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Froggy's Problem was presented in Lewis Carroll's Symbolic Logic, with a note that it "contains a marvellous trap". Because Carroll never released Part 2 of Symbolic Logic, its solution is unknown.

Achieve the strongest conclusion from these premises:

  1. When the day is fine, I tell Froggy “You’re quite the dandy, old chap!”

  2. Whenever I let Froggy forget that 10 pounds he owes me, and he begins to strut about like a peacock, his mother declares “He shall not go out a-wooing!”

  3. Now that Froggy’s hair is out of curl, he has put away his gorgeous waistcoat.

  4. Whenever I go out on the roof to enjoy a quiet cigar, I’m sure to discover that my purse is empty.

  5. When my tailor calls with his little bill, and I remind Froggy of that 10 pounds he owes me, he does not grin like a hyena.

  6. When it is very hot, the thermometer is high.

  7. When the day is fine, and I’m not in the humor for a cigar, and Froggy is grinning like a hyena, I never venture to hint that he’s quite the dandy.

  8. When my tailor calls with his little bill and finds me with an empty pocket, I remind Froggy of that 10 pounds he owes me.

  9. My railway shares are going up like anything!

  10. When my purse is empty, and when, noticing that Froggy has got his gorgeous waistcoat on, I venture to remind him of that 10 pounds he owes me, things are apt to get rather warm.

  11. Now that it looks like rain, and Froggy is grinning like a hyena, I can do without my cigar.

  12. When the thermometer is high, you need not trouble yourself to take an umbrella.

  13. When Froggy has his gorgeous waistcoat on, but is not strutting about like a peacock, I betake myself to a quiet cigar.

  14. When I tell Froggy that he’s quite a dandy, he grins like a hyena.

  15. When my purse is tolerably full, and Froggy’s hair is one mass of curls, and when he is not strutting about like a peacock, I go out on the roof.

  16. When my railways shares are going up, and when it’s chilly and looks like rain, I have a quiet cigar.

  17. When Froggy’s mother lets him go a-wooing, he seems nearly mad with joy, and puts on a waistcoat that is gorgeous beyond words.

  18. When it is going to rain, and I am having a quiet cigar, and Froggy is not intending to go a-wooing, you had better take an umbrella.

  19. When my railway shares are going up, and Froggy seems nearly mad with joy, that is the time my tailor always chooses for calling with his little bill.

  20. When the day is cool and the thermometer low, and I say nothing to Froggy about his being quite the dandy, and there’s not the ghost of a grin on his face, I haven’t the heart for my cigar!

Beyond this, it is only known that Carroll created a "problem dictionary" which was found in unpublished notes:

Universe: "Cosmophases";
*eta* = this;
a = Froggy's hair is out of curl;
b = Froggy intends to go a-wooing;
c = Froggy is grinning like a hyena;
d = Froggy's mother permits him to go a-wooing;
e = Froggy seems nearly mad with joy;
h = Froggy is strutting about like a peacock;
k = Froggy is wearing a waistcoat that is gorgeous beyond words;
l = I go out on my roof;
m = I remind Froggy of the 10 pounds he owes me;
n = I take a quiet cigar;
r = I tell Froggy that he's quite the dandy;
s = It is going to rain,
t = It is very hot;
v = My purse is empty;
w = My railway shares are going up;
z = My tailor calls with his little bill;
A = The thermometer is high;
B = You had better take an umbrella.

Attempting to write the problem in Carroll's terms using the dictionary yields:

1. All !s are r.
2. All !mh are !d.
3. Eta is a!k.
4. All ln are v.
5. All zm are !c.
6. All t are A.
7. All !s!nc are !r.
8. All zv are m.
9. Eta is w.
10. All vkm are t (?)
11. Eta is sc!n.
12. All A are !B.
13. All k!h are n.
14. All r are c.
15. All !v!a!h are l.
16. All w!ts are n.   
17. All d are ek.
18. All sn!b are B.
19. All we are z.
20. All !t!A!r!c are !n.

There is a proposed solution by Hawker which gives the conclusion "Froggy does not have permission to go out a-wooing", but little else.

