Towards a theory of ground-theoretic content

Abstract

A lot of research has recently been done on the topic of ground, and in particular on the logic of ground. According to a broad consensus in that debate, ground is hyperintensional in the sense that even logically equivalent truths may differ with respect to what grounds them, and what they ground. This renders pressing the question of what we may take to be the ground-theoretic content of a true statement, i.e. that aspect of the statement’s overall content to which ground is sensitive. I propose a novel answer to this question, namely that ground tracks how, rather than just by what, a statement is made true. I develop that answer in the form of a formal theory of ground-theoretic content and show how the resulting framework may be used to articulate plausible theories of ground, including in particular a popular account of the grounds of truth-functionally complex truths that has proved difficult to accommodate on alternative views of content.

Appendix

We begin with the definition of a mode-space.

Definition 1

(Mode-spaces) A mode-space is a pair $\langle M,\,V\rangle$ such that

• 1. M is non-empty

• 2. V is a non-empty, partial function taking non-empty sequences of non-empty subsets of M into members of M

• 3. the domain of V is closed under non-empty subsequences and countable concatenations of sequences

• 4. $V(\gamma _1\,^\frown \gamma _2\,^\frown \ldots ) = V(\delta _1^\frown \delta _2^\frown \ldots )$ whenever $V(\gamma _1) = V(\delta _1), V(\gamma _2) = V(\delta _2), \ldots$

Given a fixed mode-space $\langle M,\, V\rangle$, we call a mode any member of M. Any mode which is the value of V for some argument is called derivative, every other mode is called fundamental. We write $M^D$ ($M^F$) for the set of derivative (fundamental) modes. Any subset of M will be called a content, and their set will be denoted by ${\mathscr {C}}$. The contents containing only derivative modes will themselves be called derivative, and the other contents will be called fundamental. We write ${\mathscr {C}}^D$ (${\mathscr {C}}^F$) for the set of derivative (fundamental) contents. Any sequence for which V is defined is called a via-sequence. Since the domain of V is closed under non-empty subsequences, for any content P that is an element of some via-sequence, there is also the via-sequence $\langle P\rangle$ corresponding to the mode of verifying via P. We shall call any such content raisable and denote their set by ${\mathscr {R}}$. We call a ground-set of a derivative mode m any set of contents that underlies a content-sequence $\gamma$ with $V(\gamma ) = m$.

We say that a derivative mode m is a fusion of the sequence of derivative modes $\langle m_1,\, m_2,\, \ldots \rangle$ iff there are via-sequences $\gamma _1,\, \gamma _2, \ldots$ such that $V(\gamma _1) = m_1,~V(\gamma _2) = m_2,~\ldots$, and $m = V(\gamma _1\,^\frown \gamma _2\,^\frown \ldots )$.⁴² By the third constraint on mode-spaces, fusions will always exist, and by the fourth constraint, they will be unique. For the sequence of derivative modes $\langle m_1,\, m_2,\, \ldots \rangle$, we write its fusion as $\bigsqcup \langle m_1, \,m_2,\, \ldots \rangle$ and also as $m_1 \sqcup m_2 \sqcup \ldots$ We note a first central lemma (the proof is elementary):

Lemma 1

If $\varGamma _1,\, \varGamma _2, \ldots$ are ground-sets of $m_1,\, m_2, \ldots$, respectively, then $\varGamma _1 \cup \varGamma _2 \cup \ldots$ is a ground-set of $m_1 \sqcup m_2 \sqcup \ldots$

For convenience, we repeat the definitions of the grounding relationships and of the truth-functions.

Definition 2

Let $\varGamma \subseteq {\mathscr {C}}$ and $P \in {\mathscr {C}}$. Then

As before, we write $\varGamma < \{P_1,\, P_2,\, \ldots \}$ to abbreviate that for some $\varGamma _1, \,\varGamma _2,\, \ldots$, $\varGamma = \varGamma _1 \cup \varGamma _2 \cup \ldots$ and $\varGamma _1< P_1,~\varGamma _2 < P_2,~\ldots$, and similarly for the case of $\varGamma \le \{P_1,\, P_2,\, \ldots \}$.

