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{\mathunder }[1]{#1}\) \(\newcommand {\mathaccent }[1]{#1}\) \(\newcommand {\mathbotaccent }[1]{#1}\) \(\newcommand {\mathalpha }[1]{\mathord {#1}}\) \(\def\upAlpha{\unicode{x0391}}\) \(\def\upBeta{\unicode{x0392}}\) \(\def\upGamma{\unicode{x0393}}\) \(\def\upDigamma{\unicode{x03DC}}\) \(\def\upDelta{\unicode{x0394}}\) \(\def\upEpsilon{\unicode{x0395}}\) \(\def\upZeta{\unicode{x0396}}\) \(\def\upEta{\unicode{x0397}}\) \(\def\upTheta{\unicode{x0398}}\) \(\def\upVartheta{\unicode{x03F4}}\) \(\def\upIota{\unicode{x0399}}\) \(\def\upKappa{\unicode{x039A}}\) \(\def\upLambda{\unicode{x039B}}\) \(\def\upMu{\unicode{x039C}}\) \(\def\upNu{\unicode{x039D}}\) \(\def\upXi{\unicode{x039E}}\) \(\def\upOmicron{\unicode{x039F}}\) \(\def\upPi{\unicode{x03A0}}\) \(\def\upVarpi{\unicode{x03D6}}\) \(\def\upRho{\unicode{x03A1}}\) \(\def\upSigma{\unicode{x03A3}}\) \(\def\upTau{\unicode{x03A4}}\) \(\def\upUpsilon{\unicode{x03A5}}\) \(\def\upPhi{\unicode{x03A6}}\) \(\def\upChi{\unicode{x03A7}}\) 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Exam version: exam3sample

David W. Agler

August 20, 2026

The following exam consists of 34 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 5 of the course text, lectures, videos, handouts, and discussion. You may write on the test itself, but place final answers on the ``answer sheet'' (last page) provided.

0.1 Concepts and terminology

Q1. What is a derivation of \(\phi \) from \(\Gamma \) in the language of propositional logic (PL)?

  • 1. A derivation of \(\phi \) is a finite string of wffs starting with some premises \(\mathbf {A, B, C, \ldots }\) and ending with \(\phi \).

  • 2. A derivation of \(\phi \) is a finite string of wffs starting with some premises \(\mathbf {A, B, C, \ldots }\) or assumptions and ending with \(\phi \).

  • 3. A derivation of \(\phi \) is an infinite string of formulas from a set \(\Gamma \) of PL wffs where (i) the last formula in the string is \(\phi \) and (ii) each wff in the set is either a premise, an assumption, or is the result of the preceding wffs and the deductive apparatus.

  • 4. *A derivation of \(\phi \) is a finite string of formulas from a set \(\Gamma \) of PL wffs where (i) the last formula in the string is \(\phi \) and (ii) each wff in the set is either a premise, an assumption, or is the result of the preceding wffs and the deductive apparatus.

Q2. What is a deductive apparatus for PL?

  • 1. It is a set of rules that allow individuals to reason from facts (experience) to general laws, e.g. laws of nature.

  • 2. It is a set of rules of reason that all people use to reason from one proposition to another, including, but not limited to, hypothetical and probabilistic reasoning.

  • 3. a set of rules that state that the rows in a proof need to be numbered.

  • 4. *a set of derivation rules that express which wffs \(\phi \) can be written after which wffs \(\psi \) in a derivation.

Q3. In logic, if an argument is said to be "good", then "its conclusion follows from its premises". One way to make this idea precise is to say that "it is impossible for the premises to be true and the conclusion false". What is the other way?

  • 1. *if you can prove that the conclusion follows from the premises

  • 2. if you can construct a truth table showing that the conclusion follows from the premises.

0.2 Proofs

What derivation rule that permits the step in the proof below.

KEY: (A) = \(\land E\), (B) = \(\land I\), (C) = \(\to E\), (D) \(\to I\), (E) = \(R\)

Q5. Under the assumption of \(A\), \(B\) follows. Therefore, if A, then B. --- Answer: \(\to I\)

Q6. \(A\wedge \neg P\vdash \neg P\) --- Answer: \(\wedge E\)

Q7. \(\neg (A\vee \neg X)\wedge \neg Q\vdash \neg (A\vee \neg X)\) --- Answer: \(\wedge E\)

Q8. \(\neg X\vdash \neg X\) --- Answer: \(R\)

Q9. \(\neg Z\rightarrow S, \neg Z \vdash S\) --- Answer: \(\to E\)

KEY: (A) = \(\leftrightarrow E\), (B) = \(\leftrightarrow I\), (C) = \(\lor E\), (D) \(\lor I\), (E) = \(R\)

Q11. A if and only if B. A. Therefore, B. --- Answer: \(\leftrightarrow E\)

Q12. A or B. Assuming A is the case, it follows that C is the case.. And, assuming B is the case, it follows that C is the case. Therefore, C is the case. --- Answer: \(\vee E\)

--- Answer: \(\leftrightarrow E\)

Q13. What rule allows you to derive \(\lnot P\) from (1) an assumption \(P\) and (2) \(Z\) and \(\neg Z\) within the subproof started by \(P\)? --- Answer: \(\neg I\)

