On the possibility of fundamental facts

Abstract

Some facts are grounded, and some are fundamental (ungrounded). Roberto Loss has recently presented an argument (which is a counterpart of the famous Fitch's paradox of knowability) that there are no ungrounded facts. I argue that this argument is not unrestrictedly valid, and one of the central premises of Loss’s argument - distributivity of grounding (PG) - is not unrestrictedly true. Also, I offer a reason to think that all facts of the form “A is not a grounded fact” are fundamental.

Keywords:
Facts; Grounding; Fundamentality

1. Loss’s argument

In the literature concerning metaphysical grounding, it is common to distinguish between grounded facts¹ and fundamental (that is, ungrounded) facts. A fact P is said to be grounded if there is another fact R such that R is a ground for P. A fact F is said to be fundamental if it has no ground. It is not clear whether there are such entities as fundamental facts. For instance, following what Correia (2021) calls ‘an argument from Purity’ (where Purity is Sider’s (2011) idea that fundamental truths should include only fundamental notions), some philosophers argue that all grounding facts are grounded facts (see deRosset (2013) and Sider (2020)). A different argment (which relies on the assumption that fundamental facts should be freely recombinable) to the effect that all grounding facts are not fundamental has been presented by Bennett (2011). These arguments thus deny that the grounding chain terminates.

In a recent paper, Loss (2021) presented a new argument (hereafter ‘(LA)’) that there are no ungrounded facts. Loss argues that the very assumption that there could be a fact A such that A is both the case and ungrounded leads to an apparent contradiction. This argument relies on the following premises:

$$\begin{gathered} \left(FG\right)\text{If}A\text{is a grounded fact, then}A \\ \mathbb{G}A\to A \end{gathered}$$

(PG) If 𝐴 ∧ 𝐵 is the case, it is logically possible that both the fact that 𝐴 and the fact that 𝐵 are grounded facts

$(A\wedge B)\to ◊(\mathbb{G}A\wedge \mathbb{G}B)$

(N) Whatever is provably false is provably impossible

$⊢~A\to ⊢~◊A$

The argument that there are no ungrounded facts runs as follows:

$$\begin{gathered} 1)\mathbb{G}A\wedge \mathbb{G}~\mathbb{G}A\text{assumption} \\ (2)\mathbb{G}A1,\wedge -\text{elimination} \\ (3)\mathbb{G}~\mathbb{G}A1,\wedge -\text{elimination} \\ (4)~\mathbb{G}A3,(FG\left) \\ \right(5)\mathbb{G}A\wedge ~\mathbb{G}2,4,\wedge -\text{introduction} \\ (6)~(\mathbb{G}\wedge \mathbb{G}~\mathbb{G}A)1,5,\text{reductio} \\ (7)~◊(\mathbb{G}A\wedge \mathbb{G}~\mathbb{G}A)6,(N\left) \\ \right(8)(A\wedge ~\mathbb{G}A)\to ◊(\mathbb{G}A\wedge \mathbb{G}~\mathbb{G})(PG\left) \\ \right(9)~(A\wedge ~\mathbb{G}A)7,8,\text{modus tollens} \\ (10)A\to \mathbb{G}A9,\text{logic} \end{gathered}$$

(LA) rests on at least two questionable (from my point of view) ideas:

(a) Such a fact as “A is an ungrounded fact ”(~𝔾𝐴) could be grounded. If 'A is an ungrounded fact' is grounded, then we get a fact 'It is a grounded fact that A is an ungrounded fact'.

(b) If we have that ”It is a grounded fact that A is an ungrounded fact”, then this entails that it is a fact that A is an ungrounded fact².

The idea (a) is necessary to get the line (3) of (LA). The idea (b) is necessary to get the line (4) of (LA). Both ideas are necessary to get the line (5) of (LA).

In what follows, I argue that there are some cases in which none of these ideas (which are vital to (LA)) are true (more clearly, I argue that there are some cases in which (a) fails, and the (LA)’s inference from (3) to (4) is in fact not an instance of (FG)) . Our argument relies on two principles - (DA) and (GG).

$\left(DA\right)\left(\right(\mathbb{G}(⌜A\to B⌝)\wedge ~\mathbb{G}⌜A⌝\wedge ~\mathbb{G}⌜B⌝\left)\right)\to (\mathbb{G}⌜A⌝\to \mathbb{G}⌜B⌝)$

$\left(GG\right)\mathbb{G}⌜A⌝\to \mathbb{G}(⌜(\mathbb{G}⌜A⌝)⌝)$

Both principles are pretty straightforward. (DA) is propositionally true. Take, in turn, (GG) - if some fact A is grounded, then the fact that A is a grounded fact must be a grounded fact as well. Assume that every fact is either grounded or ungrounded, and for any A (at least for any decidable A) the following holds: if A is ungrounded, then its complement (not-A) is grounded (which is also perfectly compatible with the view that all facts are grounded). Now suppose that A is grounded, but the fact that A is grounded is ungrounded. Given that “A is grounded” is ungrounded, it must follow then that its complement - “A is ungrounded” must be grounded. By (FG) it would follow then that also it is the case that “A is ungrounded”, contradicting our assumption.

2. Grounding: operational or predicational?

Throughout the paper, “ground” (𝔾) will be threatened as a relational predicate (contrary to (LA) which takes 𝔾 to be an operator)⁴. We will add 𝔾 to the language of Q (to the language of Robinson arithmetic), thus obtaining a unique name ⌜A⌝ for any proposition A using the standard technique of Gödel numbering. We take grounding to be a relation between true propositions (“facts”). It follows then that we will assume that the notion of ground is factive in the following sense - if A grounds B, then both A and B are the case.

