On the Behavior of True and False
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Abstract
Uzquiano (Analysis 70:39–44, 2010) showed that the Hardest Logic Puzzle Ever (HLPE) [in its amended form due to Rabern and Rabern (Analysis 68:105–112, 2008)] has a solution in only two questions. Uzquiano concludes his paper by noting that his solution strategy naturally suggests a harder variation of the puzzle which, as he remarks, he does not know how to solve in two questions. Wheeler and Barahona (J Philos Logic, to appear, 2011) formulated a three question solution to Uzquiano’s puzzle and gave an information theoretic argument to establish that a two question solution for Uzquiano’s puzzle does not exist. However, their argument crucially relies on a certain conception of what it means to answer self-referential yes–no questions truly and falsely. We propose an alternative such conception which, as we show, allows one to solve Uzquiano’s puzzle in two questions. The solution strategy adopted suggests an even harder variation of Uzquiano’s puzzle which, as we will show, can also be solved in two questions. Just as all previous solutions to versions of HLPE, our solution is presented informally. The second part of the paper investigates the prospects of formally representing solutions to HLPE by exploiting theories of truth.
Keywords
Hardest logic puzzle ever Self-reference TruthIntroduction
Recall Boolos’ formulation of the Hardest Logic Puzzle Ever (HLPE):
Rabern and Rabern (2008) point out the need to distinguish HLPE as literally formulated by Boolos from a version of HLPE which is closely related to it and, as pointed out by Rabern and Rabern, is more properly called ‘the hardest logic puzzle ever’. The distinction between the puzzle as formulated by Boolos—which we call HLPE_{sem}, for semanticHLPE—and the amended puzzle—which we call HLPE_{syn}, for syntacticHLPE—only concerns the way in which Random reacts to questions. Suppose that we address a question to Random. Depending on the version of HLPE under consideration, he reacts as follows:The Puzzle: Three gods A, B and C are called, in some order, True, False, and Random. True always speaks truly, False always speaks falsely, but whether Random speaks truly or falsely is a completely random matter. Your task is to determine the identities of A, B, and C by asking three yes–no questions; each question must be put to exactly one god. The gods understand English, but will answer all questions in their own language, in which the words for ‘yes’ and ‘no’ are ‘da’ and ‘ja’ in some order. You do not know which word means which. Before I present the somewhat lengthy solution, let me give answers to certain questions about the puzzle that occasionally arise:
(B1) It could be that some god gets asked more than one question (and hence that some god is not asked any question at all).
(B2) What the second question is, and to which god it is put, may depend on the answer to the first question (and of course similarly for the third question).
(B3) Whether Random speaks truly or not should be thought of as depending on the flip of a coin hidden in his brain: if the coin comes down heads, he speaks truly, if tails, falsely.
(B4) Random will answer ‘da’ or ‘ja’ when asked any yes–no question.
Boolos (1996), p. 62
HLPE_{sem}: Random flips a coin and then, depending on the outcome of the coin-flip, answers the question either truly or falsely.
HLPE_{syn}: Random flips a coin and then, depending on the outcome of the coin-flip, answers the question with either ‘da’ or ‘ja’.
Besides pointing out the distinction between HLPE_{sem} and HLPE_{syn}, Rabern and Rabern come up with a solution to HLPE_{sem} which exploits only two (!) questions. To realize their solution, Rabern and Rabern ask the gods self-referential questions, which, as they observe, is not prohibited by Boolos’ guidelines. However, their solution does not carry over to HLPE_{syn} and so the question arises whether HLPE_{syn} allows for a two-question solution as well.
In answering λ with either ‘yes’ or ‘no’, True can be accused of lying and so True cannot answer λ “in accordance with his nature”. Accordingly, True will remain silent when asked λ. This illustrates that in HLPE_{syn}, True and False are thought of as having three reactions to questions; besides answering with ‘da’ and ‘ja’ they may also remain silent. However, HLPE_{syn} models Random as a random variable over only two of these reactions: answering with ‘da’ or answering with ‘ja’. As Uzquiano observes, a more natural way to model Random in HLPE_{syn} then, is as a ternary random variable, the outcome of which determines whether Random answers ‘da’, ‘ja’ or remains silent. HLPE_{syn}^{2} and HLPE_{syn}^{3} will be used to denote the original version and Uzquiano’s version of HLPE_{syn} respectively. Uzquiano solves HLPE_{syn}^{2} in two questions, but with respect to HLPE_{syn}^{3}, he remarks that: ‘I, for one, do not know how to solve this puzzle in two questions.’λ: Is it the case that: your answer to λ is ‘no' ?
Indeed, just as λ is (when asked to True) an interrogative version of the Liar, so τ is (when asked to True) an interrogative version of the Truthteller. And just as the Truthteller may be valuated as either true or false, so True can answer τ with either ‘yes’ or ‘no’. However, doing so is, in both cases, completely arbitrary. Questions like τ do not have a role to play in previous solutions to HLPE, and none of the mentioned papers discusses how True answers such questions. In this paper however, questions like τ will have a crucial role to play: they give rise to a fourth answer. Exploiting a four-valued answering repertoire, we will show how to solve HLPE_{syn}^{3} in two questions.τ: Is it the case that: your answer to τ is ‘yes' ?
