Mathematical Logic

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These are the solutions to Mathematics questions on Mathematical Logic.
The TI-84 Plus CE shall be used for applicable questions.
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Formulas and Notes

Validity and Invality of Symbolic Arguments
To determine whether an argument is valid or invalid:
(1.) Identify and write the different statements.

(2.) Write the premises and the conclusion in symbolic logic.
The statements must be written in the positive sense only.

(3.) Use any or a combination of these 4 methods to determine the validity or invalidity of the argument.

(I.) Valid Forms of Arguments and Invalid Forms of Arguments.
Compare with the Valid Forms of Arguments and Invalid Forms of Arguments. See the two tables below.
If the argument is not similar to any of the forms, try any of the remaining 2 methods.

(II.) Definition
This applies to one or more premises and a conclusion.
(a.) Draw the truth table of the premises and the conclusion.
(b.) Check all the cases where "all" the premises are true.
(i.) If there is no case where "all" the premises are true, the argument is valid.
(ii.) If there is any case where "all" the premises are true AND the conclusion is true in "all" those cases, the argument is valid.
(iii.) If there is any case where "all" the premises are true AND the conclusion is false, the argument is invalid.

(III.) Formula
For one premise:
Draw a truth table for the formula: $premise\: 1 \rightarrow conclusion$

For two premises:
Draw a truth table for the formula: $[(premise\: 1) \land (premise\: 2)] \rightarrow conclusion$

For three premises:
Draw a truth table for the formula: $[(premise\: 1) \land (premise\: 2) \land (premise\: 3)] \rightarrow conclusion$

For four premises:
Draw a truth table for the formula: $[(premise\: 1) \land (premise\: 2) \land (premise\: 3) \land (premise\: 4)] \rightarrow conclusion$

(i) If the formula is a tautology, the argument is valid
(ii) If the formula is not a tautology, the argument is invalid.
In other words; if the formula is a contingency or a contradiction, the argument is invalid.

(IV.) Deduction

Laws of Logical Equivalence

p, q, r,... are logical statements
True = truth value of T
False = truth value of F

$ (1.) \text{ Identity Laws} \\[3ex] p \land T \equiv p \\[3ex] p \lor F \equiv p \\[5ex] (2.) \text{ Domination Laws} \\[3ex] p \lor T \equiv T \\[3ex] p \land F \equiv F \\[5ex] (3.) \text{ Idempotent Laws} \\[3ex] p \lor p \equiv p \\[3ex] p \land p \equiv p \\[5ex] (4.) \text{ Negation Laws (Complement Laws or Laws of Complementation)} \\[3ex] p \lor \neg p \equiv T \\[3ex] p \land \neg p \equiv F \\[5ex] (5.) \text{ Double Negation Laws} \\[3ex] \neg(\neg p) \equiv p \\[5ex] (6.) \text{ Commutative Laws} \\[3ex] p \lor q \equiv q \lor p \\[3ex] p \land q \equiv q \land p \\[5ex] (7.) \text{ Associative Laws} \\[3ex] (p \lor q) \lor r \equiv p \lor (q \lor r) \\[3ex] (p \land q) \land r \equiv p \land (q \land r) \\[5ex] (8.) \text{ Distributive Laws} \\[3ex] p \land (q \lor r) \equiv (p \land q) \lor (p \land r) \\[3ex] p \lor (q \land r) \equiv (p \lor q) \land (p \lor r) \\[5ex] (9.) \text{ De Morgan's Laws} \\[3ex] \neg(p \land q) \equiv \neg p \lor \neg q \\[3ex] \neg(p \lor q) \equiv \neg p \land \neg q \\[5ex] (10.) \text{ Absorption Laws} \\[3ex] p \lor (p \land q) \equiv p \\[3ex] p \land (p \lor q) \equiv p \\[5ex] (11.) \text{ Conditional (Implication) Equivalence} \\[3ex] p \rightarrow q \equiv \neg p \lor q \\[3ex] p \rightarrow q \equiv \neg q \rightarrow \neg p...\text{contrapositive statement} \\[5ex] (12.) \text{Biconditional Equivalence} \\[3ex] p \leftrightarrow q \equiv (p \rightarrow q) \land (q \rightarrow p) \\[3ex] p \leftrightarrow q \equiv (p \land q) \lor (\neg p \land \neg q) $