There are several reasons to prefer the predicational approach to grounding over the operational approach. The main merit of predicationalism is that it allows us to connect theories of ground with axiomatic theories of truth, thus allowing us to formalize grounding-theoretic principles in a natural way (Korbmacher, 2018, p. 163). For instance, consider the principle of Transitivity (which is widely endorsed in the literature) (where “≺” stands for “grounds” (within the predicational approach), and the range of quantifiers is the set of all true propositions). As it has been shown in the literature (for instance, Korbmacher 2018, p. 164), those theorists who endorse a predicational approach can directly formalize the principle of Transitivity in the following way:

$\left(\text{Transitivity}\right)\forall x\forall y\forall z\left(\right(x\prec y\&y\prec z)\to x\prec z)$

Instead, using the operational approach, the principle of Transitivity can be formalized only by endorsing the instances of the following schema (where “≤” stands for the operational notion of ground):

$(x\le y\&y\le z)\to x\le z$

The difference between a predicational approach and an operational approach concerning their expressive power has been nicely highlighted by Korbmacher (2018, p. 164):

“Thus, on the operator approach, we can achieve quantification over truths only by moving to quantification over sentences in the meta-language, while on the predicate approach, we can directly express quantification over truths in the object language”.

Moreover, philosophers endorsing “grounding maximalism” (the view that there are no fundamental facts) clearly equate the notions of “being grounded” and “being true”. Consider thus the following definitions:

$$\begin{gathered} \left(\text{True}\right)\forall x\left(Tr\right(x)\leftrightarrow \exists y(y\prec x\left)\right) \\ \left(\text{Grounded}\right)\forall x\left(\mathbb{G}\right(x)\leftrightarrow (Tr\left(x\right)\&\exists y(y\prec x)\left)\right) \end{gathered}$$

According to (True), for any x, x is true iff there is some y such that y is a ground for x. According to (Grounded), x is grounded iff it is true and x is grounded in some y. Notice that, again, we can define a predicate 𝔾 in the object language (as well as we can express a truth predicate as being equivalent to a ground predicate) in the predicational approach. Using the operator approach, however, there is no way to claim, in the object language, that some x is true iff it is grounded is some y: “there simply is no way to express nested universal and existential quantification over truths on that approach” (Korbmacher, 2018, p. 164).

So, here is a crucial difference between (LA) and our own argument against it. Loss endorses an operational approach to metaphysical grounding. Instead, I will take “grounded” to be a predicate, thus endorsing a predicational approach. Notice that the crucial premise of our argument against the claim that there are no fundamental facts is derivable only if “grounded” is taken to be a predicate; there is no way (at least, there is no straightforward way) to derive it within an operational approach. So, a defender of (LA) might argue that either our argument fails to be an argument against (LA), or it must be clearly shown why “grounded” must be taken as a predicate and not as an operator.

Notice, however, that the idea behind this paper does not require one to accept that a predicational treatment of ground is the only correct approach to metaphysical grounding. As a result, I do not want to say that Loss must accept a predicational approach. What this paper intends to show is, in general, the following: in fact, (LA) is not sufficient to prove that there are no fundamental facts. At best, (LA) is successful in proving that if an operational approach to metaphysical grounding is endorsed, then all facts are grounded. But since (as I will show) the existence of ungrounded facts can be proven within a predicational approach, (LA) does not achieve its main goal⁵ - to show that there are no fundamental facts (unrestrictedly speaking). In this sense, even philosophers endorsing the operational approach might find our argument equally appealing.

3. The counter-argument

Let T be an extension of Robinson arithmetic (‘Q’), extended with a predicate⁶ ‘ ’ . By assumption, T is able to encode its syntax and contains, for any sentence A, a quotation name ‘⌜A⌝’ (the numeral of the Gödel number of A). Hence, T should satisfy the diagonal lemma⁷, which is a minimal syntax condition for T.

So, let ‘𝔾 ⌜S⌝ → A’ will be a well-formed formula in T, and let ‘A’ be any sentence. By the diagonal lemma, we can find a sentence B such that

(1) $T⊢B\leftrightarrow (\mathbb{G}⌜B⌝\to A)$

Assume, by a way of contradiction, that all facts are grounded

$\left(AFG\right)A\to \mathbb{G}⌜A⌝$

From (1), by (AFG), we have

(2) $T⊢\mathbb{G}(⌜(B\leftrightarrow (\mathbb{G}⌜B⌝\to A)⌝)$

From (2), by ↔ -elimination, we have

(3) $T⊢\mathbb{G}(⌜(B\to (\mathbb{G}⌜B⌝\to A)⌝)$

From (3), under assumption that B is not grounded, we have by (DA)

(4) $T⊢\mathbb{G}⌜B⌝\to \left(\mathbb{G}\right(⌜(\mathbb{G}⌜B⌝)⌝)\to \mathbb{G}⌜A⌝)$

(In fact, that B is not grounded follows if we take A to be f (so we have that $B\leftrightarrow (\mathbb{G}⌜B⌝\to f)$ (see the step 11)). We also have from (GG)

(5) $T⊢\mathbb{G}⌜B⌝\to \mathbb{G}(⌜(\mathbb{G}⌜B⌝)⌝)$

Now, by propositional logic, we have from (4) and (5)

(6) $T⊢\mathbb{G}⌜B⌝\to \mathbb{G}⌜A⌝$

So we have from (6) (by propositional logic)