Definition 3

For $P,\,Q \in {\mathscr {C}}^D$:

Definition 4

For $P, Q \in {\mathscr {R}}$:

We shall mainly be interested in mode-spaces in which the set of raisable contents is closed under these operations.

Definition 5

A mode-space is called complete iff $P \wedge Q \in {\mathscr {R}}$, $P \vee Q \in {\mathscr {R}}$, and $\uparrow P \in {\mathscr {R}}$ whenever $P,\,Q \in {\mathscr {R}}$.

Lemma 2

(Introduction Lemma) In any complete mode-space, for $\varGamma \subseteq {\mathscr {R}}$ and $P, \,Q \in {\mathscr {R}}$:

• 1. If $\varGamma \le P$, then $\varGamma < \uparrow P$.

• 2. If $\varGamma < \{P,\, Q\}$, then $\varGamma < P \sqcup Q$.

• 3. If $\varGamma < P$ or $\varGamma < Q$ or $\varGamma < \{P,\, Q\}$, then $\varGamma < P + Q$

• 4. If $\varGamma \le \{P,\, Q\}$, then $\varGamma < P \wedge Q$

• 5. If $\varGamma \le P$ or $\varGamma \le Q$ or $\varGamma < \{P,\, Q\}$, then $\varGamma < P \vee Q$.

Proof

For 1: Suppose $\varGamma \le P$. Then either (i) $\varGamma = \{P\}$, or (ii) $\varGamma < P$, or (iii) $\varGamma \setminus \{P\} < P$. Suppose (i). By definition of $\uparrow$, $V\langle P\rangle \in \uparrow P$. Since $\{P\}$ is the set underlying $\langle P\rangle$, it is a ground-set of $V\langle P\rangle$, so $\varGamma < P$. Suppose (ii). Then $\varGamma$ is a ground-set of some mode $m \in P$. But then by definition of $\uparrow$, it follows that $m \in \,\uparrow P$, and hence $\varGamma < P$. Suppose (iii). Then for some $m \in P$, $\varGamma \setminus \{P\}$ is a ground-set of m. But then by definition of $\uparrow$, it follows that $V\langle P\rangle \sqcup m \in \,\uparrow P$. By Lemma 1, $\varGamma \setminus \{P\} \cup \{P\} = \varGamma$ is a ground-set of $V\langle P\rangle \sqcup m$, hence $\varGamma < P$.

For 2: Suppose $\varGamma < \{P,\, Q\}$. Let $\varGamma _P < P$ and $\varGamma _Q < Q$ with $\varGamma _P \cup \varGamma _Q = \varGamma$. Then $\varGamma _P$ is a ground-set of some mode $m_P \in P$ and $\varGamma _Q$ is a ground-set of some mode $m_Q \in Q$. By definition of $\sqcup$, $m_P \sqcup m_Q \in P \sqcup Q$, and by Lemma 1, $\varGamma _P \cup \varGamma _Q = \varGamma$ is a ground-set of $m_P \sqcup m_Q$, hence $\varGamma < P \sqcup Q$.

For 3: Suppose $\varGamma < P$. Then there is a mode $m \in P$ of which $\varGamma$ is a ground-set. By definition of $+$, the same mode is included in $P + Q$, hence $\varGamma < P + Q$. Suppose $\varGamma < Q$. Then by the same reasoning, $\varGamma < P + Q$. Finally, suppose $\varGamma < \{P,\, Q\}$. Then by part 2., $\varGamma < P \sqcup Q$, and hence by definition of $+$, $\varGamma < P + Q$.

