%\pagestyle{plain}


\chapter[Answers to selected exercises]{Answers to selected exercises}
\label{app.solutions}

Many of the exercises may be answered correctly in different ways. Where that is the case, the solution here represents one possible correct answer.


\solutionsection{ch.intro}{pr.MartianGiraffes}
\begin{earg}
%\nextSeq %G2, G3, and G4
\item consistent
%\noSeq %G1, G3, and G4
\item inconsistent
%\nextSeq %G1, G2, and G4
\item consistent
%\lastSeq %G1, G2, and G3
\item consistent
%are consistent.
\end{earg}

\solutionsection{ch.intro}{pr.EnglishCombinations}
%\begin{earg}
%\item A valid argument that has one false premise and one true premise
%possible
\nextSeq
%\item A valid argument that has a false conclusion
%possible
\nextSeq
%\item A valid argument, the conclusion of which is a contradiction
%possible
\nextSeq
%\item An invalid argument, the conclusion of which is a tautology
%not possible
\noSeq
%\item A tautology that is contingent
%not possible
\noSeq
%\item Two logically equivalent sentences, both of which are tautologies
%possible
\nextSeq
%\item Two logically equivalent sentences, one of which is a tautology and one of which is contingent
%not possible
\noSeq
%\item Two logically equivalent sentences that together are an inconsistent set
%possible
\nextSeq
%Consider two contradictions. As a matter of logic, they are both necessarily false. So they have the same truth value: false. This makes them logically equivalent.
%\item A consistent set of sentences that contains a contradiction
%not possible
\noSeq
%\item An inconsistent set of sentences that contains a tautology
%possible
\lastSeq
%\end{earg}
are possible.

\solutionsection{ch.SL}{pr.monkeysuits}
\begin{earg}
\item $\enot M$
\item $M \eor \enot M$
\item $G \eor C$
\item $\enot C \eand \enot G$
\item $C \eif (\enot G \eand \enot M)$
\item $M \eor (C \eor G)$
\end{earg}


\solutionsection{ch.SL}{pr.avacareer}
\begin{earg}
\item $E_1 \eand E_2$
\item $F_1 \eif S_1$
\item $F_1 \eor E_1$
\item $E_2 \eand \enot S_2$
\item $\enot E_1 \eand \enot E_2$
\item $E_1 \eand E_2 \eand \enot(S_1 \eor S_2)$
\item $S_2 \eif F_2$
\item $(\enot E_1 \eif \enot E_2) \eand (E_1 \eif E_2)$
\item $S_1 \eiff \enot S_2$
\item $(E_2 \eand F_2) \eif S_2$
\item $\enot(E_2 \eand F_2)$
\item $(F_1 \eand F_2) \eiff (\enot E_1 \eand \enot E_2)$
\end{earg}

\solutionsection{ch.SL}{pr.spies}
\begin{ekey}
\item[A:] Alice is a spy.
\item[B:] Bob is a spy.
\item[C:] The code has been broken.
\item[G:] The German embassy will be in an uproar.
\end{ekey}
\begin{earg}
\item %Alice and Bob are both spies.
$A \eand B$
\item %If either Alice or Bob is a spy, then the code has been broken.
$(A \eor B) \eif C$
\item %If neither Alice nor Bob is a spy, then the code remains unbroken.
$\enot(A \eor B) \eif \enot C$
\item %The German embassy will be in an uproar, unless someone has broken the code.
$G \eor C$
\item %Either the code has been broken or it has not, but the German embassy will be in an uproar regardless.
$(C \eor \enot C) \eand G$
\item %Either Alice or Bob is a spy, but not both.
$(A \eor B) \eand \enot(A \eand B)$
\end{earg}


\solutionsection{ch.SL}{pr.wiffSL}
\begin{earg}
\item %$(A)$
(a) no (b) no
\item %$J_{374} \eor \enot J_{374}$
(a) no (b) yes
\item %$\enot \enot \enot \enot F$
(a) yes (b) yes
\item %$\enot \eand S$
(a) no (b) no
\item %$(G \eand \enot G)$
(a) yes (b) yes
\item %$\script{A} \eif \script{A}$
(a) no (b) no
\item %$(A \eif (A \eand \enot F)) \eor (D \eiff E)$
(a) no (b) yes
\item %$[(Z \eiff S) \eif W] \eand [J \eor X]$
(a) no (b) yes
\item %$(F \eiff \enot D \eif J) \eor (C \eand D)$
(a) no (b) no
\end{earg}