There is a clear intuitive sense in which answering λ and τ as such is speaking truly. False will answer the mentioned questions as follows.λ can neither be answered with ‘yes’ nor with ‘no’
τ can be answered both with ‘yes’ and ‘no’
Again, there is a clear intuitive sense in which answering λ and τ as such is speaking falsely. A possible justification of working with a four-valued (in contrast to a three-valued) answering repertoire is that the four (linguistic) answers allow us, even in the presence of self-reference, to respect Boolos’ instructions, which state that ‘True always speaks truly’ and ‘False always speaks falsely’. On the other hand, we can also interpret our two non-standard answers in non-linguistic terms, along the following lines: on questions like λ, the algorithm which describes True’s behavior yields no solutions, while on questions like τ it yields two solutions. In such cases, True does not answer with ‘yes’ or ‘no’, but its two (non-linguistic) answers reflect, respectively, the lack and abundance of solutions. Although we will work with the linguistic version of our non-standard answers, i.e., with ‘both’ and ‘neither’, we will return to the two distinct justifications of a four-valued answering repertoire (cf. section “Critical Remarks on Formalizations”).λ can both be answered with ‘yes’ and with ‘no’
τ can be answered neither with ‘yes’ nor with ‘no’
An answering repertoire of four answers naturally suggests an “even harder” variation of the hardest logic puzzle ever, HLPE_{syn}^{4}, in which Random is modeled as a four-valued random variable over the possible answers. As we will see, HLPE_{syn}^{4} can also be solved in two questions. In fact, we will only show how to solve HLPE_{syn}^{4} in two questions as our solution to HLPE_{syn}^{4} is easily seen to solve HLPE_{syn}^{3} as well. ^{2}
All previous solutions to HLPE are presented (informally) in natural language and our solution to HLPE_{syn}^{4}, as presented in section "Solving the Puzzles", is no exception. However, given the nature of the gods True and False, one would expect that solutions to HLPE allow for a formal representation that is based on a (formal) theory of truth. In section "Formalizations via Theories of Truth", we explore the prospects of such a formal representation, exploiting (Kripkean) fixed point theories of truth. We will see that, using a restricted formal language, the previous solutions to HLPE as well as the solution put forward in section "Solving the Puzzles", can be given a formal representation. The formal representations are illuminative as they clearly lay bare the differences between the previous solutions to HLPE and the present one. Although our formalization allows us to represent the (informal) solutions to HLPE, nevertheless there are some reasons for not being completely satisfied with it, as will be explained in section "Formalizations via Theories of Truth". Section "Formalizations via Theories of Truth" concludes by discussing the information theoretic argument of Wheeler and Barahona (2011), which establishes that (given a three-valued answering repertoire) HLPE_{syn}^{3} cannot be solved in less than three questions. Section "Concluding Remarks" concludes the paper.
Solving the Puzzles
Gods Who Answer with ‘Yes’ and ‘No’
In this section, we solve HLPE_{syn}^{4} under the assumption that the gods speak English: they use ‘yes’ and ‘no’ to answer positively and negatively respectively. In the next section we give up this simplifying assumption and show how to solve HLPE_{syn}^{4} itself, in which the gods answer with ‘da’ and ‘ja’.
We use the following abbreviations. A, B and C will be used as in Boolos’ guidelines and T, F and R will be used to denote True, False and Random respectively. With x an arbitrary question, N(x) reads as ‘your answer to x is ‘no”, while Y(x) reads as ‘your answer to x is ‘yes”. ^{3} Before we state our solution to (the English version of) HLPE_{syn}^{4}, we first briefly comment on the algorithm that gives rise to the answers of True and False. First, True and False calculate how their yes/no answers to a question Q influence the truth-value ^{4} of Q, in light of which they judge these yes/no answers to be correct (✓) or incorrect (X). Exploiting the correctness/incorrectness of their yes–no answers with respect to Q, they then determine which of the four possible answers (‘yes’, ‘no’, ‘both’, ‘neither’) they give to Q. The process is illustrated by the Table 1.
Reactions of True and False
Q(uestion) | Y/N | \(\mathcal{V}(Q)\) | ✓/X | True | False |
---|---|---|---|---|---|
sw: snow is white | Y(sw) | True | ✓ | Yes | No |
N(sw) | True | X | |||
sb: snow is black | Y(sb) | False | X | No | Yes |
N(sb) | False | ✓ | |||
λ: N(λ) | Y(λ) | False | X | Neither | Both |
N(λ) | True | X | |||
τ: Y(τ) | Y(τ) | True | ✓ | Both | Neither |
N(τ) | False | ✓ |
Our two-question solution has the following structure. First, we ask a question which allows us to identify a god which is not Random. Then, we ask a follow up question to the god which we know not to be Random, and use the answer we get to determine the identity of all three gods.
Finding a god that is not Random
Reactions of True and False on α_{1}
World | Y/N | \(\mathcal{V}(\alpha_1)\) | ✓/X | True | False |
---|---|---|---|---|---|
A = R | Y(α_{1}) | False | X | Neither | Both |
N(α_{1}) | True | X | |||
B = R | Y(α_{1}) | True | ✓ | Both | Neither |
N(α_{1}) | False | ✓ | |||
C = R | Y(α_{1}) | True | ✓ | Yes | No |
N(α_{1}) | True | X |
Let’s explain the first two rows. When A is Random and α_{1} is answered with ‘yes’, α_{1} is false—as all its three disjuncts are—and so answering α_{1} with ‘yes’ is incorrect when A is Random. Similarly, when A is Random and α_{1} is answered with ‘no’, α_{1} is true and so when A is Random, answering α_{1} with ‘no’ is incorrect as well. So, when A is Random, True will answer α_{1} with ‘neither’, while False will answer it with ‘both’. The other entries in the table are explained similarly. We address α_{1} to A and extract the following information from his answers.
