Save Dr. Lucky is a new Cheapass game which is a wonderfully clever meta sequel to Kill Dr. Lucky. The mechanics are essentially identical, except you're on the sinking Titanic and you want to save instead of kill Dr. Lucky. To save him you must be seen by another player (as opposed to Kill Dr. Lucky where to kill him you must not be seen by another player). The game's pace is faster than Kill Dr. Lucky since the ship is sinking, which actually destroys successive levels of the board, eventually concentrating all players on the same deck, or actually sinking the Titanic, and so gameplay can't go on indefinitely!

The 10 F-2 failure cards each contain part of a logic puzzle. Here are the clues:

1. Aboard the Ship were four couples: the Andersons, the Browns, the Carvers, and the Dawsons.
2. The husbands' names, in no particular order, were Frank, Gary, Henry, and Isaac.
3. The wives' names, in no particular order, were Selma, Thelma, Ursula, and Victoria.
4. At 1:15 AM on the fateful day, Frank was jolted awake by the Browns, arriving home drunk to their stateroom.
5. At 4:15 AM, Thelma left her stateroom and descended three decks for a secret rendezvous with Mr. Dawson, whom her husband never met.
6. At 8:17 AM, over breakfast in the dining room, Isaac informed his wife that he was not on speaking terms with the Carvers.
7. At 9:25 PM, Henry and Frank joined the Andersons in the Observation Lounge for a friendly game of Bridge.
8. At 11:40 PM, three women heard a terrible scraping sound from their neighboring staterooms. They were Selma, Isaac's wife, and Mrs. Carver.
9. At 12:15 AM on the morning after the incident, Ursula and Mrs. Dawson were the last to board Lifeboat number five.
10. At 2:24 AM, as the stern slipped effortlessly in to the icy sea, Victoria's husband remarked aloud that he had no patience for games of chance.

We have 8 entities: F, G, H, I and S, T, U, V. There are 4 properties A, B, C, D. I write A(F) to mean Frank is an Anderson. I write m(F,S) to mean Frank and Selma are married.

Here's what we know from each rule in isolation:
4: not B(F)
5: not D(T)
6: not C(I)
7: not A(H) and not A(F)
8: not m(S,I) and not C(S) and not C(I)
9: not D(U)

In isolation, clue 6 tells us nothing clue 8 doesn't also tell us. But clue 6 is important since it interacts with clue 7: Isaac wouldn't play bridge with Mr. Carver if he's not on speaking terms with the Carvers, so if Isaac played bridge, Henry and Frank are not Carvers:
6&7: A(I) implies not C(H) and not C(F)
Equivalently, if Henry or Frank is a Carver, then Isaac did not play bridge:
6&7: C(H) or C(F) implies not A(I)

We deduce from clue 10 that Victoria's husband would not play bridge, so
7&10: not m(V,F) and not m(V,H) and not A(V)

We also know from clue 5 that Thelma's husband never met Mr. Dawson, therefore it's not possible that the 2 men played bridge (clue 7). So if Thelma's husband was a bridge player, Dawson was not:
5&7: m(T,H) implies not D(F)
5&7: m(T,F) implies not D(H)
5&7: A(T) implies not D(H) and not D(F)
Equivalently, If Dawson was a bridge player, Thelma's husband was not:
5&7: D(F) implies not m(T,H) and not A(T)
5&7: D(H) implies not m(T,F) and not A(T)

These puzzles get a little annoying with things like clues 5 & 6; can we assume that Isaac wouldn't talk about whether he's on speaking terms with the Carvers if he hadn't met Mr. Carver? I'll assume so:
5&6: D(I) implies not C(T)
Equivalently,
5&6: C(T) implies not D(I)
As it turns out, I don't actually use this in what follows...

Similarly, when Frank was jolted awake by the Browns, is he considered to have met them?

Further nitpicking: the other cards are careful to word things so that seemingly different entities are in fact different (e.g. "three women" on clue 8), but clue 9 could allow Ursula and Mrs. Dawson to be the same person if one were inclined to stretch the conventions of English to the breaking point; the card could have said "...were the last two to board...".

Here's a table showing what we know (X means an impossible combination; the subscript tells the rule(s) implying it). (The table can't show the implications like 6&7.)