\solutionsection{ch.TruthTables}{pr.TT.TTorC}
\begin{earg}
\item tautology
\item contradiction
\item contingent
\item tautology
\item tautology
\item contingent
\item tautology
\item contradiction
\item tautology
\item contradiction
\item tautology
\item contingent
\item contradiction
\item contingent
\item tautology
\item tautology
\item contingent
\item contingent
\end{earg}

\solutionsection{ch.TruthTables}{pr.TT.equiv}
%\begin{earg}
%\item not equivalent
\noSeq
%\item equivalent
\nextSeq
%\item not equivalent
\noSeq
%\item not equivalent
\noSeq
%\item equivalent
\nextSeq
%\item equivalent
\nextSeq
%\item not equivalent
\noSeq
%\item equivalent
\nextSeq
%\item equivalent
\lastSeq
%\item not equivalent
\noSeq
%\end{earg}
are logically equivalent.

\solutionsection{ch.TruthTables}{pr.TT.consistent}
%\item $A\eif A$, $\enot A \eif \enot A$, $A\eand A$, $A\eor A$ %consistent
\nextSeq
%\item $A \eand B$, $C\eif \enot B$, $C$ %inconsistent
\noSeq
%\item $A\eor B$, $A\eif C$, $B\eif C$ %consistent
\nextSeq
%\item $A\eif B$, $B\eif C$, $A$, $\enot C$ %inconsistent
\noSeq
%\item $B\eand(C\eor A)$, $A\eif B$, $\enot(B\eor C)$  %inconsistent
\noSeq
%\item $A \eor B$, $B\eor C$, $C\eif \enot A$ %consistent
\nextSeq
%\item $A\eiff(B\eor C)$, $C\eif \enot A$, $A\eif \enot B$ %consistent
\nextSeq
%\item $A$, $B$, $C$, $\enot D$, $\enot E$, $F$ %consistent
\lastSeq
are consistent.

\solutionsection{ch.TruthTables}{pr.TT.valid}
%\item $A\eif A$, \therefore\ $A$ %invalid
\noSeq
%\item $A\eor\bigl[A\eif(A\eiff A)\bigr]$, \therefore\ A %invalid
\noSeq
%\item $A\eif(A\eand\enot A)$, \therefore\ $\enot A$ %valid
\nextSeq
%\item $A\eiff\enot(B\eiff A)$, \therefore\ $A$ %invalid
\noSeq
%\item $A\eor(B\eif A)$, \therefore\ $\enot A \eif \enot B$ %valid
\nextSeq
%\item $A\eif B$, $B$, \therefore\ $A$ %invalid
\noSeq
%\item $A\eor B$, $B\eor C$, $\enot A$, \therefore\ $B \eand C$ %invalid
\noSeq
%\item $A\eor B$, $B\eor C$, $\enot B$, \therefore\ $A \eand C$ %valid
\nextSeq
%\item $(B\eand A)\eif C$, $(C\eand A)\eif B$, \therefore\ $(C\eand B)\eif A$ %invalid
\noSeq
%\item $A\eiff B$, $B\eiff C$, \therefore\ $A\eiff C$ %valid
\lastSeq
are valid.