Conclusions based on A’s answer to α_{1}
A’s answer | Conclusion 1 | Conclusion 2 |
---|---|---|
Yes | (A = T and C = R) or A = R | \(B\not = R\) |
No | (A = F and C = R) or A = R | \(B\not = R\) |
Neither | (A = R and A = T) or (A = F and B = R) or A = R | \(C\not = R\) |
Both | (A = R and A = F) or (A = T and B = R) or A = R | \(C\not = R\) |
Determining the identity ofA, BandCby a follow up question
Reactions of True and False on α_{2}
World | Y/N | \(\mathcal{V}(\alpha_2)\) | Status Y/N | True | False |
---|---|---|---|---|---|
p_{1} | Y(α_{2}) | False | X | Neither | Both |
N(α_{2}) | True | X | |||
p_{2} | Y(α_{2}) | True | ✓ | Both | Neither |
N(α_{2}) | False | ✓ | |||
p_{3} | Y(α_{2}) | True | ✓ | Yes | No |
N(α_{2}) | True | X | |||
p_{4} | Y(α_{2}) | False | X | No | Yes |
N(α_{2}) | False | ✓ |
Conclusions based on B’s answer to α_{2}
B’s answer | Conclusion 1 | Conclusion 2 |
---|---|---|
Yes | (B = T and p_{3}) or (B = F and p_{4}) | p_{4} |
No | (B = T and p_{4}) or (B = F and p_{3}) | p_{3} |
Neither | (B = T and p_{1}) or (B = F and p_{2}) | p_{1} |
Both | (B = F and p_{2}) or (B = F and p_{1}) | p_{2} |
Gods Who Answer with ‘da’ and ‘ja’
We will now solve HLPE_{syn}^{4}, in which the gods answer positively and negatively by using, in some order, the words ‘da’ and ‘ja’. The methods of the previous section easily carry over to this slightly more complicated puzzle. Let M(d, y) and M(d, n) abbreviate “da’ means ‘yes” and “da’ means ‘no” respectively. Further, with x an arbitrary question, D(x) reads as ‘your answer to x is ‘da”, while J(x) reads as ‘your answer to x is ‘ja”.
Finding a god that is not Random
Reactions of True and False on β_{1}
World | Language | D/J | \(\mathcal{V}(\beta_1)\) | ✓/X | True | False |
---|---|---|---|---|---|---|
A = R | M(d, y) | D(β_{1}) | True | ✓ | Both | Neither |
J(β_{1}) | False | ✓ | ||||
B = R | M(d, y) | D(β_{1}) | False | X | Neither | Both |
J(β_{1}) | True | X | ||||
C = R | M(d, y) | D(β_{1}) | True | ✓ | da | ja |
J(β_{1}) | True | X | ||||
A = R | M(d, n) | D(β_{1}) | False | ✓ | Both | Neither |
J(β_{1}) | True | ✓ | ||||
B = R | M(d, n) | D(β_{1}) | True | X | Neither | Both |
J(β_{1}) | False | X | ||||
C = R | M(d, n) | D(β_{1}) | False | ✓ | da | ja |
J(β_{1}) | False | X |
Determining the identity ofA, BandCby a follow up question
Reactions of True and False to β_{2}
World | Language | D/J | \(\mathcal{V}(\beta_2)\) | ✓/X | True | False |
---|---|---|---|---|---|---|
p_{1} | M(d, y) | D(β_{2}) | True | ✓ | Both | Neither |
J(β_{2}) | False | ✓ | ||||
p_{2} | M(d, y) | D(β_{2}) | False | X | Neither | Both |
J(β_{2}) | True | X | ||||
p_{3} | M(d, y) | D(β_{2}) | True | ✓ | da | ja |
J(β_{2}) | True | X | ||||
p_{4} | M(d, y) | D(β_{2}) | False | X | ja | da |
J(β_{2}) | False | ✓ | ||||
p_{1} | M(d, n) | D(β_{2}) | False | ✓ | Both | Neither |
J(β_{2}) | True | ✓ | ||||
p_{2} | M(d, n) | D(β_{2}) | True | X | Neither | Both |
J(β_{2}) | False | X | ||||
p_{3} | M(d, n) | D(β_{2}) | False | ✓ | da | ja |
J(β_{2}) | False | X | ||||
p_{4} | M(d, n) | D(β_{2}) | True | X | ja | da |
J(β_{2}) | True | ✓ |
Formalizations via Theories of Truth
As noted in the introduction, all the previous solutions to HLPE are presented informally using natural language. In the previous section, we likewise introduced our four-valued conception of True and False informally by showing how it can be applied to solve HLPE_{syn}^{3}. In this section, we discuss the prospects of formally representing the present and previous solutions to HLPE. The behavior of the gods True and False in HLPE suggests that a formalization of their behavior can fruitfully be based upon a formal theory of truth. In this section, we follow this suggestion by basing ourselves upon Strong Kleene (Kripkean) fixed point theories of truth. To be sure, there are various theories of truth; we could also work with an account of True and False that is based on say, a revision theory of truth (cf. Gupta and Belnap 1993) or on fixed points that are constructed in accordance with the Supervaluation schema. We choose to work with Strong Kleene theories because such theories are very well-known, easy to present and, importantly, they allow us to represent the solutions to HLPE in a sense that will be made clear below.^{5}
In fact, we will not apply our formal modeling to HLPE itself, but rather to the four roads riddle, presented in section "The Four Roads Riddle". The four roads riddle may be considered as a simplified version of HLPE while containing HLPE’s essential features: our formalization of the four roads riddle is easily seen to carry over to (versions of) HLPE. The formal language in which we will study the four roads riddle contains a ‘yes’ and a ‘no’ predicate, but no “non-standard” answer predicates, such as predicates for ‘both’, ‘neither’, ‘silence’ or what have you. As none of the solutions to HLPE involves questions that are formed using non-standard answer predicates, the expressive limitations of our language do not prevent us from representing these solutions. To be sure, ultimately one wants an account of the behavior of True and False in a more expressive language which does contain non-standard answer predicates. In section “Critical Remarks on Formalizations”, we will briefly comment on the prospects of such an account.
After presenting the four roads riddle in sections "The Four Roads Riddle", and "Formalizations" is concerned with formalizations of the riddle. Section "Critical Remarks on Formalizations" critically looks back at what has been achieved in section "Formalizations". Section "The Wheeler and Barahona Argument" discusses the information theoretic argument of Wheeler and Barahona (2011) that was mentioned in the introduction.
