\(\newcommand{\footnotename}{footnote}\) \(\def \LWRfootnote {1}\) \(\newcommand {\footnote }[2][\LWRfootnote ]{{}^{\mathrm {#1}}}\) \(\newcommand {\footnotemark }[1][\LWRfootnote ]{{}^{\mathrm {#1}}}\) \(\let \LWRorighspace \hspace \) \(\renewcommand {\hspace }{\ifstar \LWRorighspace \LWRorighspace }\) \(\newcommand {\TextOrMath }[2]{#2}\) \(\newcommand {\mathnormal }[1]{{#1}}\) \(\newcommand \ensuremath [1]{#1}\) \(\newcommand {\LWRframebox }[2][]{\fbox {#2}} \newcommand {\framebox }[1][]{\LWRframebox } \) \(\newcommand {\setlength }[2]{}\) \(\newcommand {\addtolength }[2]{}\) \(\newcommand {\setcounter }[2]{}\) \(\newcommand {\addtocounter }[2]{}\) \(\newcommand {\arabic }[1]{}\) \(\newcommand {\number }[1]{}\) \(\newcommand {\noalign }[1]{\text {#1}\notag \\}\) \(\newcommand {\cline }[1]{}\) \(\newcommand {\directlua }[1]{\text {(directlua)}}\) \(\newcommand {\luatexdirectlua }[1]{\text {(directlua)}}\) \(\newcommand {\protect }{}\) \(\def \LWRabsorbnumber #1 {}\) \(\def \LWRabsorbquotenumber "#1 {}\) \(\newcommand {\LWRabsorboption }[1][]{}\) \(\newcommand {\LWRabsorbtwooptions }[1][]{\LWRabsorboption }\) \(\def \mathchar {\ifnextchar "\LWRabsorbquotenumber \LWRabsorbnumber }\) \(\def \mathcode #1={\mathchar }\) \(\let \delcode \mathcode \) \(\let \delimiter \mathchar \) \(\def \oe {\unicode {x0153}}\) \(\def \OE {\unicode {x0152}}\) \(\def \ae {\unicode {x00E6}}\) \(\def \AE {\unicode {x00C6}}\) \(\def \aa {\unicode {x00E5}}\) \(\def \AA {\unicode {x00C5}}\) \(\def \o {\unicode {x00F8}}\) \(\def \O {\unicode {x00D8}}\) \(\def \l {\unicode {x0142}}\) \(\def \L {\unicode {x0141}}\) \(\def \ss {\unicode {x00DF}}\) \(\def \SS {\unicode {x1E9E}}\) \(\def \dag {\unicode {x2020}}\) \(\def \ddag {\unicode {x2021}}\) \(\def \P {\unicode {x00B6}}\) \(\def \copyright {\unicode {x00A9}}\) \(\def \pounds {\unicode {x00A3}}\) \(\let \LWRref \ref \) \(\renewcommand {\ref }{\ifstar \LWRref \LWRref }\) \( \newcommand {\multicolumn }[3]{#3}\) \(\require {textcomp}\) \(\newcommand {\toprule }[1][]{\hline }\) \(\let \midrule \toprule \) \(\let \bottomrule \toprule \) \(\def \LWRbooktabscmidruleparen (#1)#2{}\) \(\newcommand {\LWRbooktabscmidrulenoparen }[1]{}\) \(\newcommand {\cmidrule }[1][]{\ifnextchar (\LWRbooktabscmidruleparen \LWRbooktabscmidrulenoparen }\) \(\newcommand {\morecmidrules }{}\) \(\newcommand {\specialrule }[3]{\hline }\) \(\newcommand {\addlinespace }[1][]{}\) \(\def \LWRpagenote {1}\) \(\newcommand {\pagenote }[2][\LWRpagenote ]{{}^{\mathrm {#1}}}\) \(\require {colortbl}\) \(\let \LWRorigcolumncolor \columncolor \) \(\renewcommand {\columncolor }[2][named]{\LWRorigcolumncolor [#1]{#2}\LWRabsorbtwooptions }\) \(\let \LWRorigrowcolor \rowcolor \) \(\renewcommand {\rowcolor }[2][named]{\LWRorigrowcolor [#1]{#2}\LWRabsorbtwooptions }\) \(\let \LWRorigcellcolor \cellcolor \) \(\renewcommand {\cellcolor }[2][named]{\LWRorigcellcolor [#1]{#2}\LWRabsorbtwooptions }\) \(\require {cancel}\) \(\newcommand {\intertext }[1]{\text {#1}\notag \\}\) \(\let \Hat \hat \) \(\let \Check \check \) \(\let \Tilde \tilde \) \(\let \Acute \acute \) \(\let \Grave \grave \) \(\let \Dot \dot \) \(\let \Ddot \ddot \) \(\let \Breve \breve \) \(\let \Bar \bar \) \(\let \Vec \vec \) \(\require {mathtools}\) \(\newcommand {\vcentcolon }{\mathrel {\unicode {x2236}}}\) \(\newcommand {\approxcolon }{\approx \vcentcolon }\) \(\newcommand {\Approxcolon }{\approx \dblcolon }\) \(\newcommand {\simcolon }{\sim \vcentcolon }\) \(\newcommand {\Simcolon }{\sim \dblcolon }\) \(\newcommand {\dashcolon }{\mathrel {-}\vcentcolon }\) \(\newcommand {\Dashcolon }{\mathrel {-}\dblcolon }\) \(\newcommand {\colondash }{\vcentcolon \mathrel {-}}\) \(\newcommand {\Colondash }{\dblcolon \mathrel {-}}\) \(\newenvironment {crampedsubarray}[1]{}{}\) \(\newcommand {\smashoperator }[2][]{#2\limits }\) \(\newcommand {\SwapAboveDisplaySkip }{}\) \(\newcommand {\LaTeXunderbrace }[1]{\underbrace {#1}}\) \(\newcommand {\LaTeXoverbrace }[1]{\overbrace {#1}}\) \(\Newextarrow \xLongleftarrow {10,10}{0x21D0}\) \(\Newextarrow \xLongrightarrow {10,10}{0x21D2}\) \(\let \xlongleftarrow \xleftarrow \) \(\let \xlongrightarrow \xrightarrow \) \(\newcommand {\LWRmultlined }[1][]{\begin {multline*}}\) \(\newenvironment {multlined}[1][]{\LWRmultlined }{\end {multline*}}\) \(\let \LWRorigshoveleft \shoveleft \) \(\renewcommand {\shoveleft }[1][]{\LWRorigshoveleft }\) \(\let \LWRorigshoveright \shoveright \) \(\renewcommand {\shoveright }[1][]{\LWRorigshoveright }\) \(\newcommand {\shortintertext }[1]{\text {#1}\notag \\}\) \(\newcommand {\LWRnicearrayarray }[1]{\begin {array}{#1}}\) \(\def \LWRnicearrayarrayopt #1[#2] {\begin {array}{#1}}\) \(\newenvironment {NiceArray}[2][]{\ifnextchar [{\LWRnicearrayarrayopt {#2}}{\LWRnicearrayarray {#2}}}{\end {array}}\) \(\newcommand {\LWRnicearraywithdelimtwo }[2][]{\ifnextchar [{\LWRnicearrayarrayopt {#2}}{\LWRnicearrayarray {#2}}}\) \(\newenvironment {NiceArrayWithDelims}[2]{\def \LWRnicearrayrightdelim {\right #2}\left #1\LWRnicearraywithdelimtwo }{\end {array}\LWRnicearrayrightdelim }\) \(\newenvironment {pNiceArray} {\begin {NiceArrayWithDelims}{(}{)}} {\end {NiceArrayWithDelims}} \) \(\newenvironment {bNiceArray} {\begin {NiceArrayWithDelims}{[}{]}} {\end {NiceArrayWithDelims}} \) \(\newenvironment {BNiceArray} {\begin {NiceArrayWithDelims}{\{}{\}}} {\end {NiceArrayWithDelims}} \) \(\newenvironment {vNiceArray} {\begin {NiceArrayWithDelims}{\vert }{\vert }} {\end {NiceArrayWithDelims}} \) \(\newenvironment {VNiceArray} {\begin {NiceArrayWithDelims}{\Vert }{\Vert }} {\end {NiceArrayWithDelims}} \) \(\newenvironment {NiceMatrix}[1][]{\begin {matrix}}{\end {matrix}}\) \(\newenvironment {pNiceMatrix}[1][]{\begin {pmatrix}}{\end {pmatrix}}\) \(\newenvironment {bNiceMatrix}[1][]{\begin {bmatrix}}{\end {bmatrix}}\) \(\newenvironment {BNiceMatrix}[1][]{\begin {Bmatrix}}{\end {Bmatrix}}\) \(\newenvironment {vNiceMatrix}[1][]{\begin {vmatrix}}{\end {vmatrix}}\) \(\newenvironment {VNiceMatrix}[1][]{\begin {Vmatrix}}{\end {Vmatrix}}\) \(\newcommand {\LWRnicematrixBlock }[1]{#1}\) \(\def \LWRnicematrixBlockopt <#1>#2{#2}\) \(\newcommand {\Block }[2][]{\ifnextchar <\LWRnicematrixBlockopt \LWRnicematrixBlock }\) \(\newcommand {\diagbox }[2]{\begin {array}{l}\hfill \quad #2\\\hline #1\quad \hfill \end {array}}\) \(\let \hdottedline \hdashline \) \(\newcommand {\Hline }[1][]{\hline }\) \(\newcommand {\CodeBefore }{}\) \(\newcommand {\Body }{}\) \(\newcommand {\CodeAfter }{}\) \(\newcommand {\line }[3][]{}\) \(\newcommand {\RowStyle }[2][]{}\) \(\newcommand {\LWRSubMatrix }[1][]{}\) \(\newcommand {\SubMatrix }[4]{\LWRSubMatrix }\) \(\newcommand {\OverBrace }[4][]{}\) \(\newcommand {\UnderBrace }[4][]{}\) \(\newcommand {\HBrace }[3][]{}\) \(\newcommand {\VBrace }[3][]{}\) \(\newcommand {\ShowCellNames }{}\) \(\newcommand {\tabularnote }[2][]{}\) \(\newcommand {\cellcolor }[3][]{}\) \(\newcommand {\rowcolor }[3][]{}\) \(\newcommand {\LWRrowcolors }[1][]{}\) \(\newcommand {\rowcolors }[4][]{\LWRrowcolors }\) \(\newcommand {\rowlistcolors }[3][]{\LWRrowcolors }\) \(\newcommand {\columncolor }[3][]{}\) \(\newcommand {\rectanglecolor }[4][]{}\) \(\newcommand {\arraycolor }[2][]{}\) \(\newcommand {\chessboardcolors }[3][]{}\) \(\newcommand {\ldots }[1][]{\dots }\) \(\newcommand {\Cdots }[1][]{\cdots }\) \(\newcommand {\Vdots }[1][]{\vdots }\) \(\newcommand {\Ddots }[1][]{\ddots }\) \(\newcommand {\Iddots }[1][]{\mathinner {\unicode {x22F0}}}\) \(\newcommand {\Hdotsfor }[1]{\ldots }\) \(\newcommand {\Vdotsfor }[1]{\vdots }\) \(\newcommand {\AutoNiceMatrix }[2]{\text {(AutoNiceMatrix #1)}}\) \(\let \pAutoNiceMatrix \AutoNiceMatrix \) \(\let \bAutoNiceMatrix \AutoNiceMatrix \) \(\let \BAutoNiceMatrix \AutoNiceMatrix \) \(\let \vAutoNiceMatrix \AutoNiceMatrix \) \(\let \VAutoNiceMatrix \AutoNiceMatrix \) \(\newcommand {\tcbset }[1]{}\) \(\newcommand {\tcbsetforeverylayer }[1]{}\) \(\newcommand {\tcbox }[2][]{\boxed {\text {#2}}}\) \(\newcommand {\tcboxfit }[2][]{\boxed {#2}}\) \(\newcommand {\tcblower }{}\) \(\newcommand {\tcbline }{}\) \(\newcommand {\tcbtitle }{}\) \(\newcommand {\tcbsubtitle [2][]{\mathrm {#2}}}\) \(\newcommand {\tcboxmath }[2][]{\boxed {#2}}\) \(\newcommand {\tcbhighmath }[2][]{\boxed {#2}}\)