\solutionsection{ch.TruthTables}{pr.TT.concepts}
\begin{earg}
\item %Suppose that \script{A} and \script{B} are logically equivalent. What can you say about $\script{A}\eiff\script{B}$?
\script{A} and \script{B} have the same truth value on every line of a complete truth table, so $\script{A}\eiff\script{B}$ is true on every line. It is a tautology.
\item %Suppose that $(\script{A}\eand\script{B})\eif\script{C}$ is contingent. What can you say about the argument ``\script{A}, \script{B}, \therefore\script{C}''?
The sentence is false on some line of a complete truth table. On that line, \script{A} and \script{B} are true and \script{C} is false. So the argument is invalid.
\item %Suppose that $\{\script{A},\script{B}, \script{C}\}$ is inconsistent. What can you say about $(\script{A}\eand\script{B}\eand\script{C})$?
Since there is no line of a complete truth table on which all three sentences are true, the conjunction is false on every line. So it is a contradiction.
\item %Suppose that \script{A} is a contradiction. What can you say about the argument ``\script{A}, \script{B}, \therefore\script{C}''?
Since \script{A} is false on every line of a complete truth table, there is no line on which \script{A} and \script{B} are true and \script{C} is false. So the argument is valid.
\item %Suppose that \script{C} is a tautology. What can you say about the argument ``\script{A}, \script{B}, \therefore\script{C}''?
Since \script{C} is true on every line of a complete truth table, there is no line on which \script{A} and \script{B} are true and \script{C} is false. So the argument is valid.
\item %Suppose that \script{A} and \script{B} are logically equivalent. What can you say about $(\script{A}\eor\script{B})$?
Not much. $(\script{A}\eor\script{B})$ is a tautology if \script{A} and \script{B} are tautologies; it is a contradiction if they are contradictions; it is contingent if they are contingent.
\item %Suppose that \script{A} and \script{B} are \emph{not} logically equivalent. What can you say about $(\script{A}\eor\script{B})$?
\script{A} and \script{B} have different truth values on at least one line of a complete truth table, and $(\script{A}\eor\script{B})$ will be true on that line. On other lines, it might be true or false. So $(\script{A}\eor\script{B})$ is either a tautology or it is contingent; it is \emph{not} a contradiction.
\end{earg}


\solutionsection{ch.TruthTables}{pr.altConnectives}
\begin{earg}
\item %$A\eor B$
$\enot A \eif B$
\item %$A\eand B$
$\enot(A \eif \enot B)$
\item %$A\eiff B$
$\enot [(A\eif B) \eif \enot(B\eif A)]$
%\item %$A \eand B$
%$\enot(\enot A \eor \enot B)$
%\item %$A \eif B$
%$\enot A \eor B$
%\item %$A \eiff B$
%$\enot(\enot A \eor \enot B) \eor \enot(A \eor B)$
\end{earg}



\solutionsection{ch.QL}{pr.QLalligators}
\begin{earg}
\item %Amos, Bouncer, and Cleo all live at the zoo.
$Za \eand Zb \eand Zc$
\item %Bouncer is a reptile, but not an alligator. 
$Rb \eand \enot Ab$
\item %If Cleo loves Bouncer, then Bouncer is a monkey. 
$Lcb \eif Mb$
\item %If both Bouncer and Cleo are alligators, then Amos loves them both.
$(Ab \eand Ac)\eif(Lab \eand Lac)$
\item %Some reptile lives at the zoo. 
$\exists x(Rx \eand Zx)$
\item %Every alligator is a reptile. 
$\forall x(Ax \eif Rx)$
\item %Any animal that lives at the zoo is either a monkey or an alligator. 
$\forall x\bigl[Zx \eif (Mx \eor Ax)\bigr]$
\item %There are reptiles which are not alligators.
$\exists x(Rx \eand \enot Ax)$
\item %Cleo loves a reptile.
$\exists x(Rx \eand Lcx)$
\item %Bouncer loves all the monkeys that live at the zoo.
$\forall x\bigl[(Mx \eand Zx) \eif Lbx\bigr]$
\item %All the monkeys that Amos loves love him back.
$\forall x\bigl[(Mx \eand Lax) \eif Lxa\bigr]$
\item %If any animal is an reptile, then Amos is.
$\exists x Rx \eif Ra$
\item %If any animal is an alligator, then it is a reptile.
$\forall x(Ax \eif Rx)$
\item %Every monkey that Cleo loves is also loved by Amos.
$\forall x\bigl[(Mx \eand Lcx) \eif Lax\bigr]$
\item %There is a monkey that loves Bouncer, but sadly Bouncer does not reciprocate this love.
$\exists x(Mx \eand Lxb \eand \enot Lbx)$
\end{earg}


\solutionsection{ch.QL}{pr.QLcandies}
\begin{earg}
\item %Boris has never tried any candy.
$\enot\exists x Tx$
\item %Marzipan is always made with sugar.
$\forall x(Mx \eif Sx)$
\item %Some candy is sugar-free.
$\exists x \enot Sx$
\item %The very best candy is chocolate.
$\exists x[Cx \eand \enot\exists y Byx]$
\item %No candy is better than itself.
$\enot \exists x Bxx$
\item %Boris has never tried sugar-free chocolate.
$\enot \exists x(Cx \eand \enot Sx \eand Tx)$
\item %Boris has tried marzipan and chocolate, but never together.
$\exists x(Cx \eand Tx) \eand \exists x(Mx \eand Tx) \eand \enot \exists x(Cx \eand Mx \eand Tx)$
%\item Boris has tried nothing that is better than sugar-free marzipan.
%$\enot\exists x[Bx \eand \exists y(
\item %Any candy with chocolate is better than any candy without it.
$\forall x[Cx \eif \forall y(\enot Cy \eif Bxy)]$
\item %Any candy with chocolate and marzipan is better than any candy without either.
$\forall x\bigl((Cx \eand Mx) \eif \forall y[(\enot Cy \eand \enot My) \eif Bxy]\bigr)$
\end{earg}