The Four Roads Riddle
The Riddle
You arrive at a cross roads at which you can head either north, south, east or west. You know that only one of the four roads, call it the good road, leads to your destination. Unfortunately, you have no clue as to which road is good. However, two gods, call them a and b, are situated at the cross roads. You know that one of these gods is True while the other god is False, but you have no clue as to whether a or b is True. The four roads riddle is as follows. Given the circumstances just sketched, can you come up with a single question that, when posed to either one of the gods, allows you to determine which road is good?
The Language L_{B} and its Ground Models
We start out by introducing a restricted formal language in which we will study the four roads riddle. Our basic formal language is a quantifier free^{6} predicate language with identity L_{B}, consisting of the following non-logical vocabulary.^{7}
a and b, which denote, in some order, True and False.
g_{T} and g_{F}, which denote, respectively, True and False.
n, w, e, s, which denote, respectively, the north, west, east and south road.
\(\{ [\sigma] \mid \sigma \in Sen(L_B) \}:\)quotational constant symbols;^{8} for each \(\sigma \in Sen(L_B) , [\sigma]\) denotes σ.
\(C = \{ c_1, c_2, \ldots, c_n\}:\)non-quotational constant symbols, which denote (arbitrary) elements of Sen(L_{B}) and which can be used to define self-referential sentences. ^{9}
G(x), interpreted as ‘x is the good road’.
Y(x, y) and N(x, y), interpreted as ‘the answer of x to y is ‘yes” and ‘the answer of x to y is ‘no” respectively.
- 1.
D = {Tr, Fa, no, ea, so, we} ∪ Sen(L_{B})
- 2.
I(g_{T}) = Tr, I(g_{F}) = Fa, I(n) = no, I(e) = ea, I(w) = we, I(s) = so
- 3.
I([σ]) = σ for all \(\sigma \in Sen(L_B), I(c_i) \in Sen(L_B)\) for all \(c_i \in C\)
- 4.
Either (I(a) = Tr and I(b) = Fa) or (I(b) = Tr and I(a) = Fa)
- 5.
Either I(G) = {no} or I(G) = {ea} or I(G) = {so} or I(G) = {we}.
Formalizations
A Three-Valued Answering Function for L_{B}
- A
(The construction of) the answering function allows us to represent the previous (three-valued) solutions to HLPE.
- B
The answering function gives the intuitive correct verdict with respect to L_{B} questions that are not considered in those solutions.
- 1.
\(\mathcal{K}_M(\sigma) = \mathcal{C}_M(\sigma)\) for all \(\sigma \in Sen(L_B^-) \)
\( \mathcal{K}_M\ \)respects the ground modelM.
- 2.
\(\mathcal{K}_M(Y(g_T, \overline{\sigma})) = \mathcal{K}_M(\sigma), \mathcal{K}_M(N(g_T, \overline{\sigma})) = 1 - \mathcal{K}_M(\sigma)\)Fixed point condition for\(Y(g_T, \cdot)\)and\(N(g_T, \cdot).\)
- 3.
\(\mathcal{K}_M(Y(g_F, \overline{\sigma})) = 1 - \mathcal{K}_M(\sigma), \mathcal{K}_M(N(g_F, \overline{\sigma})) = \mathcal{K}_M(\sigma)\)Fixed point condition for\(Y(g_F, \cdot)\)and\(N(g_F, \cdot).\)
- 4.
\(\mathcal{K}_M(Y(t_1, t_2)) = \mathcal{K}_M(N(t_1, t_2)) = 0, \) when \(I(t_1) \not \in \{ {\bf Tr}, {\bf Fa} \}\) or \(I(t_2) \not \in Sen(L_B). \)Only questions receive answers and only gods answer questions.
- 5.
- (a)
\(\mathcal{K}_M(\neg \sigma) = 1 - \mathcal{K}_M(\sigma)\)
- (b)
\(\mathcal{K}_M(\alpha \wedge \beta) = {min}\{ \mathcal{K}_M(\alpha), \mathcal{K}_M(\beta) \}\)
- (c)
\(\mathcal{K}_M(\alpha \vee\beta) = {max}\{ \mathcal{K}_M(\alpha), \mathcal{K}_M(\beta) \}\)\(\mathcal{K}_M\)is Strong Kleene.
- (a)
\(\mathcal{K}_M^\star(\sigma) = 1 \Leftrightarrow \exists \mathcal{K}_M : \mathcal{K}_M(\sigma) = 1 \;\& \not \exists \mathcal{K}_M : \mathcal{K}_M(\sigma) = 0\)
\(\mathcal{K}_M^\star(\sigma) = \frac{1}{2} \Leftrightarrow \not \exists \mathcal{K}_M : \mathcal{K}_M(\sigma) = 1\; \& \not \exists \mathcal{K}_M : \mathcal{K}_M(\sigma) = 0\)
\(\mathcal{K}_M^\star(\sigma) = 0 \Leftrightarrow \exists \mathcal{K}_M : \mathcal{K}_M(\sigma) = 0\)
- i.
True (False) answers σ with ‘yes’ just in case \(\mathcal{K}^\star_M(\sigma) = 1\,({\mathcal{K}}^\star_M(\sigma) = 0).\)
- ii.
True (False) answers σ with ‘no’ just in case \(\mathcal{K}^\star_M(\sigma) = 0\,({\mathcal{K}}^\star_M(\sigma) = 1).\)
- iii.
True and False remain silent on σ just in case \(\mathcal{K}^\star_M(\sigma) = \frac{1}{2}.\)
θ: Is your answer to θ ‘yes’ or ‘no’?
λ: Is your answer to λ ‘no’?
τ: Is your answer to τ ‘yes’?