Part of the problem (which the dictionary might resolve) is the unclear terms used (do chilly and cool mean the same thing?) and the scope of the problem over items and time. Since Lewis Carroll also produced the Raw Meat problem and the Jack Sprat problem, both of these could well be involved.

My personal incomplete analysis runs as follows:

  • 3, 9, and 11 are the only statements that give facts rather than implications.
  • 1 and 7 are redundant on current information because, per 11, the day is not fine.
  • 10 and 13 are redundant on current information because, per 3, Froggy is not wearing his waistcoat.
  • 15 is redundant on current information because, per 3, Froggy's hair is out of curl.
  • 4 and 18 are redundant on current information because, per 11, "I" am not having a cigar.
  • 20 is unnecessary because it is already established that "I" am not having a cigar.

Hawker's solution is found in a single inference:

  • 17 modus tollens 3: Froggy's mother does not currently let him go a-wooing.
  • 2 is now redundant because it is established Froggy's mother does not let him go a-wooing.

Inferences on a number of the remaining propositions simply result in a proposition with no further development:

  • 16 modus tollens 11, or-elim 11: it is not currently chilly. (21)
  • 19 and-elim 9: if Froggy is now mad with joy, my tailor will call with his little bill. (22)
  • 5 modus tollens 11: either my tailor is not calling with his little bill, or I did not remind Froggy of the 10 pounds he owes me. (23)
  • 23 modus tollens 22, either Froggy is not now mad with joy or I did not remind him of the 10 pounds he owes me. (24)
  • 8 modus ponens 22: If Froggy is mad with joy and my pocket is empty, I remind Froggy of the 10 pounds he owes me. (25)
  • 24 modus tollens 25: either Froggy is not now mad with joy or froggy is not mad with joy or my pocket is empty. (26)
  • 26 deduplication: either Froggy is not now mad with joy or my pocket is empty.

The only remaining propositions are 6, 12 and 14 which both have preconditions that cannot be established. If we make the decidedly iffy judgement that "it is not chilly" is equivalent to "it is very hot", then 21 modus ponens 6 gives us that the thermometer is high; modus ponens 12 gives us that we need not take an umbrella; but modus tollens 18 gives us only that "it is not going to rain, or I am not having a quiet cigar, or Froggy is intending to go a-wooing" which just confirms 11 (I am not having a cigar)."

14 is going nowhere, since "telling Froggy that he is quite the dandy" is mentioned only in 1 and 7. For 1, the day is known not to be fine, but we have no grounds for inferring that this means we do not tell Froggy he is a dandy; for 7, the day is also known not to be fine, but we have no ground for inferring that this means we do tell Froggy this.

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  • $\begingroup$ On second thought, there may be an issue with 17. 18 introduces "Froggy is intending to go a-wooing" as a proposition and the dictionary confirms this. It seems reasonable that Froggy would not ask his mother's permission if he wasn't intending to go out, nor would he dress up having been given permission if he wasn't going to use it. Also, 3 Froggy "putting away his waistcoat" may indicate his intention, not his state of permission. This would obstruct Hawker's solution but I don't know if it leads anywhere useful. $\endgroup$ – Mark Green Sep 9 '17 at 13:39
  • $\begingroup$ There's also a problem between 1 and 7. 1 implies that I always tell Froggy he is quite the dandy when the day is fine, but 7 states that there are some extra circumstances where I don't. If we treat this as an exception then by De Morgan 1 can become "When the day is fine and Froggy is not grinning like a hyena, I tell him.." but this then contradicts 14. This may either be time based or a Barbershop Paradox. $\endgroup$ – Mark Green Sep 10 '17 at 14:52
  • $\begingroup$ I have no authority to declare this the correct answer; only Carroll could have done that. Also, the answer given below was actually already present in the question analysis text ("modus ponens 12 gives us that we need not take an umbrella"). While it is true that this fact is logically compatible with the propositions of the puzzle, the large number of redundant propositions make it uncertain that it is Carroll's complete conclusion, especially since at least some of Carroll's problems (Jack Sprat/Raw Meat) require non-logical steps to complete. $\endgroup$ – Mark Green Nov 10 '17 at 19:24
  • $\begingroup$ If a logic problem requires non-logical steps to complete there is no point to this. My suggestion is for the reader to decide based upon all that is presented here. $\endgroup$ – Bob Bixler Nov 11 '17 at 15:23
  • $\begingroup$ “Non logical” was the wrong phrase, true. But those problems require more than applying predicate logic. Proving a correct answer would likely require some historical analysis. Do any of Carroll’s other sorites problems have that much redundancy?Still, even if I was 100% confident in your answer, it would only be my opinion. I’m just a random guy who posted this because I thought it might be interesting, I have no particular qualification to declare what is a final correct answer to an open historical problem. $\endgroup$ – Mark Green Nov 11 '17 at 16:02
11
$\begingroup$