Q14. \((L\rightarrow Q)\leftrightarrow M\), \(M \vdash L\rightarrow Q\) --- Answer: \(\leftrightarrow E\)

Q15. \(\neg F\wedge R\vdash (\neg F\wedge R)\vee L\) --- Answer: \(\vee I\)

Q16. From \(\neg Z\vee C\) and two subproofs \(D\) is derived. The first subproof is where \(\neg Z\) is assumed and \(D\) is derived. The second subproof is where \(C\) is assumed and \(D\) is derived. --- Answer: \(\vee E\)

Q17. \(W\vee M\vdash W\vee \neg \neg M\) --- Answer: \(DN\)

KEY: (A) = \(MT\), (B) = \(IMP\), (C) = \(DeM\), (D) \(DS\), (E) = \(HS\)

Q18. A or B. Not B. Therefore A. --- Answer: \(DS\)

Q19. \(Z\rightarrow Q, \neg Q\vdash \neg Z\) --- Answer: \(MT\)

Q20. \(Z\rightarrow B\vdash \neg Z\vee B\) --- Answer: \(IMP\)

Q21. If A, then B. Not-B. Therefore, not-A. --- Answer: \(MT\)

Q22. \(\neg (Z\vee B)\vdash \neg Z\wedge \neg B\) --- Answer: \(DeM\)

Q23. \(\neg M\vee P, \neg P\vdash \neg M\) --- Answer: \(DS\)

Q24. Not both A and B. Therefore, not-A or not-B. --- Answer: \(DeM\)

Q25. \(Q\rightarrow \neg Z, \neg Z\rightarrow M\vdash Q\rightarrow M\) --- Answer: \(HS\)

Q26. \((S\wedge T)\rightarrow L, \neg L\vdash \neg (S\wedge T)\) --- Answer: \(MT\)

Q27. If A then B. If B then C. Therefore, if A then C. --- Answer: \(HS\)

Provide proofs for the following syntactic entailments. Be sure to setup the proof correctly, number all lines, and clearly indicate how each line is justified using the rules from the deductive apparatus.

Q28. \(Q\vee Z, M\wedge (Q\wedge R), \neg B\wedge C \vdash (R\wedge \neg B)\wedge M\)

--- Answer: --- Answer: \(Q\vee Z, M\wedge (Q\wedge R), \neg B\wedge C \vdash (R\wedge \neg B)\wedge M\)

(A natural deduction proof with nine lines. Line 1: Q or Z, premise. Line 2: M and open parenthesis Q and R close parenthesis, premise. Line 3: not B and C, premise. Line 4: not B, from line 3 by conjunction elimination. Line 5: Q and R, from line 2 by conjunction elimination. Line 6: R, from line 5 by conjunction elimination. Line 7: M, from line 2 by conjunction elimination. Line 8: R and not B, from lines 6 and 4 by conjunction introduction. Line 9: open parenthesis R and not B close parenthesis and M, from lines 8 and 7 by conjunction introduction.)

Q29. \((A\wedge \neg Z)\rightarrow Q, S\wedge \neg Z, A \vdash Q\vee \neg \neg W\) --- Answer: --- Answer: \((A\wedge \neg Z)\rightarrow Q, S\wedge \neg Z, A \vdash Q\vee \neg \neg W\)

(A natural deduction proof with seven lines. Line 1: open parenthesis A and not Z close parenthesis implies Q, premise. Line 2: S and not Z, premise. Line 3: A, premise. Line 4: not Z, from line 2 by conjunction elimination. Line 5: A and not Z, from lines 3 and 4 by conjunction introduction. Line 6: Q, from lines 1 and 5 by conditional elimination. Line 7: Q or not not W, from line 6 by disjunction introduction.)

Q30. \(C\vee S, C\rightarrow (G\wedge F), S\rightarrow (G\wedge \neg L)\vdash G\) --- Answer: --- Answer: \(C\vee S, C\rightarrow (G\wedge F), S\rightarrow (G\wedge \neg L)\vdash G\)

(A natural deduction proof with ten lines. Line 8 by conjunction elimination. Line 10: G, from line 1 and lines 4 through 6 and lines 7 through 9 by disjunction elimination.)

Q32. \(\vdash (R\rightarrow M)\vee \neg (P\wedge \neg P)\) --- Answer: --- Answer: \(\vdash (R\rightarrow M)\vee \neg (P\wedge \neg P)\)

(A natural deduction parenthesis P and not P close introduction.)

Q33. \(\vdash (P\wedge \neg S)\rightarrow ((\neg Q\vee S)\rightarrow \neg Q)\) --- Answer: --- Answer: \(\vdash (P\wedge \neg S)\rightarrow ((\neg Q\vee S)\rightarrow \neg Q)\)

(A natural deduction proof with six lines using nested subproofs. Line 1: P and not S, assumption for conditional introduction. Line 2: not Q or S, assumption for conditional introduction. Line 3: not S, from line 1 by conjunction elimination. Line 4: not Q, from lines 2 and 3 by disjunctive syllogism. Line 5: open parenthesis not Q or S close parenthesis implies not Q, from lines 2 through 4 by conditional introduction. Line 6: open parenthesis P and not S close parenthesis implies open parenthesis open parenthesis not Q or S close parenthesis implies not Q close parenthesis, from lines 1 through 5 by conditional introduction.)