The Ref Proof-Checker and Its “Common Shared Scenario”

Abstract

In his later years, Jack Schwartz devoted much energy to the implementation of a proof-checker based on set theory and to the preparation of a large script file to be fed into it. His goal was to attain a verified proof of the Cauchy integral theorem of complex analysis.

This contribution to the memorial volume for Jack reflects that effort: it briefly reports the chronicle of his proof-checking project and highlights some features of the system as implemented; in an annex, it presents a proof scenario leading from bare set theory to two basic theorems about claw-free graphs.

Appendix: Claw-Free Graphs as Sets

• Eugenio G. Omodeo &
• Alexandru I. Tomescu 

1. Dipartimento di Matematica e Informatica, Università di Trieste, Via Valerio, 12/1, 34127, Trieste, Italy

   Eugenio G. Omodeo

2. Dipartimento di Matematica e Informatica, Università di Udine, Via delle Scienze, 206, 33100, Udine, Italy

   Alexandru I. Tomescu

3. Faculty of Mathematics and Computer Science, University of Bucharest, Str. Academiei, 14, 010014, Bucharest, Romania

   Alexandru I. Tomescu

This scenario contains the formal proofs, checked by J. T. Schwartz’s proof-verifier Ref, of two classical results on connected claw-free graphs; namely, that any such graph:

• owns a perfect matching if its number of vertices is even,

• has a Hamiltonian cycle in its square if it owns three or more vertices.

  The original proofs (cf. [14, 15] and [11]) referred to undirected graphs, the ones to be presented refer to a special class of digraphs whose vertices are hereditarily finite sets and whose edges reflect the membership relation. Ours is a legitimate change of perspective in the light of [12], as we will briefly explain at the end.

  To make our formal development self-contained, we proceed from the bare set-theoretic foundation built into Ref (cf. [13]). The lemmas exploited without proof in what follows are indeed very few, and their full proofs are available in [13].

1.1 A.1 Basic Laws on the Union-Set Global Operation

$\hphantom{i}\begin{array}{lllcl}{ \!\!\!\textsc {Def}\ {\mathsf{unionset}}{:}\ {\mbox {[Members of members of a set]}}}&{}&{ \mbox {\boldmath $\bigcup $}\mathsf{X}}&{ \mathop {=}\nolimits _{\mathrm {Def}}}&{\left \{\mathsf{u} :\:\mathsf{v} \in \mathsf{X} ,\mathsf{u} \in \mathsf{v}\right \}}\end{array}$

The proof of the following claim, that the union set of a set s is the set-theoretic ‘least upper bound’ of all its elements, can be found in [13, p. 387].

Ref has the built-in ability to reduce $\{\mathsf{f}(v):\:v \in \emptyset \}$ to ∅ and $\{\mathsf{f}(v):\:v \in \{\mathsf{x}_{0}\}\}$ to {f(x ₀)}; hence we are left with only the doubleton to consider. Let c belong to one of $\{\mathsf{f}(v):\:v \in \{\mathsf{x}_{0},\mathsf{x}_{1}\}\}$ and {f(x ₀),f(x ₁)} but not to the other. After excluding, through variable-substitution, the case $\mathsf{c} \notin \{\mathsf{f}(v):\:v \in \{\mathsf{x}_{0},\mathsf{x}_{1}\}\}$, we easily exclude both possibilities c=f(x ₀) and c=f(x ₁), through variable-substitution and equality propagation.

Under the assumption that $\mathsf{z}_{0} \mbox {\bf $\:$=$\:$}\left \{\mathsf{x}_{0} ,\mathsf{y}_{0}\right \} \mbox {$\;${\footnotesize $\&$}$\;$}\mbox {\boldmath $\bigcup $}\mathsf{z}_{0} \mathbf {\mbox {\bf $\:\not $=$\:$}}\mathsf{x}_{0} \mbox {$\;${\footnotesize $\cup $}$\;$}\mathsf{y}_{0}$ can hold, two citations of Theorem 2 readily yield $\mathsf{x}_{0} \mbox {\boldmath $\:\subseteq \:$}\mbox {\boldmath $\bigcup $}\mathsf{z}_{0}$ and $\mathsf{y}_{0} \mbox {\boldmath $\:\subseteq \:$}\mbox {\boldmath $\bigcup $}\mathsf{z}_{0}$.