For 4: Suppose $\varGamma \le \{P,\, Q\}$. Let $\varGamma _P \le P$ and $\varGamma _Q \le Q$ with $\varGamma _P \cup \varGamma _Q = \varGamma$. By part 1., $\varGamma _P < \uparrow P$ and $\varGamma _Q < \uparrow Q$, hence $\varGamma < \{\uparrow P, \,\uparrow Q\}$, and by part 2., $\varGamma < P \wedge Q$.

For 5: If $\varGamma \le P$ or $\varGamma \le Q$, then by part 2., $\varGamma < \uparrow P$ or $\varGamma <\uparrow Q$. If $\varGamma \le \{P,\, Q\}$, then by the reasoning in part 4., $\varGamma < \{\uparrow P,\, \uparrow Q\}$. So by part 3., $\varGamma < P \vee Q$. $\square$

Definition 6

A mode-space is called constrained iff $V(\gamma ) = V(\delta )$ only if the same ground-set corresponds to $\gamma$ and $\delta$.

In a constrained mode-space, every derivative mode m corresponds to a unique ground-set, which we denote by |m|.

Lemma 3

(Elimination Lemma) In any constrained and complete mode-space, for $\varGamma \subseteq {\mathscr {R}}$ and $P,\, Q \in {\mathscr {R}}$:

• 1. If $\varGamma < \,\uparrow P$, then $\varGamma \le P$

• 2. If $\varGamma < P \sqcup Q$, then $\varGamma < \{P,\, Q\}$

• 3. If $\varGamma < P + Q$, then $\varGamma < P$ or $\varGamma < Q$ or $\varGamma < \{P,\, Q\}$

• 4. If $\varGamma < P \wedge Q$, then $\varGamma \le \{P,\, Q\}$

• 5. If $\varGamma < P \vee Q$, then $\varGamma \le P$ or $\varGamma \le Q$ or $\varGamma \le \{P,\, Q\}$

Proof

For 1: Suppose $\varGamma < \,\uparrow P$. Let $m \in \,\uparrow P$ with $\varGamma = |m|$. By definition of $\uparrow$, there are three cases. (i) $m \in P\cap M^D$. Then $\varGamma < P$ and hence $\varGamma \le P$. (ii) $m = V\langle P\rangle$. Then $\varGamma = |V\langle P\rangle | = \{P\}$, so again $\varGamma \le P$. (iii) $m = V\langle P\rangle \sqcup n$ for some $n \in P\cap M^D$. Since $n \in P$, $|n| < P$. By Lemma 1, $\varGamma = |n| \cup \{P\}$. Then either $|n| = \varGamma$ or $|n| = \varGamma \setminus \{P\}$, so either $\varGamma < P$ or $\varGamma \setminus \{P\} < P$, and hence $\varGamma \le P$.

For 2: Suppose $\varGamma < P \sqcup Q$. Let $|m| = \varGamma$ and $m \in P \sqcup Q$. Then $m = m_P \sqcup m_Q$ for some $m_P \in P$ and $m_Q \in Q$. So $|m_P| < P$ and $|m_Q| < Q$. But $\varGamma = |m_P| \cup |m_Q|$, and hence $\varGamma < \{P,\, Q\}$.

For 3: Suppose $\varGamma < P + Q$. Then it is immediate from the definition of $+$ that $\varGamma < P$ or $\varGamma < Q$ or $\varGamma < P \sqcup Q$, which by part 2. implies $\varGamma < \{P,\, Q\}$.

For 4: Suppose $\varGamma < P \wedge Q$. By part 2., $\varGamma < \{\uparrow P,\, \uparrow Q\}$. So let $\varGamma _P < \uparrow P$ and $\varGamma _Q < \uparrow Q$ with $\varGamma = \varGamma _P \cup \varGamma _Q$. By part 1., $\varGamma _P \le P$ and $\varGamma _Q \le Q$, hence $\varGamma \le \{P,\, Q\}$.