Valid Forms of Symbolic Arguments
Valid Argument	Name
$ p \rightarrow q \\[2ex] p \\ \rule{0.9in}{0.3pt} \\ \therefore q $	Modus Ponens or Law of Detachment or Direct Reasoning or Affirming the Hypothesis
$ {p \rightarrow q \\[2ex] \hspace{1.5em} \neg q \\ \rule{0.9in}{0.3pt} \\ \therefore \neg p} \hspace{-1mm} {p \rightarrow \neg q \\[2ex] \hspace{2.3em} q \\ \rule{0.9in}{0.3pt} \\ \therefore \neg p} $	Modus Tollens or Law of Contraposition or Contrapositive Reasoning or Denying the Conclusion
$ p \rightarrow q \\[2ex] q \rightarrow r \\ \rule{0.9in}{0.3pt} \\ \therefore p \rightarrow r $	Law of Hypothetical Syllogism or Transitive Reasoning
$ {p \lor q \\[2ex] \hspace{1.3em} \neg q \\ \rule{0.9in}{0.3pt} \\ \therefore p} \hspace{-1mm} {p \lor q \\[2ex] \neg p \\ \rule{0.9in}{0.3pt} \\ \therefore q} $	Law of Disjunctive Syllogism or Disjunctive Reasoning
$ p \lor q \\[2ex] \neg p \lor r \\ \rule{0.9in}{0.3pt} \\ \therefore q \lor r $	Resolution
$ p \\ \rule{1in}{0.5pt} \\ \therefore p \lor q $	Addition
$ {p \land q \\ \rule{1in}{0.5pt} \\ \therefore p} \hspace{-1mm} {p \land q \\ \rule{1in}{0.5pt} \\ \therefore q} $	Simplification
$ p \\[2ex] q \\ \rule{0.9in}{0.3pt} \\ \therefore p \land q $	Conjunction
$ (p \rightarrow q) \land (r \rightarrow s) \\ \rule{2.5in}{0.5pt} \\ \therefore (p \lor r) \rightarrow (q \lor s) \\[5ex] (p \rightarrow q) \land (r \rightarrow s) \\ \rule{2.5in}{0.5pt} \\ \therefore (p \land r) \rightarrow (q \land s) \\[5ex] (p \rightarrow q) \land (r \rightarrow s) \\[2ex] p \lor r \\ \rule{2.5in}{0.5pt} \\ \therefore q \land s $	Constructive Dilemmas
$ (p \rightarrow q) \land (r \rightarrow s) \\[2ex] \neg q \lor \neg s \\ \rule{2.5in}{0.5pt} \\ \therefore \neg p \lor \neg r $	Destructive Dilemmas

Invalid Forms of Symbolic Arguments
Invalid Argument	Name
$ p \rightarrow q \\[2ex] \hspace{1.6em} q \\ \rule{0.9in}{0.3pt} \\ \therefore p $	Fallacy of the Converse or Affirming the Conclusion
$ p \rightarrow q \\[2ex] \neg p \\ \rule{0.9in}{0.3pt} \\ \therefore \neg q $	Fallacy of the Inverse or Denying the Hypothesis
$ p \rightarrow q \\[2ex] q \rightarrow r \\ \rule{0.9in}{0.3pt} \\ \therefore r \rightarrow p $	Misuse of Transitive Reasoning
$ {p \lor q \\[2ex] p \\ \rule{0.9in}{0.3pt} \\ \therefore \neg q} \hspace{-10mm} {p \lor q \\[2ex] \hspace{1.5em} q \\ \rule{0.9in}{0.3pt} \\ \therefore \neg p} \hspace{-14mm} {p \lor \neg q \\[2ex] p \\ \rule{0.9in}{0.3pt} \\ \therefore q} \hspace{-3mm} $	Misuse of Disjunctive Reasoning

(1.) Determine the validity or invalidity of the symbolic argument.
Support your work with applicable reasons.
If it is valid, prove it.
If it is invalid, use a counterexample to show that the argument is invalid by assigning truth values.