Values of \(\mathcal{K}_M^{\star}\) for I(θ), I(τ), I(λ)
World | \(\mathcal{K}_M^{\star}(I(\theta))\) | \(\mathcal{K}_M^{\star}(I(\tau))\) | \(\mathcal{K}_M^{\star}(I(\lambda))\) |
---|---|---|---|
a = g_{T} | 1 | 0 | \(\frac{1}{2}\) |
a = g_{F} | 1 | \(\frac{1}{2}\) | 0 |
Consider question θ. First note that the \(\mathcal{K}_M^\star\) account of True and False prescribes that True answer θ with ‘yes’ and that False answers θ with ‘no’. I take it that this is how it, intuitively, should be.^{11} This provides a reason for preferring the \(\mathcal{K}_M^\star\) account of True and False above an account that is based on the minimal fixed point; as θ is ungrounded, the minimal fixed point will valuate it as \(\frac{1}{2},\) implying that both True and False must remain silent on θ according to the minimal fixed point. To see how θ obtains its \(\mathcal{K}_M^\star\) value, note that \(Y(g_T, \cdot)\) and \(N(g_F, \cdot)\) are truth predicates in disguise, whereas \(Y(g_F, \cdot)\) and \(N(g_T, \cdot)\) are disguised falsity predicates. Thus, when posed to True, question θ allows for the alethic paraphrase ‘this very sentence is true or false’, whereas, when addressed to False, the paraphrase becomes ‘this very sentence is false or true’. Clearly then, there is a fixed point in which these sentences are true while there is no fixed point in which they are false; \(\mathcal{K}_M^\star(I(\theta)) = 1,\) irrespective of whether we address θ to True or False.
The maximal intrinsic fixed point also valuates θ as 1 and so an account of True and False based on it would prescribe the same answers to θ as the \(\mathcal{K}_M^\star\) account. We prefer the \(\mathcal{K}_M^\star\) account over the account based on the maximal intrinsic fixed point due to the answers that are prescribed to question τ. According to the \(\mathcal{K}_M^\star\) account of True and False, True answers question τ with ‘no’, whereas False remains silent on τ. Intuitively—as also remarked in Rabern and Rabern (2008)—False must indeed remain silent on τ, as he cannot answer it “in accordance with his nature”, which is to speak falsely. Although the previous solutions to HLPE do not discuss how True should answer τ, their authors do state that the gods remain silent on a question when they cannot answer that question “in accordance with their nature”. But True clearly can answer τ with either ‘yes’ or ‘no’ “in accordance with his nature”—although doing so is completely arbitrary—and so the question arises how True should answer τ. Now, one may take the arbitrariness of a yes/no answer to τ as a further reason for True to remain silent. However, this is not what the authors of previous solutions seem to have in mind.^{12} So an account of True and False which prescribes that True answers τ with a yes/no answer seems more in line with the spirit of the previous solutions to HLPE. The \(\mathcal{K}_M^\star\) account^{13} is such an account, whereas an account based on the maximal intrinsic fixed point is not.
Note that, due to the relations between yes/no predicates and truth/falsity predicates, τ behaves like a Truthteller (‘this very sentence is true’) when addressed to True while it behaves like a Liar (‘this very sentence is false’) when addressed to False. As there is a fixed point in which the Trutheller is false, we get that \(\mathcal{K}^\star_M(I(\tau)) = 0\) when a is True. As there is no fixed point in which the Liar is true and no fixed point in which the Liar is false, we get that \(\mathcal{K}^\star_M(I(\tau)) = \frac{1}{2}\) when a is False. The \(\mathcal{K}^\star_M\) valuation of question λ receives a dual explanation.
Putting \(\mathcal{K}^\star_M\) to Work
Reactions of a to Y(a, [G(n)])
World | \(\mathcal{K}_M^{\star}(Y(a, [G(n)]))\) | Answer of a |
---|---|---|
a = g_{T}, G(n) | 1 | Yes |
\(a = g_T, \neg G(n)\) | 0 | No |
a = g_{F}, G(n) | 0 | Yes |
\(a = g_F, \neg G(n)\) | 1 | No |
The table explains that ‘yes’ indicates that the north road is good and that ‘no’ indicates that the north road is not good. Question (1) is an instance of what is called the Embedded Question Lemma (EQL) in Rabern and Rabern (2008).
EQL Let E be the function that takes a question Q to the question ‘Is your answer to the question ‘Q’ ‘yes’?’ When either True or False are asked E(Q), an answer of ‘yes’ indicates that Q whereas an answer of ‘no’ indicates that not Q.
Proof
Both a double positive and a double negative make a positive. □
Reactions of a to I(ρ)
World | \(\mathcal{K}_M^{\star}(I(\rho))\) | Answer of a |
---|---|---|
a = g_{T}, G(w) | 1 | Yes |
a = g_{T}, G(e) | 0 | No |
a = g_{T}, G(s) | \(\frac{1}{2}\) | Silence |
a = g_{F}, G(w) | 0 | Yes |
a = g_{F}, G(e) | 1 | No |
a = g_{F}, G(s) | \(\frac{1}{2}\) | Silence |
Table 10 reveals the sense in which the \(\mathcal{K}_M^{\star}\) account of True and False allows us to give a formal representation of the informal solution to the “three roads riddle”. The principles at work in the solution to the “three roads riddle” are similar to the principles at work in the previous self-referential solutions to HLPE. Accordingly, the \(\mathcal{K}_M^{\star}\) account of True and False can be used to represent these solutions as well.
Putting a Four-Valued Answering Function for L_{B} to Work
We start by defining a four-valued valuation function of \(L_B, \mathcal{K}^\bullet_M : Sen(L_B) \rightarrow \{ 0, +, -, 1 \},\) in a similar manner as we defined \(\mathcal{K}_M^\star,\) i.e., by quantifying over all Strong Kleene fixed point valuations. We then define a four-valued answering function for True and False based on \(\mathcal{K}^\bullet_M\) and show how it can be invoked to give a formal representation of a solution to the four roads riddle. The principles at work in our solution to the four roads riddle are similar to the principles at work in our solution to HLPE_{syn}^{4} that was presented in section "Solving the Puzzles". Hence, the \(\mathcal{K}^\bullet_M\) account can also be used to give a formal representation of our solution to HLPE_{syn}^{4}.