\solutionsection{ch.QL}{pr.QLballet}
\begin{earg}
\item %All of Patrick's children are ballet dancers.
$\forall x(Cxp \eif Dx)$
\item %Jane is Patrick's daughter.
$Cjp \eand Fj$
\item %Patrick has a daughter.
$\exists x(Cxp \eand Fx)$
\item %Jane is an only child.
$\enot\exists x Sxj$
\item %All of Patrick's daughters dance ballet.
$\forall x\bigl[(Cxp \eand Fx)\eif Dx\bigr]$
\item %Patrick has no sons.
$\enot\exists x(Cxp \eand Mx)$
\item %Jane is Elmer's niece.
$\exists x(Cjx \eand Sxe \eand Fj)$
\item %Patrick is Elmer's brother.
$Spe \eand Mp$
\item %Patrick's brothers have no children.
$\forall x\bigl[(Sxp \eand Mx) \eif \enot\exists y Cyx\bigr]$
\item %Jane is an aunt.
$\exists x(Sxj \eand \exists y Cyx \eand Fj)$
\item %Everyone who dances in the ballet has a sister who also dances in the ballet.
$\forall x\bigl[Dx \eif \exists y(Sxy \eand Fy \eand Dy)\bigr]$
\item %Every man who dances in the ballet is the child of someone who dances in the ballet.
$\forall x\bigl[(Mx \eand Dx) \eif \exists y(Cxy \eand Dy)\bigr]$
\end{earg}



\solutionsection{ch.QL}{pr.QLcards}
\begin{earg}
\item %All clubs are black cards.
$\forall x(Cx \eif Bx)$
\item %There are no wild cards.
$\enot\exists x Wx$
\item %There are at least two clubs.
$\exists x \exists y(Cx \eand Cy \eand x\neq y)$
\item %There is more than one one-eyed jack.
$\exists x \exists y(Jx \eand Ox \eand Jy \eand Oy \eand x\neq y)$
\item %There are at most two one-eyed jacks.
$\forall x\forall y\forall z\bigl[(Jx \eand Ox \eand Jy \eand Oy \eand Jz \eand Oz)\eif(x=y \eor x=z \eor y=z)\bigr]$
\item %There are two black jacks.
$\exists x\exists y\bigl(Jx \eand Bx \eand Jy \eand By \eand \forall z[(Jz \eand Bz) \eif (x=z \eor y=z)]\bigr)$
\item %There are four deuces.
$\exists x_1\exists x_2\exists x_3\exists x_4\bigl[Dx_1 \eand Dx_2 \eand Dx_3 \eand Dx_4 \eand x_1 \neq x_2 \eand x_1 \neq x_3 \eand x_1 \neq x_4 \eand x_2 \neq x_3 \eand x_2 \neq x_4 \eand x_3 \neq x_4  \eand \enot\exists y(Dy \eand y\neq x_1 \eand y\neq x_2 \eand y\neq x_3 \eand y\neq x_4)\bigr]$
\item %The deuce of clubs is a black card.
$\exists x\bigl(Dx \eand Cx \eand \forall y[(Dy\eand Cy) \eif x=y] \eand Bx\bigr)$
\item %One-eyed jacks and the man with the axe are wild.
$\forall x\bigl[(Ox \eand Jx) \eif Wx\bigr] \eand \exists x\bigl[Mx \eand \forall y(My \eif x=y) \eand Wx\bigr]$
\item %If the deuce of clubs is wild, then there is exactly one wild card.
$\exists x\bigl(Dx \eand Cx \eand \forall y[(Dy\eand Cy) \eif x=y] \eand Wx\bigr)\eif \exists x(Wx \eiff x=y)$
\item %The man with the axe is not a jack.
wide scope: $\enot \exists x\bigl[Mx \eand \forall y(My \eif x=y) \eand Jx\bigr]$\\
narrow scope: $\exists x\bigl[Mx \eand \forall y(My \eif x=y) \eand \enot Jx\bigr]$
\item %The deuce of clubs is not the man with the axe.
wide scope: $\enot \exists x\exists z\bigl(Dx \eand Cx \eand Mz \eand \forall y[(Dy\eand Cy) \eif x=y]  \eand \forall y[(My \eif z=y) \eand x=z]\bigr)$\\
narrow scope: $\exists x\exists z\bigl(Dx \eand Cx \eand Mz \eand \forall y[(Dy\eand Cy) \eif x=y]  \eand \forall y[(My \eif z=y) \eand x\neq z]\bigr)$
\end{earg}





