T$ $T$ $T$ $T$ $T$ $T$
$F$ $T$ $F$ $T$ $F$ $T$ $F$ $T$
$F$ $F$ $T$ $T$ $T$ $T$ $T$ $T$
$F$ $F$ $F$ $T$ $T$ $T$ $T$ $T$

The formula is a tautology.
Argument is valid by Formula
(38.)


(39.)


(40.)


(41.)


(42.)


(43.)


(44.)