\(\mathcal{K}_M^\bullet(\sigma) = 1 \Leftrightarrow \exists \mathcal{K}_M : \mathcal{K}_M(\sigma) = 1\, \& \,\not \exists \mathcal{K}_M : \mathcal{K}_M(\sigma) = 0\)
\(\mathcal{K}_M^\bullet(\sigma) = - \Leftrightarrow \not \exists \mathcal{K}_M : \mathcal{K}_M(\sigma) = 1\, \& \, \not \exists \mathcal{K}_M : \mathcal{K}_M(\sigma) = 0\)
\(\mathcal{K}_M^\bullet(\sigma) = + \Leftrightarrow \exists \mathcal{K}_M : \mathcal{K}_M(\sigma) = 1 \, \& \, \exists \mathcal{K}_M : \mathcal{K}_M(\sigma) = 0\)
\(\mathcal{K}_M^\bullet(\sigma) = 0 \Leftrightarrow \not \exists \mathcal{K}_M : \mathcal{K}_M(\sigma) = 1 \, \& \,\exists \mathcal{K}_M : \mathcal{K}_M(\sigma) = 0\)
Values of \(\mathcal{K}_M^{\bullet}\) for I(θ), I(τ), I(λ)
World | \(\mathcal{K}_M^{\bullet}(I(\theta))\) | \(\mathcal{K}_M^{\bullet}(I(\tau))\) | \(\mathcal{K}_M^{\bullet}(I(\lambda))\) |
---|---|---|---|
a = g_{T} | 1 | + | − |
a = g_{F} | 1 | − | + |
According to the four-valued conception of True and False put forward in this paper, True answers τ with ‘both’, whereas False answers τ with ‘neither’. Similarly, according to this conception, True answers λ with ‘neither’, whereas False answers λ with ‘both’. This suggests the following answering function:
- i.
True (False) answers σ with ‘yes’ just in case \(\mathcal{K}^\bullet_M(\sigma) = 1 (\mathcal{K}^\bullet_M(\sigma) = 0)\).
- ii.
True (False) answers σ with ‘no’ just in case \(\mathcal{K}^\bullet_M(\sigma) = 0 (\mathcal{K}^\bullet_M(\sigma) = 1)\).
- iii.
True and False answer σ with ‘neither’ just in case \(\mathcal{K}^\bullet_M(\sigma) = -.\)
- iv.
True and False answer σ with ‘both’ just in case \(\mathcal{K}^\bullet_M(\sigma) = +.\)
Note that our solution to the four roads riddle, i.e., question γ, is not a self-embedded question. Likewise, none of the questions discussed in section "Solving the Puzzles" are self-embedded. In contrast, our solution to the three roads riddle, i.e., question ρ, is a self-embedded question and so are the (crucial) questions invoked in previous self-referential solutions to HLPE. In the next section, we will explain, amongst others, in which sense self-embedded questions give rise to a problem for the intuitive interpretation of the answers ‘both’ and ‘neither’ that was sketched in the introduction.
Critical Remarks on Formalizations
‘Both’, ‘Neither’ and Self-Embedded Questions
- 1.
Stick to the intended interpretation of ‘both’ and ‘neither’ and extend the method of section “Solving the Puzzles" such that it becomes applicable to self-embedded questions and such that, in particular, the answer given by False to ρ_{1} is ‘both’. This option is naturally associated with an informative conception of True and False in which, in answering our questions, they intend to convey information. For instance, in answering a Liar question λ with ‘neither’ True intends to convey the information that he can’t answer λ with ‘yes’ or ‘no’.
- 2.
Use the \(\mathcal{K}^\bullet_M\) account of True and False and give up the interpretation of ‘both’ and ‘neither’. For instance, say that if \(\mathcal{K}^\bullet_M(\sigma) = -,\) True and False reply to σ with an explosion, while if \(\mathcal{K}^\bullet_M(\sigma) = +,\) they remain silent. On this account, the non-linguistic actions of exploding and remaining silent have their origin in the “paradoxality” and the “arbitrariness” of the possible yes/ no answers. Being non-linguistic actions, exploding and remaining silent are not to be evaluated in terms of ‘speaking truly’ and ‘speaking falsely’. This option is naturally associated with an algorithmic conception of True and False in which, in answering our questions, they do not intend to convey any information, but rather, they follow an algorithm. Explosions and silences, on this conception, are best thought of as two distinct ways in which the algorithm can fail, due to the non-existence of solutions (paradoxality) and the abundance of solutions (arbitrariness) respectively. On the algorithmic conception of True and False, explosions and silences are not genuine answers, but rather, states that the gods end up in due to their processing of certain questions.
In a sense, it is more natural to speak of the failure of an algorithm due to the lack of solutions (paradoxality) than due to the lack of the abundance of solutions (arbitrariness). As such, the algorithmic conception of True and False is, arguably, more naturally associated with the 3-valued account of True and False that is underlying the previous solutions to HLPE.^{14}
I take it that option 1 is preferable; I take it that an account of True and False according to which these gods can be understood as always speaking, respectively, truly and falsely, is preferable over an account on which they sometimes do not speak at all. Such an account simply seems to do more justice to Boolos’ remarks that ‘True always speaks truly’ and ‘False always speaks falsely’. To be sure, Boolos may not have envisioned the possibility to ask self-referential questions. Then again, I take it that an account of True and False which manages to respect Boolos’ instructions, even in the presence of self-reference, is preferable to an account which does not.
Reactions of True and False on ρ_{1}
Y/N | \(\mathcal{V}(N(a, \rho_1))\) | \(\mathcal{V}(Y(a, [N(a, \rho_1)]))\) | ✓/X | True | False |
---|---|---|---|---|---|
Y(a, ρ_{1}) | False | False | X | Neither | Both |
N(a, ρ_{1}) | True | True | X |
Although these remarks on self-embedded questions do not define a rigorous algorithm for calculating the answers of True and False, I take it that they illustrate that, despite our possibility to ask self-embedded questions, there are hopes for developing a formal account of True and False according to which they can be understood in accordance with Boolos’ instructions. However, self-embedded questions aside, a satisfying formal account of True and False faces more issues that have to be resolved. Below we discusses two such issues.