	Frank	Gary	Henry	Isaac	Anderson	Brown	Carver	Dawson
Selma				X8			X8
Thelma								X5
Ursula								X9
Victoria	X7&10		X7&10		X7&10
Anderson	X7		X7
Brown	X4
Carver				X6,8
Dawson

Let's assume A(I) and see what happens. This tells us not A(G) and not B(I) and not D(I). Using 6&7 we know not C(F) and not C(H). That tells us D(F), which tells us not D(G) and not D(H), which tells us B(H), which tells us not B(G), which tells us C(G). This completes the lower left portion, telling us all the mens' surnames:

	Frank	Gary	Henry	Isaac	Anderson	Brown	Carver	Dawson
Selma				X8			X8
Thelma								X5
Ursula								X9
Victoria	X7&10		X7&10		X7&10
Anderson	X7	XA(I)	X7	YESassume
Brown	X4	XB(H)	YES	XA(I)
Carver	X6&7	YES	X6&7	X6,8
Dawson	YES	XD(F)	XD(F)	XA(I)

Using 5&7 and D(F), we know not m(T,H) and not A(T). A(I), therefore not m(I,T) and not m(T,V) (since in these puzzles married couples have the same last name). Thus m(I,U) and m(G,V), so not m(H,U) and not m(G,U) and not m(G,S) and not m(G,T). Also by the marriage last name rule, Frank Dawson can't be married to Thelma or Ursula, so m(F,S). But now we reach a contradiction: Henry can only be married to Selma, but she's already married to Frank. And poor Thelma isn't married to anyone. Therefore our initial assumption A(I) has led to a contradiction, and so it must be that not A(I). Here's the table leading to the contradiction:

	Frank	Gary	Henry	Isaac	Anderson	Brown	Carver	Dawson
Selma	YES	Xm(G,V)		X8			X8
Thelma	XD	Xm(G,V)	X5&7	XA	X5&7			X5
Ursula	XD	Xm(I,U)	Xm(I,U)	YES				X9
Victoria	X7&10	YES	X7&10	XA	X7&10
Anderson	X7	XA(I)	X7	YESassume
Brown	X4	XB(H)	YES	XA(I)
Carver	X6&7	YES	X6&7	X6,8
Dawson	YES	XD(F)	XD(F)	XA(I)

So let's backtrack now that we know not A(I). That implies A(G), which implies not B(G) and not C(G) and not D(G). Now married couples have the same last name in these puzzles, so A(G) and not A(V) imply not m(G,V). Thus m(I,V) and not m(I,T) and not m(I,U). So we have:

	Frank	Gary	Henry	Isaac	Anderson	Brown	Carver	Dawson
Selma				X8			X8
Thelma				Xm(I,V)				X5
Ursula				Xm(I,V)				X9
Victoria	X7&10	XA	X7&10	YES	X7&10
Anderson	X7	YES	X7	Xproved
Brown	X4	XA(G)
Carver		XA(G)		X6,8
Dawson		XA(G)

Now let's assume m(T,H) since we'll be able to use 5&7 and this gives us lots of Xs in the upper left: not m(H,S) and not m(H,U) and not m(F,T) and not m(G,T). 5&7 tells us not D(F). That gives us C(F), which gives us not C(H). Thus we have:

	Frank	Gary	Henry	Isaac	Anderson	Brown	Carver	Dawson
Selma			Xm(T,H)	X8			X8
Thelma	Xm(T,H)	Xm(T,H)	YESassume	Xm(I,V)				X5
Ursula			Xm(T,H)	Xm(I,V)				X9
Victoria	X7&10	XA	X7&10	YES	X7&10
Anderson	X7	YES	X7	Xproved
Brown	X4	XA(G)
Carver	YES	XA(G)	XC(F)	X6,8
Dawson	X5&7	XA(G)

Now again use that married people have the same last name. Frank Carver can't be married to Selma, so m(F,U), which tells us not m(G,U), thus m(G,S). This lets us fill in the upper right (women's last names). Frank and Ursula Carver give us C(U) and thus not A(U) and not B(U) and not C(T) and not C(V). Gary and Selma Anderson give us A(S) and thus not B(S) and not D(S) and not A(T). Then D(V) is the only choice left in the D column, so not B(V), so it must be B(T). From there, we see it must be Henry and Thelma Brown, and Isaac and Victoria Dawson:

	Frank	Gary	Henry	Isaac	Anderson	Brown	Carver	Dawson
Selma	XC	YES	Xm(T,H)	X8	YESA	XA(S)	X8	XA(S)
Thelma	Xm(T,H)	Xm(T,H)	YESassume	Xm(I,V)	XA(S)	YES	XC(U)	X5
Ursula	YES	Xm(F,U)	Xm(T,H)	Xm(I,V)	XC(U)	XC(U)	YESC	X9
Victoria	X7&10	XA	X7&10	YES	X7&10	XD(V)	XC(U)	YES
Anderson	X7	YES	X7	Xproved
Brown	X4	XA(G)	YES	X
Carver	YES	XA(G)	XC(F)	X6,8
Dawson	X5&7	XA(G)	X	YES

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