For 5: Suppose $\varGamma < P \vee Q$. By part 3., there are three cases. (i) $\varGamma < \uparrow P$. Then by part 1., $\varGamma \le P$. (ii) $\varGamma < \uparrow Q$. Then by part 1. again, $\varGamma \le Q$. (iii) $\varGamma \le \{\uparrow P,\, \uparrow Q\}$. Then by the reasoning in part 4., $\varGamma \le \{P,\, Q\}$. $\square$

We move on to the case of bilateral contents. We define the truth-functional operations on bilateral contents as well as the notion of ground-theoretic equivalence ($\approx$).

Definition 7

For ${\mathbf {P}},\, {\mathbf {Q}}\in {\mathscr {R}}\times {\mathscr {R}}$:

Definition 8

For ${\mathbf {P}},\, {\mathbf {Q}}\in {\mathscr {R}}\times {\mathscr {R}}$: ${\mathbf {P}}\approx {\mathbf {Q}}:\leftrightarrow$ for all ${\varvec{\varGamma }}$: ${\varvec{\varGamma }}< {\mathbf {P}}$ iff ${\varvec{\varGamma }}< {\mathbf {Q}}$, and for all ${\varvec{\varDelta }}$ and ${\mathbf {R}}$: ${\varvec{\varDelta }}, {\mathbf {P}}< {\mathbf {R}}$ iff ${\varvec{\varDelta }}, {\mathbf {Q}}< {\mathbf {R}}$.

Recall that ground between bilateral contents is defined simply as ground between the positive components. As a result, bilateral contents will be ground-theoretically equivalent (written $\approx$) provided their positive components are the same.

Lemma 4

(DeMorgan) For ${\mathbf {P}},\, {\mathbf {Q}}\in {\mathscr {R}}\times {\mathscr {R}}$:

• 1. $\lnot ({\mathbf {P}}\wedge {\mathbf {Q}}) \approx \lnot {\mathbf {P}}\vee \lnot {\mathbf {Q}}$

• 2. $\lnot ({\mathbf {P}}\vee {\mathbf {Q}}) \approx \lnot {\mathbf {P}}\wedge \lnot {\mathbf {Q}}$

Proof

By application of the definitions.

For 1: $(\lnot ({\mathbf {P}}\wedge {\mathbf {Q}}))^+ = ({\mathbf {P}}\wedge {\mathbf {Q}})^- = {\mathbf {P}}^- \vee {\mathbf {Q}}^- = (\lnot {\mathbf {P}})^+ \vee (\lnot {\mathbf {Q}})^+ = (\lnot {\mathbf {P}}\vee \lnot {\mathbf {Q}})^+$

For 2: $(\lnot ({\mathbf {P}}\vee {\mathbf {Q}}))^+ = ({\mathbf {P}}\vee {\mathbf {Q}})^- = {\mathbf {P}}^- \wedge {\mathbf {Q}}^- = (\lnot {\mathbf {P}})^+ \wedge (\lnot {\mathbf {Q}})^+ = (\lnot {\mathbf {P}}\wedge \lnot {\mathbf {Q}})^+$ $\square$

As an immediate consequence of the definition of ground on bilateral contents and the previous lemmata, we obtain

Theorem 5

(Truth-functions and ground, introduction) In any complete mode-space, for ${\varvec{\varGamma }}\subseteq {\mathscr {R}}\times {\mathscr {R}}$ and ${\mathbf {P}},\, {\mathbf {Q}}\in {\mathscr {R}}\times {\mathscr {R}}$:

• 1. If ${\varvec{\varGamma }}\le \{{\mathbf {P}},\, {\mathbf {Q}}\}$, then ${\varvec{\varGamma }}< {\mathbf {P}}\wedge {\mathbf {Q}}$

• 2. If ${\varvec{\varGamma }}\le {\mathbf {P}}$ or ${\varvec{\varGamma }}\le {\mathbf {Q}}$ or ${\varvec{\varGamma }}< \{{\mathbf {P}},\, {\mathbf {Q}}\}$, then ${\varvec{\varGamma }}< {\mathbf {P}}\vee {\mathbf {Q}}$.