(7) $T⊢(\mathbb{G}⌜A⌝\to A)\to (\mathbb{G}⌜B⌝\to A)$

We also have from (1) that

(8) $T⊢(\mathbb{G}⌜B⌝\to A)\to B$

So from (7) and (8), by conditional proof, we have

(9) $T⊢(\mathbb{G}⌜A⌝\to A)\to B$

Since we are assuming that $~\mathbb{G}⌜B⌝$ (see the comment to the step 4), and $\mathbb{G}⌜A⌝\to A$ is the case by (FG), from (DA) we can derive (10)⁹

(10) $T⊢\left(\mathbb{G}\right(⌜(\mathbb{G}⌜A⌝)\to A)⌝)\to \mathbb{G}⌜B⌝$

From (10) and (6), by conditional proof, we have

(11) $T⊢\left(\mathbb{G}\right(⌜(\mathbb{G}⌜A⌝)\to A)⌝)\to \mathbb{G}⌜A⌝$

Since A could be any sentence, let A be “f”(where ‘f’ means ‘false’). Thus, we have from (11)

(12) $T⊢\left(\mathbb{G}\right(⌜(\mathbb{G}⌜f⌝)\to f)⌝)\to \mathbb{G}⌜f⌝$

Since ‘$⌜\mathbb{G}⌜f⌝\to f$’ is propositionally equivalent to ‘$~\mathbb{G}⌜f⌝$’, (12) is equivalent to

(13) $T⊢\mathbb{G}(⌜(~\mathbb{G}⌜f⌝)⌝)\to \mathbb{G}⌜f⌝$

Assume now the following tautology

(14) $f\to A$

Since f has no ground, it is vacuously true that

(15) $T⊢\mathbb{G}⌜f⌝\to \mathbb{G}⌜A⌝$

From (15), by contraposition, we have

(16) $T⊢~\mathbb{G}⌜A⌝\to ~\mathbb{G}⌜f⌝$

From (16) it is easy to derive

(17) $T⊢\mathbb{G}(⌜(~\mathbb{G}⌜A⌝)⌝)\to \mathbb{G}(⌜(~\mathbb{G}⌜f⌝)⌝)$

Finally, from (17) and (13), by conditional proof, we have

(18) $T⊢\mathbb{G}(⌜(~\mathbb{G}⌜A⌝)⌝)\to \mathbb{G}⌜f⌝$

Let us now return to (LA). By (LA), every fact is grounded. Suppose that there is a fact of the form ‘$A\wedge ~\mathbb{G}A$’. Then, by (PG), it is possible for ‘$\mathbb{G}⌜A⌝\wedge \mathbb{G}(⌜(~\mathbb{G}⌜A⌝)⌝)$’ (‘A is grounded and that A is not grounded is grounded’) to exist. Suppose that this conjunctive fact exists. Given that, by (18), ‘that A is not grounded is grounded’ entails ‘f is grounded’ , we have such a ‘fact’ as ‘$\mathbb{G}⌜A⌝\wedge \mathbb{G}⌜f⌝$’. But since it is not possible for f to be a fact, then ‘$\mathbb{G}⌜A⌝\wedge ⌜f⌝$’ is indeed not a fact. Hence, if ‘$\mathbb{G}⌜f⌝$’ is false and thus not a fact (once again, we identify facts with true propositions by definition), then ‘$\mathbb{G}⌜A⌝\wedge \mathbb{G}⌜f⌝$’ is also false and thus not a fact (because this conjunctive ‘fact’ has a false conjunct).

We then have not a contradiction of the form ‘$\mathbb{G}A\wedge ~\mathbb{G}A$’ ((LA), step (5)), but just a false construction. By reductio, we conclude that if there is such a fact as ‘$~\mathbb{G}⌜A⌝$’, then ‘$~\mathbb{G}⌜A⌝$’ is not even a groundable fact.

We have that '$~\mathbb{G}⌜A⌝$’ is not even a groundable fact, because ‘$\mathbb{G}⌜f⌝$’ is not a fact at all¹¹. However, ‘$~\mathbb{G}⌜f⌝$’ is indeed a fact. But then, we have from (18), by contraposition

(19) $T⊢~\mathbb{G}⌜f⌝\to ~\mathbb{G}(⌜(~\mathbb{G}⌜A⌝)⌝)$

By (19), if f is not grounded, then it is not grounded that A is not grounded. But it is a necessary truth that f is not grounded. Hence, it is a necessary truth that the fact that A is not grounded is not grounded. Thus, again, ‘$~\mathbb{G}⌜A⌝$’ is not even a groundable fact.

It might be argued that (LA) does not require the existence of such conjunctive facts as ' $A\wedge ~\mathbb{G}A$’. On the contrary, (LA) demonstrates that there are no such facts - if such a fact existed, then there would be such a fact as ‘$\mathbb{G}A\wedge ~\mathbb{G}A$’. Suppose now that it is true that ‘$\mathbb{G}(⌜(~\mathbb{G}⌜A⌝)⌝)\to \mathbb{G}⌜f⌝$’, so if there is such a fact as ‘$A\wedge ~\mathbb{G}⌜A⌝$’, then instead of having such a fact as ‘$\mathbb{G}⌜A⌝\wedge ~\mathbb{G}⌜A⌝$’ we have such a fact as ‘$\mathbb{G}⌜A⌝\wedge \mathbb{G}⌜f⌝$’. Of course, there is no such fact as ‘$\mathbb{G}⌜A⌝\wedge \mathbb{G}⌜f⌝$’. Why then can’t we conclude that if ‘$A\wedge ~\mathbb{G}⌜A⌝$’ entails ‘$\mathbb{G}⌜A⌝\wedge \mathbb{G}⌜f⌝$’, and there is no such a fact as ‘$\mathbb{G}⌜A⌝\wedge \mathbb{G}⌜f⌝$’, then (by modus tollens) there is no such a fact as ‘$A\wedge ~\mathbb{G}⌜A⌝$’?