$9 \implies w$

$11 \implies s $

$11 \implies \neg n$

$16 \implies (w \space \wedge \space \neg t \space \wedge \space s \implies n) $

$ \neg n \implies \neg(w \wedge \neg t \wedge s) \implies \neg \neg t \implies t$

$6 \wedge t \implies A$

$12 \wedge A \implies \neg B $

Answer:

You do not need to take an umbrella. (The trap is statement 11 that says it now looks like rain.)

(Note: Numbers refer to problem statement number.

Letters are as described by Carroll and above in original post.
$\neg $ means not or false.
Arrow ($\implies$) means implies or therefore.
$\wedge$ means "and".

The above solution can be proven by a computer simulation that tests all combinations of true and false for each premise or condition and applying each combination to all problem statements. If all problem statements for the given combination are true then the true/false condition of each premise is noted. The end result will show that some conditions are always true and some are always false. This end result shows that conditions d,k,n, and B(bb) are always false. Conditions a,c,s,t,w, and A(aa) are always true. Here's that program in Power Basic:

FUNCTION PBMAIN
DEFINT a-z
FOR a=1 TO 1 'eq 3
FOR b=0 TO 1
FOR c=1 TO 1 'eq 11
FOR d=0 TO 1
FOR e=0 TO 1
FOR h=0 TO 1
FOR k=0 TO 0 'eq 3
FOR l=0 TO 1
FOR m=0 TO 1
FOR n=0 TO 0 'eq 11
FOR r=0 TO 1
FOR s=1 TO 1 'eq 11
FOR t=0 TO 1
FOR v=0 TO 1
FOR w=1 TO 1 'eq 9
FOR z=0 TO 1
FOR aa=0 TO 1
FOR bb=0 TO 1
'********************************************************
 IF s=0 AND r=0 THEN bad=1 'eq 1
 IF m=0 AND h=1 AND d=1 THEN bad=1  'eq 2
 IF l=1 AND n=1 AND v=0 THEN bad=1 'eq 4
 IF z=1 AND m=1 AND c=1 THEN bad=1   'eq 5  xx
 IF t=1 AND aa=0 THEN bad=1 'eq 6  xxx
 IF s=0 AND n=0 AND c=1 AND r=1 THEN bad=1  'eq 7
 IF z=1 AND v=1 AND m=0 THEN bad=1  'eq 8
 IF v=1 AND k=1 AND m=1 AND t=0 THEN bad=1   'eq 10
 IF aa=1 AND bb=1 THEN bad=1   'eq 12  critical ***** d shows and bb gone when
 removed
 IF k=1 AND h=0 AND n=0 THEN bad=1  'eq 13
 IF r=1 AND c=0 THEN bad=1  'eq 14
 IF v=0 AND a=0 AND h=0 AND l=0 THEN bad=1  'eq 15
 IF w=1 AND t=0 AND s=1 AND n=0 THEN bad=1   'eq 16
 IF d=1 AND (e=0 OR k=0) THEN bad=1  'eq 17  critical ***** bb shows and d gone
when removed
 IF s=1 AND n=1 AND b=0 AND bb=0 THEN bad=1  'eq 18
 IF w=1 AND e=1 AND z=0 THEN bad=1  'eq 19
 IF t=0 AND aa=0 AND r=0 AND c=0 AND n=1 THEN bad=1 'eq 20
'************************************************************
IF bad=0 THEN
 tot=tot+1
 asum=asum+a
 bsum=bsum+b
 csum=csum+c
 dsum=dsum+d
 esum=esum+e
 hsum=hsum+h
 ksum=ksum+k
 lsum=lsum+l
 msum=msum+m
 nsum=nsum+n
 rsum=rsum+r
 ssum=ssum+s
 tsum=tsum+t
 vsum=vsum+v
 wsum=wsum+w
 zsum=zsum+z
 aasum=aasum+aa
 bbsum=bbsum+bb
END IF