Exam 3 - Practice Exam

David W. Agler

July 26, 2025

This exam has 26 questions, for a total of 100 points. Place your name on the answersheet (last page). Place proofs on the blank space on the answersheet.

0.1 Multiple Choice: Concepts and terminology

Q1. What is a derivation of \(\phi \) from \(\Gamma \) using \(\mathbf {PD}\)?

  • 1. *A derivation of \(\phi \) is a finite string of formulas from a set \(\Gamma \) of \(\mathbf {PL}\) 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.

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

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

  • 4. A derivation of \(\phi \) is an infinite string of formulas from a set \(\Gamma \) of \(\mathbf {PL}\) 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 \(\mathbf {PL}\)?

  • 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, there are two notions of logical consequence (entailment). The first notion is the semantic (model-theoretic) notion, which says that \(\phi \) is a logical consequence of \(\Gamma \) if and only if there is no interpretation of the members of \(\Gamma \) and \(\phi \) such that all of the members of \(\Gamma \) are true and \(\phi \) is false. What is the other notion of logical consequence?

  • 1. the semi-semantic notion of logical consequence (entailment). This notion says that \(\phi \) is a semi-semantic consequence of \(\Gamma \) if and only if it is impossible for the premises to be true and the conclusion false, defined in terms of truth tables and trees.

  • 2. the intuitive notion of logical consequence (entailment). This notion says that \(\phi \) is a logical consequence of \(\Gamma \) if and only if \(\phi \) intuitively follows from \(\Gamma \)

  • 3. the legal notion of logical consequence (entailment). This notion says that \(\phi \) is a logical consequence of \(\Gamma \) if and only if \(\phi \) would be accepted in a court of law or some practical matter.

  • 4. *the syntactic (proof-theoretic) notion of logical consequence (entailment). This notion says that \(\phi \) is a logical consequence of \(\Gamma \) if and only if there is a derivation of \(\phi \) from \(\Gamma \)

  • 5. none of the above

Write the abbreviation of the derivation rule that permits the step in the proof below.

Q4. From \(X\) to \(X\) --- Answer: \(R\)

Q5. From \(A\wedge X\) to \(X\) --- Answer: \(\wedge E\)

Q6. From \(\neg A\) and \(B\) to \(\neg A\wedge B\) --- Answer: \(\wedge I\)

Q7. From \(\neg \neg B\wedge \neg X\) to \(\neg \neg B\) --- Answer: \(\wedge E\)

Q8. From \(Z\rightarrow M\) and \(\neg Q\) to \((Z\rightarrow M)\wedge \neg Q\) --- Answer: \(\wedge I\)

Q9. From \(\neg Q\rightarrow S\) and \(\neg Q\) to \(S\). --- Answer: \(\rightarrow E\)

Q10. From \((B\rightarrow Q)\leftrightarrow M\) and \(M\) to \(B\rightarrow Q\) --- Answer: \(\leftrightarrow E\)

Q11. From \(S\rightarrow L\) and \(\neg L\) to \(\neg S\) --- Answer: \(MT\)

Q12. From \(X\rightarrow \neg Z\) and \(\neg Z\rightarrow M\) to \(X\rightarrow M\) --- Answer: \(HS\)

Q13. From \(S\wedge T\) to \(X\vee (S\wedge T)\) --- Answer: \(\vee I\)

Q14. From \(V\vee P\) and \(\neg P\) to \(V\) --- Answer: \(DS\)

Q15. From \(A\rightarrow B\) to \(\neg A\vee B\) --- Answer: \(IMP\)

Q16. From \(\neg (A\vee B)\) to \(\neg A\wedge \neg B\) --- Answer: \(DeM\)

Q17. From \(W\) to \(\neg \neg W\) --- Answer: \(DN\)

Q18. What rule allows you to derive \(\neg A\) from (1) an assumption \(A\) and (2) \(B\) and \(\neg B\) within the subproof started by \(A\)? --- Answer: \(\neg I\)

Q19. What derivation rule is best described as follows: if \((P\rightarrow Q)\wedge Q\) is on a line of the proof, then it is legitimate to derive \(P\rightarrow Q\) on a line and it is legitimate to derive \(Q\) on another line. --- Answer: \(\wedge E\)

Q20. From \(A\vee B\) and two subproofs \(C\) is derived. The first subproof is where \(A\) is assumed and \(C\) is derived. The second subproof is where \(B\) is assumed and \(C\) is derived. --- Answer: \(\vee E\)

0.2 Proofs

Solve the following proofs. 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.