• 3. If ${\varvec{\varGamma }}\le {\mathbf {P}}$, then ${\varvec{\varGamma }}< \lnot \lnot {\mathbf {P}}$

• 4. If ${\varvec{\varGamma }}\le \lnot {\mathbf {P}}$ or ${\varvec{\varGamma }}\le \lnot {\mathbf {Q}}$ or ${\varvec{\varGamma }}\le \{\lnot {\mathbf {P}},\, \lnot {\mathbf {Q}}\}$, then ${\varvec{\varGamma }}< \lnot ({\mathbf {P}}\wedge {\mathbf {Q}})$

• 5. If ${\varvec{\varGamma }}\le \{\lnot {\mathbf {P}},\, \lnot {\mathbf {Q}}\}$, then ${\varvec{\varGamma }}< \lnot ({\mathbf {P}}\vee {\mathbf {Q}})$

Theorem 6

(Truth-functions and ground, elimination) In any complete and constrained mode-space, for ${\varvec{\varGamma }}\subseteq {\mathscr {R}}\times {\mathscr {R}}$ and ${\mathbf {P}},\, {\mathbf {Q}}\in {\mathscr {R}}\times {\mathscr {R}}$:

• 1. If ${\varvec{\varGamma }}< {\mathbf {P}}\wedge {\mathbf {Q}}$, then ${\varvec{\varGamma }}\le \{{\mathbf {P}},\, {\mathbf {Q}}\}$

• 2. If ${\varvec{\varGamma }}< {\mathbf {P}}\vee {\mathbf {Q}}$, then ${\varvec{\varGamma }}\le {\mathbf {P}}$ or ${\varvec{\varGamma }}\le {\mathbf {Q}}$ or ${\varvec{\varGamma }}\le \{{\mathbf {P}},\, {\mathbf {Q}}\}$

• 3. If ${\varvec{\varGamma }}< \lnot \lnot {\mathbf {P}}$, then ${\varvec{\varGamma }}\le {\mathbf {P}}$

• 4. If ${\varvec{\varGamma }}< \lnot ({\mathbf {P}}\wedge {\mathbf {Q}})$, then ${\varvec{\varGamma }}\le \lnot {\mathbf {P}}$ or ${\varvec{\varGamma }}\le \lnot {\mathbf {Q}}$ or $\varGamma \le \{\lnot {\mathbf {P}},\, \lnot {\mathbf {Q}}\}$

• 5. If ${\varvec{\varGamma }}< \lnot ({\mathbf {P}}\vee {\mathbf {Q}})$, then ${\varvec{\varGamma }}\le \{\lnot {\mathbf {P}},\, \lnot {\mathbf {Q}}\}$

We turn now to the structural properties of ground.

Definition 9

A proposition P is

• irreflexive iff: P does not occur in $\gamma$ whenever $V(\gamma ) \in P$,

• closed iff: P contains a mode with ground-set $\varGamma _1 \cup \varGamma _2 \cup \ldots$ whenever P contains modes with ground-sets $\varGamma _1, \,\varGamma _2, \ldots$,

• transitive iff: P includes a mode with ground-set $\varGamma ,\, \varDelta$ whenever P includes a mode with ground-set $\varDelta ,\, Q$ and Q includes a mode with ground-set $\varGamma$.

• normal iff: irreflexive, closed, and transitive.

Theorem 7

(Structural Principles for Normal Propositions) In any constrained mode-space, for normal propositions $P,\, P_1,\, P_2,\, \ldots ,\, Q,\, R$ and sets of normal propositions $\varGamma ,\, \varGamma _1,\, \varGamma _2,\, \ldots ,\, \varDelta$:

• 1. $P \not \prec P$

• 2. If $\varGamma _1 < P$, $\varGamma _2 <P$, ..., then $\varGamma _1,\, \varGamma _2,\, \ldots < P$.

• 3. If $\varGamma \le P$ and $Q \prec P$ for all $Q \in \varGamma$, then $\varGamma < P$.