bad=0
NEXT bb
NEXT aa
NEXT z
NEXT w
NEXT v
NEXT t
NEXT s
NEXT r
NEXT n
NEXT m
NEXT l
NEXT k
NEXT h
NEXT e
NEXT d
NEXT c
NEXT b
NEXT a
PRINT "total=";tot
PRINT
range=50
IF asum=tot OR asum=0 OR asum>tot-range OR asum<range THEN PRINT "a";asum
IF bsum=tot OR bsum=0 OR bsum>tot-range OR bsum<range THEN PRINT "b";bsum
IF csum=tot OR csum=0 OR csum>tot-range OR csum<range THEN PRINT "c";csum
IF dsum=tot OR dsum=0 OR dsum>tot-range OR dsum<range THEN PRINT "d";dsum
IF esum=tot OR esum=0 OR esum>tot-range OR esum<range THEN PRINT "e";esum
IF hsum=tot OR hsum=0 OR hsum>tot-range OR hsum<range THEN PRINT "h";hsum
IF ksum=tot OR ksum=0 OR ksum>tot-range OR ksum<range THEN PRINT "k";ksum
IF lsum=tot OR lsum=0 OR lsum>tot-range OR lsum<range THEN PRINT "l";lsum
IF msum=tot OR msum=0 OR msum>tot-range OR msum<range THEN PRINT "m";msum
IF nsum=tot OR nsum=0 OR nsum>tot-range OR nsum<range THEN PRINT "n";nsum
IF rsum=tot OR rsum=0 OR rsum>tot-range OR rsum<range THEN PRINT "r";rsum
IF ssum=tot OR ssum=0 OR ssum>tot-range OR ssum<range THEN PRINT "s";ssum
IF tsum=tot OR tsum=0 OR tsum>tot-range OR tsum<range THEN PRINT "t";tsum
IF vsum=tot OR vsum=0 OR vsum>tot-range OR vsum<range THEN PRINT "v";vsum
IF wsum=tot OR wsum=0 OR wsum>tot-range OR wsum<range THEN PRINT "w";wsum
IF zsum=tot OR zsum=0 OR zsum>tot-range OR zsum<range THEN PRINT "z";zsum
IF aasum=tot OR aasum=0 OR aasum>tot-range OR aasum<range THEN PRINT "aa";aasum
IF bbsum=tot OR bbsum=0 OR bbsum>tot-range OR bbsum<range THEN PRINT "bb";bbsum
PRINT
PRINT "finished"
SLEEP  500000000
END FUNCTION
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  • 2
    $\begingroup$ Welcome to puzzling.stackexchange.com! I'm interested in your answer, but unfortunately, it's utterly unintelligible to me. Can you add some explanation of what each line means? $\endgroup$ – phroureo Nov 3 '17 at 15:52
  • $\begingroup$ See my recent edit. $\endgroup$ – Bob Bixler Nov 3 '17 at 16:35
  • $\begingroup$ @BobBixler Welcome! I have updated the symbols to look a bit more beautiful. Hope you are okay with it and it doesn't change the answer and its meaning. Also, please have a look at the tour page $\endgroup$ – ABcDexter Nov 3 '17 at 17:31
  • 1
    $\begingroup$ Fine with me. Bob B. $\endgroup$ – Bob Bixler Nov 3 '17 at 18:00
  • 1
    $\begingroup$ Not necessarily. This problem has superfluous statements right from the start. Statement #11 makes Statements#1, #4, #7, #18 and #20 not relevant to the problem. Statement #3 makes statements #10 and #15 of no relevance. So you can through out 7 statements before you even start to seriously work the problem. It seems that Carroll wasn't interested in making all statements fundamentally important. $\endgroup$ – Bob Bixler Nov 6 '17 at 20:39

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