\solutionsection{ch.semantics}{pr.TorF1}
%\item $Bc$
\noSeq
%\item $Ac \eiff \enot Nc$
\nextSeq
%\item $Nc \eif (Ac \eor Bc)$
\nextSeq
%\item $\forall x Ax$
\nextSeq
%\item $\forall x \enot Bx$
\noSeq
%\item $\exists x(Ax \eand Bx)$
\nextSeq
%\item $\exists x(Ax \eif Nx)$
\noSeq
%\item $\forall x(Nx \eor \enot Nx)$
\nextSeq
%\item $\exists x Bx \eif \forall x Ax$
\lastSeq
are true in the model.

\solutionsection{ch.semantics}{pr.TorF2}
\noSeq%\item $\exists x(Rxm \eand Rmx)$
\nextSeq%\item $\forall x(Rxm \eor Rmx)$
\noSeq%\item $\forall x(Hx \eiff Wx)$
\nextSeq%\item $\forall x(Rxm \eif Wx)$
\nextSeq%\item $\forall x\bigl[Wx \eif(Hx \eand Wx)\bigr]$
\noSeq%\item $\exists x Rxx$
\lastSeq%\item $\exists x\exists y Rxy$
\noSeq%\item $\forall x \forall y Rxy$
\noSeq%\item $\forall x \forall y (Rxy \eor Ryx)$
\noSeq%\item $\forallx \forall y \forall z\bigl[(Rxy \eand Ryz) \eif Rxz\bigr]$
are true in the model.


\solutionsection{ch.semantics}{pr.Contingent}
\begin{earg}
\item %$Da \eand Db$
	The sentence is true in this model:
	\begin{partialmodel}
		UD & \{Stan\}\\
		\extension{D} & \{Stan\}\\
		\referent{a} & Stan\\
		\referent{b} & Stan
	\end{partialmodel}
	And it is false in this model:
	\begin{partialmodel}
		UD & \{Stan\}\\
		\extension{D} & $\emptyset$\\
		\referent{a} & Stan\\
		\referent{b} & Stan
	\end{partialmodel}
\item %$\exists x Txh$
	The sentence is true in this model:
	\begin{partialmodel}
		UD & \{Stan\}\\
		\extension{T} & \{$<$Stan, Stan$>$\}\\
		\referent{h} & Stan
	\end{partialmodel}
	And it is false in this model:
	\begin{partialmodel}
		UD & \{Stan\}\\
		\extension{T} & $\emptyset$\\
		\referent{h} & Stan
	\end{partialmodel}
\item %$Pm \eand \enot\forall x Px$
	The sentence is true in this model:
	\begin{partialmodel}
		UD & \{Stan, Ollie\}\\
		\extension{P} & \{Stan\}\\
		\referent{m} & Stan
	\end{partialmodel}
	And it is false in this model:
	\begin{partialmodel}
		UD & \{Stan\}\\
		\extension{P} & $\emptyset$\\
		\referent{m} & Stan
	\end{partialmodel}
\end{earg}