Expressive Incompleteness
Theories of truth typically do not contain predicates associated with the non-classical semantic values they employ in their meta-language. As such, we cannot expect much guidance from theories of truth in developing a formal account of how True and False answer non-standard questions. However, the methods of section "Solving the Puzzles" do give us some guidance here. Using ‘NE(x, y)’ to abbreviate x answers y with ‘neither’, we can translate questions μ_{1} and μ_{2} by letting I(μ_{1}) = N(a, μ_{1}) ∨ NE(μ_{1}) and I(μ_{2}) = Y(a, [NE(a, μ_{2})]). Tables 13 and 14 explain how True and False answer μ_{1} and μ_{2}.$$ \begin{aligned} &\mu_1: {\hbox{Is\,your\,answer\,to}}\,\mu_1\,{\hbox{`no'\,or\,`neither'?}}\\ &\mu_2: {\hbox{Do\,you\,answer\,`yes'\,to\,the\,question\,of\,whether\,you\,answer\, `neither'\,to}}\,\mu_2? \end{aligned} $$
In treating μ_{1} and μ_{1}^{′} alike, we are bound to conclude that μ_{1}′ is ‘neither true nor false’. But this exactly what μ_{1}^{′}says, and so μ_{1}′ is true and so not ‘neither true nor false’. In sum, accepting that μ_{1}^{′} is ‘neither true nor false’ seems to be tantamount to accepting a contradiction, while accepting that True answers μ_{1} with ‘neither’ has a clear rationale in the (assumed) nature of True.$$ \mu_1': \hbox{Sentence}\,\mu_1'\, {\hbox{is\,false\,or\,(neither\,true\,nor\,false)}} $$
Reactions of True and False on μ_{1}
Y/N | \(\mathcal{V}(N(a, \mu_1) \vee NE(\mu_1))\) | ✓/X | True | False |
---|---|---|---|---|
Y(a, μ_{1}) | False | X | Neither | Both |
N(a, μ_{1}) | True | X |
Reactions of True and False on μ_{2}
Y/N | \(\mathcal{V}(NE(a, \mu_2))\) | \(\mathcal{V}(Y(a, [NE(a, \mu_2)]))\) | ✓/X | True | False |
---|---|---|---|---|---|
Y(a, μ_{2}) | False | False | X | No | Yes |
N(a, μ_{2}) | False | False | ✓ |
Tokens or Types?
Question λ is familiar: True answers it with ‘neither’. Question λ_{1} asks whether True answers question λ with ‘no’. As True answers λ with ‘neither’ (hence, not with ‘no’), the truthful answer to λ_{1} is ‘no’. Hence, True should answer λ_{1} with ‘no’. Or so it seems. Yet if we base our account of True and False on, say \(\mathcal{K}^\bullet_M,\) we get different predictions. Fixed point theories of truth satisfy what is called the intersubstitutability of truth.^{16} As a consequence, \(Y(g_T, \overline{\sigma})\) and σ have the same semantic value according to \(\mathcal{K}^\bullet_M,\) whenever \(\overline{\sigma}\) denotes σ. In particular then, we have that \(\mathcal{K}_M^\bullet(Y(g_T, \lambda)) = \mathcal{K}_M^\bullet(Y(g_T, \lambda_1)) = -\) and so according to the \(\mathcal{K}^\bullet_M\) account, True will answer both λ and λ_{1} with ‘neither’.$$ \begin{aligned} &\lambda: {\hbox{Is\,your\,answer\,to}}\,\lambda\,{\hbox{`no'?}}\\ &\lambda_1: {\hbox{Is\,your\,answer\,to}}\,\lambda\, {\hbox{`no'?}} \end{aligned} $$
It is often argued that theories of truth should satisfy the intersubstitutability of truth, cf. Field (2008) or Beall (2009). In a nutshell, the argument is that if truth does not satisfy the intersubstitutability property, it can not play its stereotypical role of serving as a device of generalization. Let’s accept this argument pertaining to theories that describe the behavior of a truth (and falsity) predicate. Does it carry over to a theory that describes the behavior of a ‘answers with yes’ (and ‘answers with no’) predicate? Not necessarily. For one thing, it is not clear that the stereotypical role of a ‘answers with yes’ predicate is to serve as a device of generalization and so the typical argument for the intersubstitutablity breaks down: although the analogy between true/falsity predicates and yes/no predicates is close, it is not perfect, as we also noted in the previous subsection. I take the intuitive reasoning with respect to λ and λ_{1} that was given above convincing and I do not see why the intersubstitutablity of truth should lead us to dismiss that reasoning. Accordingly, I take it that a fully satisfying formal account of the behavior of True and False should be token-sensitive.
As the reader may have noticed, questions λ and λ_{1} constitute an interrogative version of what Gaifman (1992) calls the “two lines puzzle”. In fact, Gaifman’s pointer semantics is an example of a token-sensitive theory of truth which gives up the intersubstitutability property and which yields (in alethic terms) similar conclusions with respect to the status of λ and λ_{1} as the intuitive reasoning above. As such, it seems promising to develop a token-sensitive account of True and False on the basis of Gaifman’s work. Clearly, doing so is far beyond the scope of this paper.
The Wheeler and Barahona Argument
Wheeler and Barahona (2011) argued that HLPE_{syn}^{3} cannot be solved in less than three questions. Their argument relies on the following lemma from Information Theory.
Using QL, we see that when True and False have a three-valued answering repertoire (as in section “Critical Remarks on Formalizations”), we cannot solve the four roads riddle by asking a single question. In a similar vein—though in a more complicated setting—Wheeler and Barahona appealed to QL to argue that HLPE_{syn}^{3} cannot be solved in less than 3 questions, where they assumed that True and False have a three-valued answering repertoire.(QL) If a question has n possible answers, these answers cannot distinguish m > n different possibilities.