A third citation of the same Theorem 2 enables us to derive from $\mbox {\boldmath $\bigcup $}\mathsf{z}_{0} \mathbf {\mbox {\bf $\:\not $=$\:$}}\mathsf{x}_{0} \mbox {$\;${\footnotesize $\cup $}$\;$}\mathsf{y}_{0}$ that some element of $\mathsf{z}_{0} \mbox {\bf $\:$=$\:$}\left \{\mathsf{x}_{0} ,\mathsf{y}_{0}\right \}$ is not included in $\mathsf{x}_{0} \mbox {$\;${\footnotesize $\cup $}$\;$}\mathsf{y}_{0}$, which is manifestly absurd.

Arguing by contradiction, let x ₀,y ₀ be a counterexample, so that in either one of $\mbox {\boldmath $\bigcup $}(\mathsf{x}_{0} \mbox {$\;${\footnotesize $\cup $}$\;$}\left \{\mathsf{y}_{0}\right \})$ and $\mathsf{y}_{0} \mbox {$\;${\footnotesize $\cup $}$\;$}\mbox {\boldmath $\bigcup $}\mathsf{x}_{0}$ there is an a not belonging to the other set. Taking the definition of $\mbox {\boldmath $\bigcup $}$ into account, by monotonicity we must exclude the possibility that $\mathsf{a} \in \mbox {\boldmath $\bigcup $}\mathsf{x}_{0} \mbox {\boldmath $\setminus $}\mbox {\boldmath $\bigcup $}(\mathsf{x}_{0} \mbox {$\;${\footnotesize $\cup $}$\;$}\left \{\mathsf{y}_{0}\right \})$; through variable-substitution, we must also discard the possibility that $\mathsf{a} \in \mbox {\boldmath $\bigcup $}(\mathsf{x}_{0} \mbox {$\;${\footnotesize $\cup $}$\;$}\left \{\mathsf{y}_{0}\right \})\mbox {\boldmath $\setminus $}\mbox {\boldmath $\bigcup $}\mathsf{x}_{0} \mbox {\boldmath $\setminus $}\mathsf{y}_{0}$.

The only possibility left, namely that $\mathsf{a} \in \mathsf{y}_{0} \mbox {\boldmath $\setminus $}\mbox {\boldmath $\bigcup $}(\mathsf{x}_{0} \mbox {$\;${\footnotesize $\cup $}$\;$}\left \{\mathsf{y}_{0}\right \})$, is also manifestly absurd. This contradiction leads us to the desired conclusion.

1.2 A.2 Transitive Sets

Any strict subset of a transitive set t, owns a subset in t which does not belong to it.

The ‘(⋆)’ context restriction in the following three steps serves to hide the semantics of arb: which, to the limited extent necessary here, has been captured by the preceding Peddicord’s lemma.

Assuming that s is transitive, that t equals s deprived of some sources and that t is not transitive, there must be an element y of t which is not a subset of t, so that a z∈y exists which does not belong to t. Due to the transitivity of s, y is included in s and hence z belongs to s; hence, under the assumption that $\mathsf{s} \mbox {\boldmath $\setminus $}\mathsf{t}$ and $\mbox {\boldmath $\bigcup $}\mathsf{s}$ are disjoint, z does not belong to $\mbox {\boldmath $\bigcup $}\mathsf{s}$.

However, this is untenable.

1.3 A.3 Basic Laws on the Finitude Property

To begin developing an acceptable treatment of finiteness without much preparatory work, we adopt here the definition (reminiscent of Tarski’s 1924 paper “Sur les ensembles finis”): a set F is finite if every non-null family of subsets of F owns an inclusion-minimal element. This notion is readily specified in terms of the power-set operator, as follows:

The lemma on the monotonicity of finitude and the ${\textsc {Theory}\;}$ of finite induction displayed below are proved in full—together with various other laws on finiteness which we will not need here—in [13, pp. 405–407].