Notice however an important difference between (LA) and our argument. By (LA), if A is true but ungrounded fact, it is possible for both A and ~𝔾𝐴 to be grounded facts (this is what (PG) exactly says). But if it is possible for both A and ~𝔾𝐴 to be grounded facts, then it is possible for such a fact as ‘$\mathbb{G}A\wedge ~\mathbb{G}A$’ to exist. Hence, by (LA), it is not possible for 𝐴 to be an ungrounded fact because such a ‘fact’ as 𝔾~𝔾𝐴 is not compatible with such a fact as 𝔾𝐴, but there is nothing wrong with assuming that ~𝔾𝐴, taken separately from A, could be a grounded fact. Contrary to (LA), I argue that such a fact as ‘$~\mathbb{G}⌜A⌝$’ is not a groundable fact by itself (and not because a possibility that this fact is grounded contradicts a possibility that A is grounded). Thus, if we have a possibility that a certain fact A is both the case and ungrounded, it is wrong to assume that it is even possible for ‘$~\mathbb{G}⌜A⌝$’ to be a grounded fact. But if ‘$~\mathbb{G}⌜A⌝$’ is not even groundable, then (PG) is false. If (PG) is false, it is not possible for (LA) to derive such a ‘fact’ as ‘$\mathbb{G}A\wedge \mathbb{G}f$’ (and, of course, it is not possible to derive ‘$\mathbb{G}A\wedge ~\mathbb{G}A$’) and, finally, it is not possible to conclude that ‘$~(A\wedge ~\mathbb{G}A)$’. Hence, the fact that 'A is not grounded is not a groundable fact’ is not very helpful for (LA).

Finally, let us show in a slightly different way that (PG) is not true. From (1), we have a construction ‘$B\leftrightarrow (\mathbb{G}⌜B⌝\to A)$’. From (12), we let A be f. Thus, we get ‘$B\leftrightarrow (\mathbb{G}⌜B⌝\to f)$’. Since B is assumed to be true, B is a fact. So, if we have a sentence ‘$B\leftrightarrow (\mathbb{G}⌜B⌝\to f)$’ (where B is a fact), we clearly have ‘$B\wedge ~\mathbb{G}⌜B⌝$’. Suppose now that (PG) is true and thus it is possible for ‘$~\mathbb{G}⌜B⌝$’ to be a grounded fact. If so, then we have 𝔾⌜f⌝. Thus, by reductio, ‘$~𝔾(⌜(~𝔾⌜B⌝)⌝)$’ (‘that B is not a grounded fact is not a grounded fact’). But since B is equivalent to ‘$~\mathbb{G}⌜B⌝$’, then if ‘$~\mathbb{G}⌜B⌝$’ is not grounded, B is also not grounded.

Hence, if we have a construction ‘$B\leftrightarrow (\mathbb{G}⌜B⌝\to f)$’, and it is true that B is a fact, then it is true that we have ‘$B\wedge ~\mathbb{G}⌜B⌝$’. If (PG) is true, it should be therefore possibly true that ‘$\mathbb{G}⌜B⌝\wedge \mathbb{G}(⌜(~\mathbb{G}⌜B⌝)⌝)$’. But it is not possibly true. If we have ‘$B\wedge ~\mathbb{G}⌜B⌝$’ then, under the assumption that ‘$B\leftrightarrow ~\mathbb{G}⌜B⌝$’, we have

(20) $(B\wedge ~\mathbb{G}⌜B⌝)\to (~\mathbb{G}⌜B⌝\wedge ~\mathbb{G}(⌜(~\mathbb{G}⌜B⌝)⌝\left)\right)$

(If B is the case and ungrounded, then B is ungrounded, and that B is ungrounded is ungrounded)

It is indeed true that if '$B\leftrightarrow ~\mathbb{G}⌜B⌝$’, then we have

(21) $(B\leftrightarrow ~\mathbb{G}⌜B⌝)\to (\mathbb{G}⌜B⌝\leftrightarrow (⌜\mathbb{G}(~\mathbb{G}⌜B⌝)⌝\left)\right)$

But we never have the following

(22) $(B\leftrightarrow ~\mathbb{G}⌜B⌝)\to (\mathbb{G}⌜B⌝\wedge (⌜\mathbb{G}(~\mathbb{G}⌜B⌝)⌝\left)\right)$

We never have (22) because neither ‘$\mathbb{G}⌜B⌝$’ nor ‘$\mathbb{G}(⌜(~\mathbb{G}⌜B⌝)⌝$’ are facts (even though both ‘B’ and ‘$~\mathbb{G}⌜B⌝$’ are facts). Thus, if ‘$B\leftrightarrow ~\mathbb{G}⌜B⌝$’ is a fact, B is an ungrounded fact. So, if a predicational approach to grounding is accepted, (PG) is not true¹².