When True and False have a four-valued answering repertoire however, QL tells us that we may be able to solve the four roads riddle by asking a single question. Of course, whether or not we are actually able to do so depends on our ability to find questions such that their four possible answers are correlated to the four relevant states of affairs in a suitable way. Section "The Wheeler and Barahona Argument" showed how to solve the four roads riddle by a single question. So, by moving from a three- to a four-valued answering repertoire, we can escape the QL based conclusion that “the four roads riddle cannot be solved in a single question”. Similarly, by moving from a three- to a four-valued answering repertoire, we escaped the conclusion of Wheeler and Barahona (2011) that HLPE_{syn}^{3} cannot be solved in less than three questions.
Let me conclude this section by stating a puzzle, HLPE_{syn}^{4*}, which I do not know how to solve (in two questions) and which is not unsolvable on the basis of QL. HLPE_{syn}^{4*} is defined just like HLPE_{syn}^{4}, apart from the following difference. The gods react to your questions with ‘huh’, ‘duh’, ‘da’ and ‘ja’. As before, ‘da’ and ‘ja’ mean, in some order, ‘yes’ and ‘no’. But now ‘huh’ and ‘duh’ mean, in some order, ‘neither’ and ‘both’. Can we solve HLPE_{syn}^{4*} in two questions?
Concluding Remarks
We put forward an alternative conception of how True and False answer yes–no questions, resulting in a four-valued answering repertoire. We then showed how this conception could be invoked to solve HLPE_{syn}^{4} (and HLPE_{syn}^{3}) in two questions. Our four-valued (in contrast to a three-valued) answering repertoire allowed us to escape the argument of Wheeler and Barahona (2011), which established that HLPE_{syn}^{3} cannot be solved in less than three questions.
The second part of the paper was concerned with formalizations of (the present and previous) solutions to versions of HLPE, that were all presented informally. We showed how—by appealing to Strong Kleene fixed point theories of truth and by working in a restricted setting—to give a formal representation of the solutions to HLPE. Although in an important sense our formalization “gets the job done”, we discussed some desiderata of a formalization of the behavior of True and False that were not met by the one that was presented. To develop a formal account of True and False that meets these desiderata—i.e., a token-sensitive account for an expressive complete language in which True and False can be understood as, respectively, “always speaking truly” and “always speaking falsely”—is postponed to future work.
Footnotes
- 1.
Actually, Uzquiano distinguishes two versions of HLPE_{syn} and gives two-question solutions for both versions. The versions differ with respect to the abilities of True and False to predict the answers of Random. The first version assumes that True and False cannot predict Random’s answers (which seems reasonable given that Random answers randomly), while the second version assume that True and False can predict Random’s answers (which seems reasonable as True and False are omniscient). Uzquiano’s solution to the second version is also a solution to the first version but not vice versa. For our purposes, the distinction does not matter: we give a solution that works for both versions.
- 2.
Note that HLPE_{syn}^{3} and HLPE_{syn}^{4} (deliberately) violate Boolos’ instruction (B4).
- 3.
We could use two place answering predicates and remove the indexical “your”. However, as this results in a less streamlined presentation, we chose not to do so.
- 4.
We treat yes–no questions on par with their associated yes–no statements. That is sloppy, but also very convenient.
- 5.
Which is not to say that other theories of truth do not allow such representation.
- 6.
We do so for sake of simplicity: the definition of the three- and four-valued answering functions below are easily seen to carry over to quantified languages.
- 7.
We will use =, ∧, ∨, ¬, → and ↔ and as logical symbolism, the interpretation of which is as expected.
- 8.
The set of quotational constant symbols has a joint recursive definition together with Sen(L_{B}), the set of sentences of L_{B}. The definition of these sets can safely be left to the reader.
- 9.
For instance, when posed to god a, the sentence Y(a, c_{1}) may be paraphrased as: ‘Is it the case that: your answer to this question is ‘yes’?’, provided that the denotation of c_{1} is Y(a, c_{1}).
- 10.
In the present setting, the number of fixed point valuations over M depends on the denotations of the members of C; if, say, I(c) = (g_{T} = g_{T}) for every \(c \in C,\) there is a unique fixed point valuation over M.
- 11.
I take it that question θ reveals an interesting dissimilarity between positively answering a yes–no question and asserting its alethic counterpart: while ‘yes’ is clearly a truthful answer to θ, the ungroundedness of ‘this very sentence is true or false’ may deems its assertion inappropriate. More concretely, answering θ with ‘yes’ makes it true, while asserting ‘this very sentence is true or false’ does not render the asserted sentence true.
- 12.
Rabern and Rabern (2009) comment on the answering function that they had in mind in their published paper: according to this function, True gives a classical (yes/no) answer to questions like τ.
- 13.
Although the \(\mathcal{K}_M^\star\) account prescribes that True answers τ with ‘no’, we do not think that there is any further reason to prefer such an account over an account according to which True answers τ with ‘yes’. Further, some obvious modifications to \(\mathcal{K}_M^\star\) will yield just such an account.
- 14.
As pointed out by an anonymous referee, it can be argued that a 3-valued algorithmic conception of True and False does not require that we introduce a predicate in our language that represents failures of the algorithm as such failures do not belong to the language of the gods. In other words, it can be argued that the problem of expressive completeness (see section"Expressive Incompleteness") does not arise on a 3-valued algorithmic conception of True and False.
- 15.
For further dissimilarities, see footnote 11 and section " Tokens or Types?".
- 16.
Meaning that \(T(\overline{\sigma})\) and σ are intersubstitutable (without change of semantic value) in every non opaque context.
Notes
Acknowledgments
Many thanks to Reinhard Muskens, Harrie de Swart and two anonymous referees of this journal for their valuable comments on this paper.
Open Access
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