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Exam version: su26a

David W. Agler

June 23, 2026

The following exam consists of 37 questions, for a total of 100 points. Read each question carefully (note: answers may break onto the next page). This exam tests your knowledge over the material from Chapter 3 and Chapter 4 of the course text and lectures. You may write on the test itself, but place final answers on the ``answer sheet'' (last page) provided.

0.1 Definitions, Concepts, and Basic Mechanics

Q1. What advantage does the truth table and truth tree tests have over the imagination test for validity?

  • 1. the truth table/tree tests are poetic; they take into account the spirit of human nature

  • 2. If an argument is persuasive in English, then the table/tree methods will tell us whether we ought to be persuaded by them.

  • 3. If an argument is deductively valid in English, then the truth table/tree method will always correctly determine whether it is (in fact) valid in the language of propositional logic (PL).

  • 4. *The truth table/tree tests are mechanical (decision procedures)

Q2. What is one advantage of truth trees over truth tables?

  • 1. trees provide the user a more graphical way of seeing the truth or falsity of an argument, specifically by showing whether an argument is true or false under every interpretation

  • 2. Tree trees use a node-branch (or tree) structure to test arguments for validity while tables use a table.

  • 3. If an argument is translated into PL, a table can show that a set of wffs is consistent but a tree cannot.

  • 4. *In contrast to truth tables where the complexity of the table is a function of the number of propositional letters (more letters, more rows required), the complexity of a truth tree is not a function of the number of propositional letters.

Q3. A set of wffs \(\Gamma \) semantically entails a wff \(\phi \) if and only if what? That is, \(\Gamma \models \phi \) if and only if what?

  • 1. there is an interpretation such that each of the members (wffs) of \(\Gamma \) are true and \(\phi \) is false.

  • 2. there is at least two interpretations such that each of the members (wffs) of \(\Gamma \) are true and \(\phi \) is false.

  • 3. there is no interpretation such that each of the members (wffs) of \(\Gamma \) are false and \(\phi \) is true.

  • 4. *there is no interpretation such that each of the members (wffs) of \(\Gamma \) are true and \(\phi \) is false.

0.2 Determining the truth of wffs

Determine the truth value (write T or F on the answersheet). Note that in some cases you don't need to know all of the truth values for some (or all) of the propositional letters.

Q15. Suppose \(\mathscr {I}(R)=F\), what is \(v(P\wedge \neg Q)\vee \neg R\)? --- Answer: T

Q16. What is the truth value of \(A\vee \neg A\)? --- Answer: T

Q17. What is the truth value of \(\neg Z\rightarrow \neg Z\)? --- Answer: T

Q18. Suppose \(\mathscr {I}(Q)=F\), what is \(v(P\rightarrow (Q\rightarrow P))\)? --- Answer: T

0.3 Truth-tree decomposition rules

Write the abbreviation (e.g. \(\wedge D\)) for the decomposition rule that should be used on wffs below. Indicate only the first decomposition rule that would be used.

Q20. \(\neg B\wedge \neg Q\)? --- Answer: \(\wedge D\)

Q21. \(\neg \neg A\vee \neg Z\) --- Answer: \(\vee D\)

Q22. \((Q\vee \neg L)\wedge M\) --- Answer: \(\wedge D\)

Q23. \(\neg \neg S\) --- Answer: \(\neg \neg D\)

Q24. \(A\rightarrow \neg R\) --- Answer: \(\rightarrow D\)

Q25. \(\neg C\rightarrow \neg Z\) --- Answer: \(\rightarrow D\)

Q26. \(\neg A\leftrightarrow X\) --- Answer: \(\leftrightarrow D\)

Q27. \(\neg (A\rightarrow \neg R)\) --- Answer: \(\neg \rightarrow D\)

Q28. \(\neg \neg (B\wedge Q)\) --- Answer: \(\neg \neg D\)

Q29. \(\neg (A\wedge R)\) --- Answer: \(\neg \wedge D\)

Q30. \(\neg (B\leftrightarrow \neg \neg C)\) --- Answer: \(\neg \leftrightarrow D\)

Q31. \(\neg \neg (Z\leftrightarrow \neg \neg S)\)? --- Answer: \(\neg \neg D\)

Q32. \(\neg \neg Z\leftrightarrow \neg M\)? --- Answer: \(\leftrightarrow D\)

0.4 Truth table construction

On the answer sheet, construct a truth table that tests for the indicated property. To receive full credit, you must (1) construct the entire truth table (each row and each T and F), (2) label whether the table indicates the property in question (e.g., "tautology"), and (3) clearly explain why the table indicates the property in question (e.g., "The table shows \(\phi \) is a tautology because ..."). Be sure (1) the table is fully complete (do not skip steps) and (2) Ts and Fs are clearly distinguishable (you can use 1 or O if it is easier).