A few things should be added here. As it has been demonstrated by (LA), there are no facts of the form $\mathbb{G}A\wedge \mathbb{G}~\mathbb{G}A$ (which is just an instance of (PG)), So it may seem that our argument against (PG) - which is, in that case, an argument against the possibility of ($\mathbb{G}A\wedge \mathbb{G}~\mathbb{G}A$) - misses the target because the impossibility of ($\mathbb{G}A\wedge \mathbb{G}~\mathbb{G}A$) is also proven by (LA). Notice however that while (LA) holds that ($\mathbb{G}A\wedge \mathbb{G}~\mathbb{G}A$) is not true, it still requires the truth of (PG). What (LA) really shows is the following: if it is true that ($A\wedge ~\mathbb{G}A$) then, given that (PG) is true, it must be the case that ($\mathbb{G}A\wedge \mathbb{G}~\mathbb{G}A$); however, given that (FG) is true, it is not the case that ($\mathbb{G}A\wedge \mathbb{G}~\mathbb{G}A$). So, according to (LA), ($\mathbb{G}A\wedge \mathbb{G}~\mathbb{G}A$) is not the case not because (PG) is not valid, but because there are no facts of the form ($A\wedge ~\mathbb{G}A$), and we will get a contradiction if we apply (PG) to it. Instead, our argument, unlike (LA), shows that there are facts of the form ($A\wedge ~\mathbb{G}A$) but (i) neither such facts are groundable nor (ii) 𝔾 could be distributed over this conjunction. It follows then that there are facts of the form ($A\wedge ~\mathbb{G}A$), but there are no facts of the form ($\mathbb{G}A\wedge \mathbb{G}~\mathbb{G}A$). It follows that (PG) should be false. But (LA) requires that (PG) should be true. Obviously, Loss (2021) does not want to say that ($\mathbb{G}A\wedge \mathbb{G}~\mathbb{G}A$) is false because (PG) is false (it actually says that ($\mathbb{G}A\wedge \mathbb{G}~\mathbb{G}A$) is false because ($A\wedge ~\mathbb{G}A$) is false). But this is exactly what we want to say - we want to say that ($\mathbb{G}A\wedge \mathbb{G}~\mathbb{G}A$) is false because (PG) is false.

Moreover, in order to show that the very possibility of ($\mathbb{G}A\wedge \mathbb{G}~\mathbb{G}A$) yields a contradiction, Loss requires (FG). However, (FG) says that if there is some grounded fact, then it is true (indeed, (FG) requires that if, for some fact F, F is a grounded fact, then that F is a grounded fact is itself a fact). But, as it has been demonstrated, $\mathbb{G}~\mathbb{G}A$is not a fact. It follows that derive ~𝔾𝐴 cannot be derived from $\mathbb{G}~\mathbb{G}A$. If so, there is no way to argue that ($\mathbb{G}A\wedge ~\mathbb{G}A$) in order to prove that all facts are grounded.

4. Discussion.

It may be immediately argued that our argument relies on the existence of such facts as ‘$B\leftrightarrow (\mathbb{G}⌜B⌝\to f)$’ , but it is unclear whether such facts exist. Maybe there are no such facts, and it is impossible to replace A with f in (1)¹³ (see step (12) of the argument). One may also object that it is unclear whether there are such facts as ‘$B\leftrightarrow (\mathbb{G}⌜B⌝\to A)$’. We can show however that the existence of such facts follows from plausible principles of grounding.

(LA) relies on the principle (FG) which says that if some fact A is grounded, then it is the case that A. This principle is plausible. Suppose that A is ‘2+2=4’. It is indeed true that if A is grounded, then it is the case that 2+2=4. Suppose that A is ‘2+2=5’. Since ‘2+2=5’ is indeed not a fact, it is false that A is grounded. In this case (where A is f ), we have ‘$\mathbb{G}⌜f⌝\to f$’, which is also a highly plausible principle. Thus, it seems that ‘$\mathbb{G}A\to A$’ is true for all instances of A.

Assume now that '$\mathbb{G}⌜A⌝\to A$’ is a fact. Let B be this fact. We can show that if there is such a fact as ‘$B\leftrightarrow (\mathbb{G}⌜A⌝\to A)$’, then there is such a fact as ‘$B\leftrightarrow (\mathbb{G}⌜B⌝\to A)$’. Consider the plausible principle of ground, according to which if there is a fact of the form “$A\leftrightarrow B$” and A is not the case, then there is a fact of the form “to say that A is grounded is to say that B is grounded”:

$\left(GB\right)\left(\right(A\leftrightarrow B)\wedge ~\mathbb{G}⌜A⌝\wedge ~\mathbb{G}⌜B⌝)\to (\mathbb{G}⌜A⌝\leftrightarrow \mathbb{G}⌜B⌝)$

(notice that “$\mathbb{G}⌜A⌝\to A$ is ungrounded “ follows from the step 18 by taking A to be f, so if B is taken to be the same fact as “A is ungrounded “, B is ungrounded as well).