\solutionsection{ch.semantics}{pr.NotEquiv}
There are many possible correct answers. Here are some:
\begin{earg}
\item %$Ja$, $Ka$
	Making the first sentence true and the second false:
	\begin{partialmodel}
		UD & \{alpha\}\\
		\extension{J} & \{alpha\}\\
		\extension{K} & $\emptyset$\\
		\referent{a} & alpha
	\end{partialmodel}
\item %$\exists x Jx$, $Jm$
	Making the first sentence true and the second false:
	\begin{partialmodel}
		UD & \{alpha, omega\}\\
		\extension{J} & \{alpha\}\\
		\referent{m} & omega
	\end{partialmodel}
\item %$\forall x Rxx$, $\exists x Rxx$
	Making the first sentence false and the second true:
	\begin{partialmodel}
		UD & \{alpha, omega\}\\
		\extension{R} & \{$<$alpha,alpha$>$\}
	\end{partialmodel}
\item %$\exists x Px \eif Qc$, $\exists x (Px \eif Qc)$
	Making the first sentence false and the second true:
	\begin{partialmodel}
		UD & \{alpha, omega\}\\
		\extension{P} & \{alpha\}\\
		\extension{Q} & $\emptyset$\\
		\referent{c} & alpha
	\end{partialmodel}
\item %$\forall x(Px \eif \enot Qx)$, $\exists x(Px \eand \enot Qx)$
	Making the first sentence true and the second false:
	\begin{partialmodel}
		UD & \{iota\}\\
		\extension{P} & $\emptyset$\\
		\extension{Q} & $\emptyset$
	\end{partialmodel}
\item %$\exists x(Px \eand Qx)$, $\exists x(Px \eif Qx)$
	Making the first sentence false and the second true:
	\begin{partialmodel}
		UD & \{iota\}\\
		\extension{P} & $\emptyset$\\
		\extension{Q} & \{iota\}
	\end{partialmodel}
\item %$\forall x(Px\eif Qx)$, $\forall x(Px \eand Qx)$
	Making the first sentence true and the second false:
	\begin{partialmodel}
		UD & \{iota\}\\
		\extension{P} & $\emptyset$\\
		\extension{Q} & \{iota\}
	\end{partialmodel}
\item %$\forall x\exists y Rxy$, $\exists x\forall y Rxy$
	Making the first sentence true and the second false:
	\begin{partialmodel}
		UD & \{alpha, omega\}\\
		\extension{R} & \{$<$alpha, omega$>$, $<$omega, alpha$>$\}
	\end{partialmodel}
\item %$\forall x\exists y Rxy$, $\forall x\exists y Ryx$
	Making the first sentence false and the second true:
	\begin{partialmodel}
		UD & \{alpha, omega\}\\
		\extension{R} & \{$<$alpha, alpha$>$, $<$alpha, omega$>$\}
	\end{partialmodel}
\end{earg}





\solutionsection{ch.semantics}{pr.IdentityModels}
\begin{earg}
\item %Show that $\{{\enot}Raa, \forall x (x=a \eor Rxa)\}$ is consistent.
There are many possible answers. Here is one:
\begin{partialmodel}
UD & \{Harry, Sally\}\\
\extension{R} &\{$<$Sally, Harry$>$\}\\
\referent{a} & Harry
\end{partialmodel}
\item %Show that $\{\forall x\forall y\forall z(x=y \eor y=z \eor x=z), \exists x\exists y\ x\neq y\}$ is consistent.
There are no predicates or constants, so we only need to give a UD.
Any UD with 2 members will do.
\item %Show that $\{\forall x\forall y\ x=y, \exists x\ x \neq a\}$ is inconsistent.
We need to show that it is impossible to construct a model in which these are both true. Suppose $\exists x\ x \neq a$ is true in a model. There is something in the universe of discourse that is \emph{not} the referent of $a$. So there are at least two things in the universe of discourse: \referent{a} and this other thing. Call this other thing $\beta$--- we know $a \neq \beta$. But if $a \neq \beta$, then $\forall x\forall y\ x=y$ is false. So the first sentence must be false if the second sentence is true. As such, there is no model in which they are both true. Therefore, they are inconsistent.
\end{earg}



\solutionsection{ch.proofs}{pr.justifySLproof}

\begin{multicols}{2}
\begin{proof}
\hypo{1}{W \eif \enot B}
\hypo{2}{A \eand W}
\hypo{2b}{B \eor (J \eand K)}
\have{3}{W}\ae{2}
\have{4}{\enot B} \ce{1,3}
\have{5}{J \eand K} \oe{2b,4}
\have{6}{K} \ae{5}
\end{proof}

\begin{proof}
\hypo{1}{L \eiff \enot O}
\hypo{2}{L \eor \enot O}
\open
	\hypo{a1}{\enot L}
	\have{a2}{\enot O}\oe{2,a1}
	\have{a3}{L}\be{1,a2}
	\have{a4}{\enot L}\by{R}{a1}
\close
\have{3}{L}\ne{a1-a4}
\end{proof}