$ p \\[2ex] p \rightarrow r \\[2ex] p \rightarrow (\neg q \lor \neg r) \\[2ex] q \lor \neg s \\ \rule{2.1in}{0.3pt} \\ \therefore s \\[3ex] $

$ \underline{\text{Deduction Method}} \\[3ex] p...\text{Premise 1} \\[2ex] p \rightarrow r...\text{Premise 2} \\[2ex] p \rightarrow (\neg q \lor \neg r)...\text{Premise 3} \\[2ex] q \lor \neg s ...\text{Premise 4} \\ \rule{2.1in}{0.3pt} \\ \therefore s...\text{Conclusion} \\[5ex] \underline{\text{Premise 2 and Premise 1: Modus Ponens}} \\[3ex] p \rightarrow r \\[2ex] p \\ \rule{0.9in}{0.3pt} \\ r ...\text{output 1} \\[5ex] \underline{\text{Premise 3 and Premise 1: Modus Ponens}} \\[3ex] p \rightarrow (\neg q \lor \neg r) \\[2ex] p \\ \rule{1.7in}{0.3pt} \\ \neg q \lor \neg r ...\text{output 2} \\[5ex] \underline{\text{output 2 and output 1: Disjunctive Syllogism}} \\[3ex] \neg q \lor \neg r \\[2ex] \hspace{2.5em} r \\ \rule{1.2in}{0.3pt} \\ \neg q ...\text{output 3} \\[5ex] \underline{\text{Premise 4 and output 3: Disjunctive Syllogism}} \\[3ex] q \lor \neg s \\[2ex] \neg q \\ \rule{1.2in}{0.3pt} \\ \neg s ...\text{output 4} \\[5ex] \neg s \ne s \\[3ex] \text{output 4} \ne \text{conclusion} \\[3ex] \text{Argument is invalid by Deduction} \\[3ex] $ Definition Method/Approach:
Valid Argument: the conclusion must be true in all cases in which all the premises are true.
Invalid Argument: if there is any case in which the conclusion is false when all the premises all true.
In that regard, let us assign truth values to the statements such that the premises are true.

$ p...\text{Premise 1} \\[3ex] \text{Assign } p = T \\[5ex] p \rightarrow r...\text{Premise 2} \\[2ex] \text{Assign } r = T \text{ for Premise 2 to be true} \\[5ex] p \rightarrow (\neg q \lor \neg r)...\text{Premise 3} \\[3ex] \underline{\text{For Premise 3 to be true:}} \\[3ex] \neg r = F \\[3ex] \neg q \lor \neg r \\[3ex] \equiv \neg q \lor F \\[3ex] \text{Assign }\neg q = T \implies q = F \\[3ex] \neg q \lor \neg r \\[3ex] \equiv T \lor F \\[3ex] \equiv T \\[3ex] p \rightarrow T \\[3ex] = T \rightarrow T \\[3ex] \equiv T \\[5ex] q \lor \neg s ...\text{Premise 4} \\[3ex] \underline{\text{For Premise 4 to be true:}} \\[3ex] q \lor \neg s \\[3ex] F \lor \neg s \\[3ex] \text{Assign }\neg s = T \implies s = F \\[3ex] F \lor T \equiv T \\[5ex] \underline{\text{Assignments}} \\[3ex] p = T \\[3ex] r = T \\[3ex] q = F \\[3ex] s = F \\[3ex] $ All premises are true
The conclusion, $s$ should be true; however $s = F$
Argument is invalid by Definition.

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