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Can someone please check whether my solutions are okay? I felt a little confused with some.

If Mr. Jones is happy, Mrs. Jones is not happy, and if Mr. Jones is not happy, Mrs. Jones is not happy.

$p$: Mr. Jones is happy.

$q$: Mrs. Jones is happy.

$(p\to \neg q)\wedge (\neg p \to \neg q)$

Either Sam will come to the party and Max will not, or Sam will not come to the party and Max will enjoy himself.

$p$: Sam will come to the party.

$q$: Max will come to the party.

$r$: Max will enjoy himself.

$(p \vee \neg q) \vee (\neg p \wedge r)$

(I do not know what it means by "and Max will not" so I assumed statement $q$.)

A sufficient condition for $x$ to be odd is that $x$ is prime.

$p$: $x$ is odd.

$q$: $x$ is prime.

$q\to p$

A necessary condition for a sequence $s$ to converge is that $s$ be bounded.

$p$: The sequence $s$ converges.

$q$: The sequence $s$ is bounded.

$p\to q$

A necessary and sufficient condition for the sheikh to be happy is that he has wine, women, and song.

$p$: The sheikh is happy.

$q$: The sheikh has wine, women, and song.

$p\leftrightarrow q$

Karpov will win the chess tournament unless Kasparov wins today.

$p$: Karpov will win the chess tournament.

$q$: Kasparov wins today.

$p\leftrightarrow \neg q$

(I do not understand the logic for "unless".)

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Generally your statements are OK, there are just few errors.

Either Sam will come to the party and Max will not, or Sam will not come to the party and Max will enjoy himself.

$(p \land \lnot q) \lor (\lnot p \land r)$, and not $(p \lor \lnot q) \lor (\lnot p \land r)$ (I guess you just made a typo), where:

$p$: Sam will come to the party;

$q$: Max will come to the party;

$r$: Max will enjoy himself.

A necessary and sufficient condition for the sheikh to be happy is that he has wine, women, and song.

$p \leftrightarrow (q \land r \land s)$, where:

$p$: The sheikh is happy;

$q$: The sheikh has wine;

$r$: The sheikh has women;

$s$: The sheikh has song.

Karpov will win the chess tournament unless Kasparov wins today.

$\lnot q \to p$, or equivalently $p \lor q$, and not $p \leftrightarrow \lnot q$, where:

$p$: Karpov will win the chess tournament;

$q$: Kasparov wins today.

Someones prefer to translate "unless" by $\leftrightarrow$, but I prefer the translation given here.

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  • $\begingroup$ Thank you so much for taking the time in correcting my mistakes and letting me know what I did wrong. Some of them I definitely guessed! $\endgroup$ – numericalorange Jun 10 '18 at 0:28

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