Q33. Determine whether \(A\to B, \neg B\vdash \neg A\) is a valid sequent. --- Answer: \(A\to B, \neg B\vdash \neg A\) is a valid sequent

.
A B \(A\to B\) \(\neg B\) \((A\to B)\wedge \neg B\) \(\neg A\)
T T T F F F
T F F T F F
F T T F F T
F F T T T T

Q34. Determine whether \(B\rightarrow \neg (Z\vee \neg B)\) is a contingency, tautology, or contradiction. --- Answer: The wff is a contingency. Notice that under the main operator (the right arrow), there is at least one T and at least one F

.
B Q B \(\rightarrow \) \(\neg \) (Z \(\vee \) \(\neg \) B)
T T T F F T T F T
T F T T T F F F T
F T F T F T T T F
F F F T F F T T F

0.5 Truth tree construction

On the answer sheet, construct a truth tree that tests for the indicated property. To receive full credit, you must (1) construct the entire tree (numbering, the tree, and the node justification), (2) label whether the tree indicates the property in question (e.g., "tautology"), and (3) if an interpretation can be recovered from the tree, provide that interpretation.

Q35. Determine whether \((B\rightarrow B)\wedge \neg (D\vee \neg Q)\) is a contingency, tautology, or contradiction. --- Answer: Answer may vary, but the tree for this wff shows the wff is a contingency. It is a contingency since the tree test for contradiction shows that it is not a contradiction (all branches close), while the tree test for tautology shows that it is not a tautology (since the negated version of the wff yields a tree where all branches close). Since a wff is exactly one of the following (contingency, tautology, contradiction), the wff is a contingency.

Test for contradiction:

(A truth tree begins with the root formula (B implies B) and not (D or not Q), marked as premise and checked. The conjunction is decomposed on a single branch into B implies B and not (D or not Q). The negated disjunction is then

decomposed on the same branch into not D and not Q. Finally, B implies B branches into two leaf nodes, not B and B. The tree ends with two terminal branches: one ending in not B and the other ending in B.)

Test for tautology:

(A truth tree for testing tautology begins with the root formula not ((B implies B) and not (D or not Q)), marked as premise and checked. It splits into two branches. The left branch contains not (B implies B), which is

decomposed on the same branch into B and not B. The right branch contains not not (D or not Q), which is simplified to D or not Q and then branches into D and not Q. The left branch contains both B and not B, while the right side ends in
two branches, one with D and one with not Q.)

Q36. On the answer sheet, construct either a truth table or a truth tree for the following argument: \(A\wedge (B\wedge C), A\rightarrow B, \neg B\vee C\models C\wedge \neg D\). To receive full credit, you must (i) construct the entire truth table or tree, (ii) label whether it is a valid or invalid (that is, entailment of non-entailment), and (iii) if the argument is invalid, identify write out the interpretation (e.g. \(\mathscr {I}(P)=T, \mathscr {I}(Q)=F\)) demonstrating its invalidity.

--- Answer: The argument is invalid or a case of non-entailment. If creating a table, then a table of at least 16 rows is required and the table must provide a completed row where the premises are true and the conclusion is false. In the case of a tree, the following is a possible answer:

(A truth tree begins with four premises on a single branch: A and (B and C), A implies B, not B or C, and not (C and not D). The first conjunction is decomposed on the same branch into A and B and C, and then B and C is

decomposed into B and C. The disjunction not B or C splits the tree into two branches: the left branch contains not B and closes because it contradicts B; the right branch contains C. On the right branch, not (C and not D) splits into two
branches, one with not C and one with not not D. The not C branch closes because it contradicts C. The not not D branch simplifies to D after a further step. The conditional A implies B then splits into two branches, one with not A and
one with B. The not A branch closes because it contradicts A, and the B branch remains open, ending with D.)

Since it is invalid, an interpretation must be provided. Here is an interpretation: \(\mathscr {I}(A)=T, \mathscr {I}(B)=T, \mathscr {I}(C)=F, \mathscr {I}(D)=T\)

\(\begin {array}{c | c} P & \neg P \\ \hline T & F \\ F & T \\ \end {array} \)

\(\begin {array}{c c | c c c c} P&R&P\wedge R& P\vee R & P\rightarrow R&P\leftrightarrow R \\ \hline T& T& T & T & T & T \\ T& F& F & T & F & F \\ F& T& F & T & T & F \\ F& F& F & F & T & T \\ \end {array} \)

(A truth tree with root node phi and psi; the left child is phi with justification wedge decomposition; the left child of phi is psi with justification wedge decomposition)

(A truth tree with root node not parenthesis phi and psi; two branches below; left branch is not phi with justification not and decomposition; right branch is not psi with justification not and decomposition)

(A truth tree with root node phi or psi; two branches below; left branch has phi with justification vee D; right branch has psi with justification vee D)

(A truth tree with root node not parenthesis phi or psi; first child is not phi with justification not or decomposition; second child is not psi with justification not or decomposition)

(A truth tree with the root node phi implies psi; this branches into two children; the left child is not phi, justified by the implication decomposition rule; the right child is psi, also justified by the implication

decomposition rule)

(A truth tree with root node not parenthesis phi implies psi; first child is phi with justification not implies decomposition; second child is not parenthesis psi with justification not implies decomposition)

(A truth tree for the formula phi if and only if psi; the root is phi if and only if psi; it branches into two; the left branch has phi then psi, both justified by biconditional decomposition; the right branch has not phi then

not psi, both justified by biconditional decomposition)

(A truth tree for not phi if and only if psi; the root is not phi if and only if psi; it branches into two; the left branch has phi then not psi; the right branch has not phi then psi; each step is justified by the not

biconditional decomposition rule)

(A truth tree with root node not not phi; one branch leads to phi with justification double negation decomposition)