• 4. If $\varGamma ,\, P \le P$, then $\varGamma \le P$.

• 5. If $\varGamma _1 \le P,~\varGamma _2 \le P,~\ldots$, then $\varGamma _1 \cup \varGamma _2 \cup \ldots \le P$.

• 6. If $\varGamma < P$ and $\varDelta ,\, P < Q$, then $\varGamma ,\, \varDelta < Q$.

• 7. If $\varGamma _1 \le P_1$, $\varGamma _2 \le P_2$, ..., and $P_1,\, P_2,\, \ldots \le Q$ then $\varGamma _1,\, \varGamma _2,\, \ldots \le Q$

• 8. If $P \preceq Q$ and $Q \prec R$ then $P \prec R$

• 9. If $P \prec Q$ and $Q \preceq R$ then $P \prec R$

• 10. If $P \preceq Q$ and $Q \preceq R$ then $P \preceq R$

Proof

1–4., and 6. were established in Sect. 5.4 above.

For 5., suppose $\varGamma _1 \le P,~\varGamma _2 \le P, \ldots$ Consider all the $\varGamma _i$ which are distinct from $\{P\}$. By definition of $\le$ and 1., $\varGamma _i\setminus \{P\} < P$ for any such $\varGamma _i$. If there are any such $\varGamma _i$, let $\varGamma ^\prime$ be the result of removing P from their union. By 2, $\varGamma ^\prime < P$, so $\varGamma _1 \cup \varGamma _2 \cup \ldots = \varGamma ^\prime \cup \{P\} \le P$. If there are no such $\varGamma _i$, then $\varGamma _1 \cup \varGamma _2 \cup \ldots = \{P\} \le P$.

For 7., suppose $\varGamma _1 \le P_1$, $\varGamma _2 \le P_2$, ..., and $P_1,\, P_2,\, \ldots \le Q$. We confine ourselves to showing that $\varGamma _1 \cup \{P_2,\, \ldots \} \le Q$ follows. By applying the same reasoning repeatedly and making use of the reflexivity of $\le$, we may establish the desired conclusion. Now either (a) $\varGamma _1 = \{P_1\}$, or (b) $\varGamma _1 < P_1$, or (c) $\{P_1\} \subset \varGamma _1$ and $\varGamma _1\setminus \{P_1\} < P_1$. If (a), then our intended result $\varGamma _1 \cup \{P_2,\, \ldots \} \le Q$ follows immediately. Suppose that (b). Then if (b1) $P_1 = Q$, we have $\varGamma _1 < Q$ and hence $\varGamma _1 \le Q$. Moreover, by 4., we have $\{P_2,\, \ldots \} \le Q$. So by 5., $\varGamma _1 \cup \{P_2,\, \ldots \} \le Q$ follows. But if (b2) $P_1 \ne Q$, then $P_1 \in \{P_1,\, P_2,\, \ldots \}\setminus \{Q\}$ and $\{P_1,\, P_2,\, \ldots \}\setminus \{Q\} < Q$, so by 6., $\varGamma _1 \cup \{P_2,\, \ldots \}\setminus \{Q\} < Q$, hence $\varGamma _1 \cup \{P_2,\, \ldots \} \le Q$. Suppose finally that (c). Then if (c1) $P_1 = Q$, $\varGamma _1 \le Q$, and so $\varGamma _1 \cup \{P_2,\, \ldots \} \le Q$ follows as in case (b1). But if (c2) $P_1 \ne Q$, then by similar reasoning as in case (b2), $\varGamma _1\setminus \{P_1\} \cup \{P_1,\, P_2,\, \ldots \} = \varGamma _1 \cup \{P_2, \ldots \} \le Q$.