The argument is this. Assume that

(23) $T⊢B\leftrightarrow (\mathbb{G}⌜A⌝\to A)$

From (23), by ↔-elimination and (GB), we have

(24) $T⊢\mathbb{G}⌜B⌝\to \left(\mathbb{G}\right(⌜(\mathbb{G}⌜A⌝)\to A)⌝)$

From (24) and (11), by conditional proof, we have

(25) $T⊢\mathbb{G}⌜B⌝\to \mathbb{G}⌜A⌝$

Assume now the following tautology

(26) $T⊢A\to (\mathbb{G}⌜A⌝\to A)$

From (26) and (23), by conditional proof, we have

(27) $T⊢A\to B$

It is vacuously true that

(28) $T⊢\mathbb{G}⌜A⌝\to \mathbb{G}⌜B⌝$

Now, from (28) and (25), we have

(29) $T⊢\mathbb{G}⌜A⌝\leftrightarrow \mathbb{G}⌜B⌝$

From (29), by propositional logic, we have

(30) $T⊢(\mathbb{G}⌜A⌝\to A)\leftrightarrow (\mathbb{G}⌜B⌝\to A)$

And from (23) and (30) we have

(31) $T⊢B\leftrightarrow (\mathbb{G}⌜B⌝\to A)$

Hence, assuming that ‘$\mathbb{G}⌜A⌝\to A$’ is a fact, and $B\leftrightarrow (\mathbb{G}⌜A⌝\to A)$, we can show that there is such a fact as ‘$B\leftrightarrow (\mathbb{G}⌜B⌝\to A)$’¹⁴. Assume now that ‘$\mathbb{G}⌜f⌝\to f$’ is a fact, and let B be this fact. We will then get (by a similar argument) that if ‘$B\leftrightarrow (\mathbb{G}⌜f⌝\to f)$’, then ‘$B\leftrightarrow (\mathbb{G}⌜B⌝\to f)$’. Since both ‘$\mathbb{G}⌜A⌝\to A$’ and ‘$\mathbb{G}⌜f⌝\to f$’ are unproblematic, then both ‘$B\leftrightarrow (\mathbb{G}⌜B⌝\to A)$’ and ‘$B\leftrightarrow (\mathbb{G}⌜B⌝\to f)$' are unproblematic as well. But it follows then that ‘$~\mathbb{G}⌜B⌝$’ is an ungrounded fact. If we have '$B\leftrightarrow (\mathbb{G}⌜f⌝\to f)$’ and ‘$B\leftrightarrow (\mathbb{G}⌜B⌝\to f)$’, then substituting ‘$\mathbb{G}⌜f⌝\to f$’ for ‘B’ in ‘$B\leftrightarrow (\mathbb{G}⌜B⌝\to f)$’, we get:

(32) $(\mathbb{G}⌜f⌝\to f)\leftrightarrow \left(\mathbb{G}\right(⌜(\mathbb{G}⌜f⌝\to f)⌝\to f\left)\right)$

(If there is such a fact as ‘2+2=5 is not grounded’, then this fact is equivalent to ‘that 2+2=5 is not grounded is not a grounded fact’)¹⁵

I have argued therefore that such a fact as $B\leftrightarrow (\mathbb{G}⌜B⌝\to f)$really exists. But those who believe that all facts are grounded may respond that if being true is equivalent to being grounded (it follows from (FG) and (AFG)), it is not possible to construe such a fact as $B\leftrightarrow (\mathbb{G}⌜B⌝\to f)$. A possibility of such a fact relies on extending Robinson arithmetic with the predicate $\mathbb{G}⌜A⌝$. But since 𝔾 is equivalent to being true (as Loss believes), and thus satisfies an analogue of the T-schema, then 𝔾 cannot be consistently added to Q due to Tarski’s undefinability theorem. However, what is the reason to believe that all facts are grounded? It seems that (LA) gives such a reason. Of course, if there is such a fact as $B\leftrightarrow (\mathbb{G}⌜B⌝\to f)$, then it is not the case that all facts are grounded. But a defender of (LA) may respond that there is no such fact, so if (LA) is correct, then all facts are grounded. However, we have a good reason to believe that (LA) is not correct. (LA) relies on the assumption $\mathbb{G}A\wedge \mathbb{G}~\mathbb{G}A$ - that is, if there is such a fact as $A\wedge ~\mathbb{G}A$, then it is possible for both conjuncts of this fact to be grounded. But this is wrong, because it is not even possible for $\mathbb{G}A\wedge \mathbb{G}~\mathbb{G}A$ to be true. This follows from our argument for (GG). Suppose now, together with Loss, that $\mathbb{G}A\leftrightarrow A$. It follows then that for no true A do we have $A\wedge \mathbb{G}~\mathbb{G}A$. It simply means that if there is such a fact as $A\wedge ~\mathbb{G}A$ (that is, if A is both true and ungrounded), it is not possible for ‘A is an ungrounded fact' to be grounded, so (PG) turns out to be wrong on mere formal grounds. Thus, it is wrong to think that any (possible) fact could be grounded. I conclude therefore that (LA) gives no good reason to think that all facts are grounded. If so, there is no reason to claim that such a fact as ‘$B\leftrightarrow (\mathbb{G}⌜B⌝\to f)$' is impossible because 𝔾 cannot be added to Q.

Moreover, we can respond to those who believe that being grounded is the same as being true with the following example. Let E be some essentialist fact about our system T (in his article, Loss agrees that all essentialist facts are ungrounded (Loss, 2021, p. 37), but notes that (LA) is restricted to non-essentialist facts). It is clear then that T does not ground that E is a grounded fact. Suppose now that the grounding predicate is equivalent to the truth predicate, so we can replace 𝔾 with the truth predicate Tr. We know that, for any A, ‘It is not true that A is true' is equivalent to ‘It is true that A is not true' $(~Tr(⌜Tr⌜A⌝)⌝\leftrightarrow Tr(⌜~Tr⌜A⌝)⌝)$. However, ‘T does not ground that E is grounded' is not equivalent (and does not entail) to ‘T grounds that E is ungrounded' - this is not possible because, as I said before, it is impossible for $\mathbb{G}~\mathbb{G}A$ (if A is true) to be a fact. Moreover, if T grounded the ungroundedness of E, it would mean that E is both grounded and ungrounded, which is impossible. Here is a crucial disanalogy between Tr and 𝔾.

As it has been argued in the paper, for those who endorse a predicational approach and believe that all facts are grounded, it is not possible to prove that there are no ungrounded facts in the same way as (LA) did. So, if a predicational approach to metaphysical grounding is at least minimally plausible (so there are some facts of the form ‘$\mathbb{G}⌜A⌝$’), then (LA) does not achieve its goal - to prove that there are no ungrounded facts (speaking unrestrictedly). Notice also that this argument against (LA) must not be interpreted in the sense that (LA) is not sound because there is such a single self-referential fact as ‘$B\leftrightarrow (\mathbb{G}⌜B⌝\to f)$’. Given that Loss has restricted the scope of (LA) to non- essentialist facts (Loss, 2021, p. 37), a defender of (LA) can easily introduce another restriction and say that (LA) is valid for all facts excluding essentialist and self-referential ones. Thus, if our argument were that (LA) is not sound simply because there is a single self-referential fact B, then it would not be a very promising argument.