Q21. \((A\wedge \neg W)\wedge \neg S, T\wedge \neg R, S\vee R \vdash \neg W\wedge \neg R\)

--- Answer: \((A\wedge \neg W)\wedge \neg S, T\wedge \neg R, S\vee R \vdash \neg W\wedge \neg R\)

(A natural deduction proof with seven lines; Line 1: open parenthesis A and not W close parenthesis and not S; justification P; Line 2: T and not R; justification P; Line 3: S or R; justification P, not W and not R; Line 4: not

R; justification And-Elimination, line 2; Line 5: A and not W; justification And-Elimination, line 1; Line 6: not W; justification And-Elimination, line 5; Line 7: not W and not R; justification And-Introduction, lines 6 and 4)

Q22. \((P\vee \neg W)\rightarrow \neg B, \neg W\vdash \neg L\vee \neg B\)

--- Answer: \((P\vee \neg W)\rightarrow \neg B, \neg W\vdash \neg L\vee \neg B\)

(A five-line natural deduction proof; Line 1: (P or not W) implies not B; premise; Line 2: not W; premise; Line 3: P or not W; derived by Or-Introduction from line 2; Line 4: not B; derived by Implication-Elimination from lines 1

and 3; Line 5: not L or not B; derived by Or-Introduction from line 4)

Q23. \(W\rightarrow D, \neg C\rightarrow D, W\vee \neg C\vdash \neg \neg D\)

--- Answer: \(W\rightarrow D, \neg C\rightarrow D, W\vee \neg C\vdash \neg \neg D\)

(A natural deduction proof using the intelim system; three premises are W implies D, not C implies D, and W or not C with double negation D; W is assumed, leading to D by implies elimination; not C is assumed, leading to D by

implies elimination; D is derived by or elimination from the previous steps; double negation introduction gives not not D)

Q24. \(B\leftrightarrow \neg B\vdash \neg (A\vee C)\)

--- Answer: \(B\leftrightarrow \neg B\vdash \neg (A\vee C)\)

(A natural deduction proof using the intelim system; first line is B if and only if not B, marked as a premise; second line assumes B; third line derives not B by biconditional elimination from lines 1 and 2; fourth line repeats

B by reiteration from line 2; fifth line concludes not B by negation introduction from lines 2 to 4; sixth line assumes A or C; seventh line repeats not B by reiteration from line 5; eighth line derives B by biconditional elimination from
lines 7 and 1; ninth line concludes not (A or C) by negation introduction from lines 6 to 8)

Q25. \(\vdash \neg (W\rightarrow Q)\rightarrow \neg Q\)

--- Answer: \(\vdash \neg (W\rightarrow Q)\rightarrow \neg Q\)

(A natural deduction proof using the intelim system; line 1 assumes not (W implies Q); line 2 rewrites as not (not W or Q) by implication equivalence; line 3 rewrites as not not W and not Q by De Morgan’s law; line 4 derives not

Q by And-Elimination; line 5 concludes not (W implies Q) implies not Q by conditional introduction, discharging the assumption from line 1)

Q26. \(\vdash \neg (P\vee Q)\rightarrow (R\rightarrow \neg Q)\)

--- Answer: \(\vdash \neg (P\vee Q)\rightarrow (R\rightarrow \neg Q)\)

(A natural deduction proof; line 1 assumes not (P or Q); line 2 assumes R; line 3 derives not P and not Q by De Morgan’s Law from line 1; line 4 derives not Q by And-Elimination from line 3; line 5 derives R implies Q by

Conditional Introduction from lines 2 to 4; line 6 derives not (P or Q) implies (R implies Q) by Conditional Introduction from lines 1 to 5)