For 8., suppose $P \preceq Q$ and $Q \prec R$. If $P = Q$, then $P \prec R$ follows immediately. So suppose $P \ne Q$, and hence $P \prec Q$. Let $m \in Q$ be such that $P \in |m|$, and let $n \in R$ be such that $Q \in |n|$. Then $|m| \cup \{P\} < Q$ and $|n| \cup \{Q\} < R$, so by 6, $|n| \cup |m| \cup \{P\} < R$, so $P \prec R$.

For 9., suppose $P \prec Q$ and $Q \preceq R$. If $Q = R$, then $P \prec R$ follows immediately. So suppose $Q \ne R$, and hence $Q \prec R$. Then by the same reasoning as before, $P \prec R$.

For 10., suppose $P \preceq Q$ and $Q \preceq R$. If either $P \prec Q$ or $Q \prec R$, then it follows by the previous results that $P \prec R$, and hence $P \preceq R$. If neither $P \prec Q$ nor $Q \prec R$, then $P = Q$ and $Q = R$, hence $P = R$, and thus again $P \preceq R$. $\square$

Together with the Subsumption principles and the Identity principle $P \le P$ for weak ground established in Sect. 5.2, parts 1, 3, 7–10 correspond to the basic rules of the logic proposed in Fine (2012c), which is thereby shown to be sound with respect to the class of normal propositions in a constrained mode-space.

We call a bilateral content irreflexive, closed, transitive, or normal, just in case its positive component has the relevant property. We wish then to show that any truth-functional combinations of normal propositions are themselves normal. By the definitions of the truth-functional operations, it suffices to show that for unilateral contents, normality is preserved under $\wedge , \vee$, and $\uparrow$.

Theorem 8

For $P,\, Q \in {\mathscr {R}}$, in some constrained mode-space: If P and Q are normal, then so are $P \wedge Q$, $P \vee Q$, and $\uparrow P$.

Proof

By reducing failures for $P \wedge Q$, $P \vee Q$, $\uparrow P$ of the characteristic principles of irreflexivity, amalgamation, and transitivity (1, 2 and 6 in Theorem 7) to failures of the corresponding principles for P or Q. We give the proof for $P \wedge Q$; the other cases may be established by parallel means.

For irreflexivity, suppose that $P \wedge Q \prec P \wedge Q$. Then for some $\varGamma$: $\varGamma ,\, P \wedge Q < P \wedge Q$. Then by the Elimination Lemma, $\varGamma ,\, P \wedge Q \le \{P,\, Q\}$, and hence either (i) $P \wedge Q \preceq P$ or (ii) $P \wedge Q \preceq Q$. Suppose (i). Then since $P \prec P \wedge Q$, it follows by transitivity of P that $P \prec P$, in contradiction to P’s irreflexivity.

For amalgamation, suppose that $\varGamma _1 < P \wedge Q$, $\varGamma _2 < P \wedge Q$, .... For each $\varGamma _i$: $\varGamma _i \le \{P,\, Q\}$. So let $\varGamma _{i_P} \le P$ and $\varGamma _{i_Q} \le Q$ with $\varGamma _i = \varGamma _{i_P} \cup \varGamma _{i_Q}$ for all i. Then by weak ground amalgamation for P and Q, $\varGamma _{1_P},\, \varGamma _{2_P},\, \ldots \le P$, and $\varGamma _{1_Q},\, \varGamma _{2_Q},\, \ldots \le Q$, so $\varGamma _1,\, \varGamma _2,\, \ldots \le \{P,\, Q\}$, and hence $\varGamma _1,\, \varGamma _2,\, \ldots < P \wedge Q$.

For transitivity, suppose that $\varDelta ,\, R < P \wedge Q$ and $\varGamma < R$. Then $\varDelta ,\, R \le \{P,\, Q\}$. So we may write $\varDelta ,\, R$ as the union of weak full grounds of respectively P and Q. R will be a member of at least one of these. By replacing R with $\varGamma$, using the transitivity of P and Q, we may infer $\varGamma ,\, \varDelta < \{P,\, Q\}$, and hence $\varGamma ,\, \varDelta < P \wedge Q$. $\square$