1.4 A.4 Some Combinatorics of the Union-Set Operation

Preparatory to a technique to which we will resort for extending perfect matchings, we introduce the following trivial combinatorial lemma:

For, supposing the contrary, $\mbox {\boldmath $\bigcup $}(\mathsf{m}\mbox {$\;${\footnotesize $\cup $}$\;$}\left \{\mathsf{x}\mbox {$\;${\footnotesize $\cup $}$\;$}\left \{\mathsf{y}\right \}\right \})$ would differ from each of $\mathsf{s}\mbox {\boldmath $\setminus $}\left \{\mathsf{s}\right \}$, $\mathsf{s}\mbox {\boldmath $\setminus $}\left \{\mathsf{c}\right \}$, and $\mathsf{s}\mbox {\boldmath $\setminus $}\left \{\mathsf{v}\right \}$, the first of which equals s. Thanks to Theorem 2e, we can rewrite $\mbox {\boldmath $\bigcup $}(\mathsf{m}\mbox {$\;${\footnotesize $\cup $}$\;$}\left \{\mathsf{x}\mbox {$\;${\footnotesize $\cup $}$\;$}\left \{\mathsf{y}\right \}\right \})$ as $\mathsf{x}\mbox {$\;${\footnotesize $\cup $}$\;$}\left \{\mathsf{y}\right \}\mbox {$\;${\footnotesize $\cup $}$\;$}\mbox {\boldmath $\bigcup $}\mathsf{m}$; but then the decision algorithm for a fragment of set theory known as ‘multi-level syllogistic with singleton’ yields an immediate contradiction.

Arguing by contradiction, suppose that there are counterexamples to the claim. Then, by exploiting finite induction, we can pick a minimal counterexample, f ₀.

Momentarily supposing that $\mathsf{f}_{0} \mbox {\bf $\:$=$\:$}\left \{\mathsf{a}\right \}$, one gets $\mbox {\boldmath $\bigcup $}\mathsf{f}_{0} \mbox {\boldmath $\:\not \subseteq \:$}\mathsf{a}$, because $\mbox {\boldmath $\bigcup $}\mathsf{f}_{0} \mbox {\boldmath $\:\subseteq \:$}\mathsf{a}$ would imply $\mathsf{f}_{0} \mbox {\boldmath $\setminus $}\mbox {\boldmath $\bigcup $}\mathsf{f}_{0} \mbox {\boldmath $\:\supseteq \:$}\left \{\mathsf{a}\right \} \mbox {\boldmath $\setminus $}\mathsf{a}$ and hence would imply the emptiness of $\left \{\mathsf{a}\right \} \mbox {\boldmath $\setminus $}\mathsf{a}$, whence the manifest absurdity a∈a follows. But, on the other hand, $\mbox {\boldmath $\bigcup $}\left \{\mathsf{a}\right \} \mbox {\boldmath $\:\subseteq \:$}\mathsf{a}$ trivially holds; therefore we must exclude that f ₀ is a singleton $\left \{\mathsf{a}\right \}$.

Due to our minimality assumption, the strict non-null subset of f ₀ cannot be a counterexample to the claim; therefore it has sources and hence .

Since does not intersect f ₀, the inequality just found conflicts with the equality which one gets from Theorem 2e through equality propagation.

1.5 A.5 Claw-Free, Transitive Sets and Their Pivots

A claw is defined to be a pair Y,F such that (1) F has at least three elements, (2) no element of F belongs to any other element of F, (3) either Y belongs to all elements of F or there is a W in Y such that Y belongs to all elements of $\mathsf{F} \mbox {\boldmath $\setminus $}\left \{\mathsf{W}\right \}$.

To really interest us, a claw-free set must be transitive: we omit this requirement in the definition given here below, but we will make it explicit in the major theorems pertaining to claw-freeness (Fig. 6).

Fig. 6

The forbidden orientations of a claw in a claw-free set

By way of first approximation, we want to select from each finite transitive set s not included in $\left \{\emptyset \right \}$ a ‘pivotal pair’ consisting of an element x of maximum rank in s and an element y of maximum rank in x. To avoid introducing the recursive notion of rank of a set, we slightly generalize the idea: for any set s (not necessarily finite or transitive) we define the frontier of s to consist of those elements x of s which own elements y belonging to s such that the length of no membership chain issuing from y, ending in s, and contained in s ever exceeds 2. Any element y which is thus related to an element x of the frontier of s will be called a pivot of s.