Our argument is, in a nutshell, this. Given that there is a single self-referential ungrounded fact ‘$B\leftrightarrow (\mathbb{G}⌜B⌝\to f)$’, we can derive from this fact that there is a plurality of non- self-referential ungrounded facts of the form ‘$~\mathbb{G}⌜A⌝$’ (for any A). Indeed, to show that B is ungrounded is not the same as to show that all such facts as ‘$~\mathbb{G}⌜A⌝$’ are ungrounded. So, the argument is that (LA) is not sound because ‘$~\mathbb{G}⌜A⌝$’ is ungrounded. It is not the case that the argument is that (LA) is not sound just because B is ungrounded.

Research Data Availability:

All research data are available within the main text of the article.

References

• Bennett, K. 2011. By Our Bootstraps. Philosophical Perspectives 25 (1), pp. 27-41.
• Bondar, O. 2022. A Short Argument against Truthmaker Maximalism. Metaphysica 23 (1): 27-36. https://doi.org/10.1515/mp-2021-0017
  » https://doi.org/10.1515/mp-2021-0017
• Boolos, G., J. Burgess, and R. Jeffrey. 2007. Computability and Logic. Cambridge: Cambridge University Press.
• Correia, F. 2021. A New Argument for the Groundedness of Grounding Facts. Erkenntnis: 1-16 DOI: 10.1007/s10670-021-00416-7.
  » https://doi.org/10.1007/s10670-021-00416-7
• deRosset, I. 2013. Grounding Explanations. Philosophers’ Imprint, 13 (7): 1-26.
• Korbmacher, J. 2018. Axiomatic Theories of Partial Ground I. The Base Theory. Journal of Philosophical Logic, 47, pp. 161-191.
• Loss, R. 2021. There are no fundamental facts. Analysis, 81(1): 32-39 DOI 10.1093/analys/anaa008.
  » https://doi.org/10.1093/analys/anaa008
• Rodrigues-Pereyra, G. 2006. Truthmaker maximalism defended. Analysis, 66: 260-264. DOI 10.1111/j.1467-8284.2006.00625.x.
  » https://doi.org/10.1111/j.1467-8284.2006.00625.x
• Sider, T. 2011. Writing the Book of the World. Oxford: Oxford University Press.
• Sider, T. 2020. Ground Grounded. Philosophical Studies, 177: 2073-2096. DOI 10.1007/s11098-018-1204-6.
  » https://doi.org/10.1007/s11098-018-1204-6