\begin{proof}
\hypo{1}{Z \eif (C \eand \enot N)}
\hypo{2}{\enot Z \eif (N \eand \enot C)}
\open
	\hypo{a1}{\enot(N \eor  C)}
	\have{a2}{\enot N \eand \enot C} \by{DeM}{a1}
	\open
		\hypo{b1}{Z} 
		\have{b2}{C \eand \enot N}\ce{1,b1}
		\have{b3}{C}\ae{b2}
		\have{b4}{\enot C}\ae{a2}
	\close
	\have{a3}{\enot Z}\ni{b1-b4}
	\have{a4}{N \eand \enot C}\ce{2,a3}
	\have{a5}{N}\ae{a4}
	\have{a6}{\enot N}\ae{a2}
\close
\have{3}{N \eor C}\ne{a1-a6}
\end{proof}
\end{multicols}

\solutionsection{ch.proofs}{pr.justifyQLproof}

\begin{multicols}{2}
%$\{\forall x(\exists yLxy \eif \forall zLzx), Lab\} \vdash \forall xLxx$
\begin{proof}
\hypo{1}{\forall x(\exists yLxy \eif \forall zLzx)}
\hypo{2}{Lab}
\have{3}{\exists y Lay \eif \forall zLza}\Ae{1}
\have{4}{\exists y Lay} \Ei{2}
\have{5}{\forall z Lza} \ce{3,4}
\have{6}{Lca}\Ae{5}
\have{7}{\exists y Lcy \eif \forall zLzc}\Ae{1}
\have{8}{\exists y Lcy}\Ei{5}
\have{9}{\forall z Lzc}\ce{7,8}
\have{10}{Lcc}\Ae{9}
\have{11}{\forall x Lxx}\Ai{10}
\end{proof}

% $\{\forall x(Jx \eif Kx), \exists x\forall y Lxy, \forall x Jx\} \vdash \exists x(Kx \eand Lxx)$


\begin{proof}
\hypo{a}{\forall x(Jx \eif Kx)}
\hypo{b}{\exists x\forall y Lxy}
\hypo{c}{\forall x Jx}
\have{d}{Ja}\Ae{c}
\have{e}{Ja \eif Ka}\Ae{a}
\have{f}{Ka}\ce{e,d}
\open
	\hypo{2}{\forall y Lay}
	\have{3}{Laa}\Ae{2}
	\have{4}{Ka \eand Laa}\ai{f,3}
	\have{5}{\exists x(Kx \eand Lxx)}\Ei{4}
\close
\have{j}{\exists x(Kx \eand Lxx)}\Ee{b,2-5}
\end{proof}


%$\{\forall x(\exists y)(Rxy \eor Ryx),\forall x\enot Rmx\}\vdash\exists xRxm$

\begin{proof}
\hypo{p1}{\forall x\exists y(Rxy \eor Ryx)}
\hypo{p2}{\forall x\enot Rmx}
\have{3}{\exists y(Rmy \eor Rym)}\Ae{p1}
	\open
		\hypo{a1}{Rma \eor Ram}
		\have{a2}{\enot Rma}\Ae{p2}
		\have{a3}{Ram}\oe{a1,a2}
		\have{a4}{\exists x Rxm}\Ei{a3}
	\close
\have{n}{\exists x Rxm} \Ee{3,a1-a4}
\end{proof}

%$\vdash \exists x Mx \eor \forall x\enot Mx$

\begin{proof}
	\open
		\hypo{p1}{\enot (\exists x Mx \eor \forall x\enot Mx)}
		\have{p2}{\enot \exists x Mx \eand \enot \forall x\enot Mx} \by{DeM}{p1}
		\have{p3}{\enot \exists x Mx}\ae{p2}
		\have{p4}{\forall x\enot Mx}\by{QN}{p3}
		\have{p5}{\enot \forall x\enot Mx}\ae{p2}
	\close
\have{n}{\exists x Mx \eor \forall x\enot Mx} \ne{p1-p5}
\end{proof}
\end{multicols}


\solutionsection{ch.proofs}{pr.likes}
Regarding the translation of this argument, see p.~\pageref{likes2}.

