In each of the following we give an English sentence and a number of candidate logical expressions. For each of the logical expressions, state whether it (1) correctly expresses the English sentence; (2) is syntactically invalid and therefore meaningless; or (3) is syntactically valid but does not express the meaning of the English sentence.
a. Every cat loves its mother or father.
(i) $\forall x \operatorname{Cat}(x) \Rightarrow \operatorname{Loves}(x, \operatorname{Mother}(x) \vee$ Father $(x))$.
(ii) $\forall x \neg \operatorname{Cat}(x) \vee \operatorname{Loves}(x, \operatorname{Mother}(x)) \vee \operatorname{Loves}(x$, Father $(x))$.
(iii) $\forall x \operatorname{Cat}(x) \wedge(\operatorname{Loves}(x$, Mother $(x)) \vee$ Loves $(x$, Father $(x)))$.
b. Every dog who loves one of its brothers is happy.
(i) $\forall x \operatorname{Dog}(x) \wedge(\exists y$ Brother $(y, x) \wedge$ Loves $(x, y)) \Rightarrow \operatorname{Happy}(x)$.
(ii) $\forall x, y \operatorname{Dog}(x) \wedge B$ Brother $(y, x) \wedge \operatorname{Loves}(x, y) \Rightarrow \operatorname{Happy}(x)$.
(iii) $\forall x \operatorname{Dog}(x) \wedge[\forall y$ Brother $(y, x) \Leftrightarrow \operatorname{Loves}(x, y)] \Rightarrow \operatorname{Happy}(x)$.
c. No dog bites a child of its owner.
(i) $\forall x \operatorname{Dog}(x) \Rightarrow \neg$ Bites $(x$, Child $($ Owner $(x)))$.
(ii) $\neg \exists x, y \operatorname{Dog}(x) \wedge \operatorname{Child}(y$, Owner $(x)) \wedge \operatorname{Bites}(x, y)$.
(iii) $\forall x \operatorname{Dog}(x) \Rightarrow(\forall y \operatorname{Child}(y$, Owner $(x)) \Rightarrow \neg$ Bites $(x, y))$.
(iv) $\neg \exists x \operatorname{Dog}(x) \Rightarrow(\exists y$ Child $(y$, Ouner $(x)) \wedge \operatorname{Bites}(x, y))$.
d. Everyone's zip code within a state has the same first digit.
(i) $\forall x, s, z_1\left[\operatorname{State}(s) \wedge \operatorname{LivesIn}(x, s) \wedge Z i p(x)=z_1\right] \Rightarrow$ $\left[\forall y, z_2 \operatorname{LivesIn}(y, s) \wedge \operatorname{Zip}(y)=z_2 \Rightarrow \operatorname{Digit}\left(1, z_1\right)=\operatorname{Digit}\left(1, z_2\right)\right]$.
(ii) $\forall x, s\left[\right.$ State $\left.(s) \wedge \operatorname{LivesIn}(x, s) \wedge \exists z_1 \quad Z i p(x)=z_1\right] \Rightarrow$ $\left[\forall y, z_2 \operatorname{LivesIn}(y, s) \wedge \operatorname{Zip}(y)=z_2 \wedge \operatorname{Digit}\left(1, z_1\right)=\operatorname{Digit}\left(1, z_2\right)\right]$.
(iii) $\forall x, y, s \operatorname{State}(s) \wedge \operatorname{LivesIn}(x, s) \wedge \operatorname{LivesIn}(y, s) \Rightarrow \operatorname{Digit}(1, Z i p(x)=Z i p(y))$.
(iv) $\forall x, y, s \operatorname{State}(s) \wedge \operatorname{LivesIn}(x, s) \wedge \operatorname{LivesIn}(y, s) \Rightarrow$ $\operatorname{Digit}(1, \operatorname{Zip}(x))=\operatorname{Digit}(1, \operatorname{Zip}(y))$.