• 1
  I will follow the standard view on grounding: the expression “A grounds B” will mean that B is metaphysically dependent on A.
• 2
  By itself, (b) is simply an instance of (PG). See however the discussion at the end of the part 3 of this paper.
• 3
  I assume that both principles are valid and thus harmless. (DA) is similar to modus ponens - if it is a grounded fact that A entails B, then if A is a grounded fact, so is B. (GG) is intuitively highly plausible. It may not be the case that if some fact A is the case, then A is a grounded fact, but it seems to be highly plausible to assume that if there is such a fact as A is a grounded fact, then the fact that A is a grounded fact should be grounded as well.
• 4
  The key difference between the two approaches is as follows: on the predicational approach, 𝔾 takes terms denoting sentences as arguments, while on the operational approach, 𝔾 takes sentences as arguments.
• 5
  It follows therefore that (if our argument is valid) in order to prove that there are no ungrounded facts (unrestrictedly speaking) a defender of (LA) must provide an additional argument showing that a predicational approach to grounding is unacceptable (at least in the context of (LA)) and so “grounded” should not be considered as a predicate.
• 6
  I accept that facts are true propositions: if a proposition 2+2=4 is true, then 2+2=4 is a fact. I also accept that false propositions are ungrounded: if A is 2+2=5, then $⌜A⌝\to f$ (it is not the case that A is both false and grounded, thus it is not the case that A is a fact). I also allow (together with Loss (2021)) a “nested” occurrence of 𝔾, thus allowing expressions of the form “It is a grounded fact that A is a grounded fact” to be well-formed. Notice that this is completely unproblematic given the fact that, in the language of sufficiently expressive and consistent theory (like Q or PA), the notion of ground can be extended to talk about sentences involving themselves. For example, for some A, if “A is grounded” is grounded, then it means that, in Q, we can find some fact F such that “A is grounded” is dependent on F. Notice also that, in T, one might express “A is grounded” in terms of natural numbers, leading to results that can be analyzed using the tools of T.
• 7
  Boolos, Burgess, and Jeffrey (2007, p.221) define the diagonal lemma as follows:
  (DL) Let T be a theory containing Q. Then for any formula B(y) there is a sentence G such that $⊢TG\leftrightarrow B(⌜G⌝)$.
• 8
  Notice that if B is a grounded fact, there is no guarantee that the fact $(\mathbb{G}⌜B⌝\to A)⌝)$ should be grounded as well. This would require to assume the premise according to which if it is a grounded fact that A entails B, then if A is grounded, so is B. But this premise is clearly invalid in our context. Consider the possible scenario in which A is grounded, B is fundamental, and A entails B. Suppose that it is a grounded fact that A entails B. If so, then it should be the case that if A is grounded, then B is grounded as well, contradicting our assumption that B is fundamental. Notice also that if B is instead ungrounded, then (4) is vacuously true.
• 9
  Proof. From (6), $\mathbb{G}⌜B⌝\to \mathbb{G}⌜A⌝$. Since A is f, it is vacuously true that $\mathbb{G}⌜A⌝\to \mathbb{G}⌜B⌝$. From this, we have that $\mathbb{G}⌜B⌝\leftrightarrow \mathbb{G}⌜A⌝$, so $(\mathbb{G}⌜B⌝\to A)\leftrightarrow (\mathbb{G}⌜A⌝\to A)$. Hence, B (that is, $\mathbb{G}⌜B⌝\to A$) must be equivalent to $\mathbb{G}⌜A⌝\to A$. Thus, if $(\mathbb{G}⌜A⌝\to A)$ is grounded, so is B.
• 10
  In short, the argument is this. Assume (1), (DA), (GG), and (AFG). So, from (18), there is no such a fact as ‘$\mathbb{G}(⌜(~\mathbb{G}⌜A⌝)⌝)$’. (DA) and (GG) are both unproblematic. The sentence (1) is indeed true because it is construed by the diagonal lemma. Hence, (AFG) should be untrue. By the way, the important consequence of (1)-(18) is this - even if (AFG) is assumed, we can derive (from (1)) that there are non-self-referential ungrounded facts.
• 11
  Notice that this distinction between grounded facts and groundable facts (facts which are apt for being ground) does not entail, as one might expect, that $\mathbb{G}⌜f⌝$, being false (and thus ungrounded), still must be taken as at least groundable. For instance, take f to be any logical falsity, Then f is indeed ungroundable. But since, as we will see, it is not possible, for any A, that “A is ungrounded” is grounded, then that A is ungrounded must be also ungroundable (independently of whether “A is ungrounded” is true or not).
• 12
  Notice that (PG) is a substantial premise of (LA). Without it, it is not possible to derive a contradiction from the assumption that there is some (true) ungrounded fact (this is (LA)`s assumption for reduction) and thus to conclude that there are no ungrounded facts.
• 13
  For similar arguments, see Barrio & Rodrigues-Pereyra (2015) and Bondar (2021).
• 14
  Someone might argue that this proof relies on the assumption that such a fact as $\left(\mathbb{G}\right(⌜(\mathbb{G}⌜A⌝)\to A)⌝)\to \mathbb{G}⌜A⌝$ does not depend on its existence on such a fact as $B\leftrightarrow (\mathbb{G}⌜B⌝\to A)$.
  But this is not obvious. In such a case, we can derive the same equivalence $\left(\right(\mathbb{G}⌜A⌝\to A)\leftrightarrow (\mathbb{G}⌜B⌝\to A\left)\right)$ in a different way (without using $\left(\mathbb{G}\right(⌜(\mathbb{G}⌜A⌝)\to A)⌝)\to \mathbb{G}⌜A⌝$). So, assume $B\leftrightarrow (\mathbb{G}⌜B⌝\to A)$. We have then the following argument
  $$\begin{gathered} 1)T⊢B\leftrightarrow (\mathbb{G}⌜B⌝\to A)\text{assumption} \\ 2)T⊢\mathbb{G}⌜B⌝\leftrightarrow \left(\mathbb{G}\right(⌜\mathbb{G}⌜B⌝)⌝\to \mathbb{G}⌜A⌝)\text{from (1), by}\left(GB\right)\text{and}\left(DA\right) \\ 3)T⊢\mathbb{G}⌜B⌝\to \mathbb{G}(⌜\mathbb{G}⌜B⌝)⌝(GG\left) \\ 4\right)T⊢\mathbb{G}⌜B⌝\to \mathbb{G}⌜A⌝\text{(2), (3), logic} \\ 5)T⊢\mathbb{G}⌜A⌝\to (\mathbb{G}(⌜\mathbb{G}⌜B⌝)⌝\to \mathbb{G}⌜A⌝)\text{tautology} \\ 6)T⊢\mathbb{G}⌜A⌝\to \mathbb{G}⌜B⌝\text{(5), (2), conditional proof} \\ 7)T⊢\mathbb{G}⌜B⌝\leftrightarrow \mathbb{G}⌜A⌝\text{(4), (6), logic} \\ 8)T⊢(\mathbb{G}⌜B⌝\to A)\leftrightarrow (\mathbb{G}⌜A⌝\to A)\text{(7), logic} \end{gathered}$$
  So, $B\leftrightarrow (\mathbb{G}⌜B⌝\to A)$ is equivalent to $B\leftrightarrow (\mathbb{G}⌜A⌝\to A)$, and thus unproblematic.
• 15
  Loss (2021) has also presented a slightly different argument which relies on (PG+) instead of (PG)
  $(PG+)(◊\mathbb{G}A\wedge (A\wedge B\left)\right)\to ◊(\mathbb{G}A\wedge B)$
  (If it is the case that A ∧ 𝐵 and the fact that A is ‘groundable’, then it is possible for B to be the case and for the fact that A to be a grounded fact).
  This argument also relies on the assumption that ‘there seem to be no truths that can clearly and uncontroversially said to be fundamental’ (Loss, 2021: 38). But since we have found such a truth which seems to be fundamental - ‘$\mathbb{G}⌜f⌝\to f$’ - our argument against (PG) is equally an argument against (PG+)
• Article info
  CDD: 110
• Funding:
  No funding agency

Publication Dates

• Publication in this collection
  07 Nov 2025
• Date of issue
  2025

History

• Received
  30 Aug 2024
• Reviewed
  24 May 2025
• Accepted
  08 Sept 2025