\begin{proof}
\hypo{1}{\exists x\forall y[\forall z(Lxz \eif Lyz) \eif Lxy]}
\open
	\hypo{a}{\forall y[\forall z(Laz \eif Lyz) \eif Lay]}
	\have{b}{\forall z(Laz \eif Laz) \eif Laa} \Ae{a}
	\open
		\hypo{c1}{\enot \exists x Lxx} \by{for reductio}{}
		\have{c2}{\forall x\enot Lxx} \by{QN}{c1}
		\have{c3}{\enot Laa} \Ae{c2}
		\have{c4}{\enot \forall z(Laz \eif Laz)} \by{MT}{c2,c3}
		\open
			\hypo{d1}{Lab}
			\have{d2}{Lab} \by{R}{d1}
		\close
		\have{c5}{Lab \eif Lab} \ci{d1--d2}
		\have{c6}{\forall z(Laz \eif Laz)} \Ai{c5}
		\have{cn}{\enot \forall z(Laz \eif Laz)} \by{R}{c4}
	\close
	\have{h}{\exists x Lxx} \ne{c1--cn}
\close
\have{n}{\exists x Lxx} \Ee{1, a--h}
\end{proof}



\solutionsection{ch.proofs}{pr.QLequivornot}
%\item $\forall x Px \eif Qc$, $\forall x (Px \eif Qc)$
\noSeq
%\item $\forall x Px \eand Qc$, $\forall x (Px \eand Qc)$
\nextSeq
%\item $Qc \eor \exists x Qx$, $\exists x (Qc \eor Qx)$
\nextSeq
%\item $\forall x\forall y \forall z Bxyz$, $\forall x Bxxx$
\noSeq
%\item $\forall x\forall y Dxy$, $\forall y\forall x Dxy$
\lastSeq
%\item $\exists x\forall y Dxy$, $\forall y\exists x Dxy$
\noSeq
are logically equivalent.

\solutionsection{ch.proofs}{pr.QLvalidornot}
\noSeq%\item $\forall x\exists y Rxy$, \therefore\ $\exists y\forall x Rxy$
\nextSeq%\item $\exists y\forall x Rxy$, \therefore\ $\forall x\exists y Rxy$
\noSeq%\item $\exists x(Px \eand \enot Qx)$, \therefore\ $\forall x(Px \eif \enot Qx)$
\nextSeq%\item $\forall x(Sx \eif Ta)$, $Sd$, \therefore\ $Ta$
\nextSeq%\item $\forall x(Ax\eif Bx)$, $\forall x(Bx \eif Cx)$, \therefore\ $\forall x(Ax \eif Cx)$
\noSeq%\item $\exists x(Dx \eor Ex)$, $\forall x(Dx \eif Fx)$, \therefore\ $\exists x(Dx \eand Fx)$
\nextSeq%\item $\forall x\forall y(Rxy \eor Ryx)$, \therefore\ $Rjj$
\noSeq%\item $\exists x\exists y(Rxy \eor Ryx)$, \therefore\ $Rjj$
\noSeq%\item $\forall x Px \eif \forall x Qx$, $\exists x \enot Px$, \therefore\ $\exists x \enot Qx$
\lastSeq%\item $\exists x Mx \eif \exists x Nx$, $\enot \exists x Nx$, \therefore\ $\forall x \enot Mx$
are valid. Here are complete answers for some of them:
\begin{earg}
\item %$\forall x\exists y Rxy$, \therefore\ $\exists y\forall x Rxy$
	\begin{solutioninlist}
	\begin{partialmodel}
		UD & \{mocha, freddo\}\\
		\extension{R} & \{$<$mocha, freddo$>$, $<$freddo, mocha$>$\}
	\end{partialmodel}
	\end{solutioninlist}
\item %$\exists y\forall x Rxy$, \therefore\ $\forall x\exists y Rxy$
	\begin{solutioninlist}
	\begin{proof}
		\hypo{p}{\exists y\forall x Rxy} \by{want $\forall x\exists y Rxy$}{}
		\open
			\hypo{ass}{\forall x Rxa}
			\have{R}{Rba} \Ae{ass}
			\have{ER}{\exists y Rby} \Ei{R}
			\have{AER}{\forall x \exists y Rxy} \Ai{ER}
		\close
		\have{c}{\forall x\exists y Rxy} \Ee{p, ass-AER}
	\end{proof}
	\end{solutioninlist}
\end{earg}