Our next claim is that if we choose a pivot element y of a transitive set s from an element of the frontier of s, then removal of all predecessors of y from s leads to a transitive set t such that y is a source of t.

Arguing by contradiction, let s,x,y,t be a counterexample to the claim. Taking the definition of t into account to exploit monotonicity, we readily get $\mathsf{t} \mbox {\boldmath $\:\subseteq \:$}\mathsf{s}$ and x∈t.

Now taking the definition of front into account, we can simplify our initial assumption to the following:

Since s is transitive, if t were not transitive then by Theorem 4c $\mathsf{s} \mbox {\boldmath $\setminus $}\mathsf{t}$ would have an element z not being a source of s. But then y would belong to $\mathsf{z} \in \mbox {\boldmath $\bigcup $}\mathsf{s}$, which conflicts with y being a pivot.

Now knowing that Trans(t), we must consider the other possibility, namely that $\mathsf{y} \notin \mathsf{t} \mbox {\boldmath $\setminus $}\mbox {\boldmath $\bigcup $}\mathsf{t}$. However, after expanding t and $\mbox {\boldmath $\bigcup $}\mathsf{t}$ according to their definitions, …

… we see that neither one of the possibilities $\mathsf{y} \in \mbox {\boldmath $\bigcup $}\mathsf{t}$, y∉t is tenable.

Pivotal elements, in a transitive claw-free set such as the one treated in this Theory, own at most two predecessors.

Suppose that y,x constitute a counter-example, so that y has, in addition to x, at least two predecessors z and w in s ₀.

The transitivity of s ₀, since y∈x and x∈s ₀, implies that y∈s ₀; therefore, in view of the claw-freeness of s ₀, y and $\mathsf{e} \mbox {\bf $\:$=$\:$}\left \{\mathsf{x} ,\mathsf{z} ,\mathsf{w}\right \}$ do not form a claw.

It readily follows from the definition of claw that $\left \{\mathsf{x} ,\mathsf{z} ,\mathsf{w}\right \}$ and $\mbox {\boldmath $\bigcup $}\left \{\mathsf{x} ,\mathsf{z} ,\mathsf{w}\right \}$ intersect; therefore, we can pick an element a common to the two.

But then y∈a, $\mathsf{a} \mbox {\boldmath $\:\subseteq \:$}\mbox {\boldmath $\bigcup $}\mbox {\boldmath $\bigcup $}\left \{\mathsf{x} ,\mathsf{z} ,\mathsf{w}\right \}$, and $\mbox {\boldmath $\bigcup $}\mbox {\boldmath $\bigcup $}\left \{\mathsf{x} ,\mathsf{z} ,\mathsf{w}\right \} \mbox {\boldmath $\:\subseteq \:$}\mbox {\boldmath $\bigcup $}\mbox {\boldmath $\bigcup $}\mathsf{s}_{0}$ must hold, implying that $\mathsf{y} \in \mbox {\boldmath $\bigcup $}\mbox {\boldmath $\bigcup $}\mathsf{s}_{0}$; but we have started with the assumption that $\mathsf{y} \notin \mbox {\boldmath $\bigcup $}\mbox {\boldmath $\bigcup $}\mathsf{s}_{0}$. This contradiction proves the claim.

Via Skolemization, we give a name to the third item in a pivotal tripleton:

The removal of the predecessors of a pivot from a claw-free, transitive non-trivial set such as the one treated by this theory does not disrupt transitivity.

1.6 A.6 Hanks, Cycles, and Hamiltonian Cycles

The following notion approximately models the concept of a graph where every vertex has at least two incident edges. However, we neither require that (1) edges be doubletons, nor that (2) the set H of edges and the one of vertices—which is understood to be $\mbox {\boldmath $\bigcup $}\mathsf{H}$—be disjoint.

For, assuming that h ₀ is a hank, non-null, and a subset of a doubleton $\left \{\mathsf{x}_{0} ,\mathsf{u}_{0}\right \}$, we will reach a contradiction arguing as follows. If a is one of the (at most two) elements of h ₀, since $\emptyset \mathbf {\mbox {\bf $\:\not $=$\:$}}\mathsf{a}\mbox {$\;${\footnotesize $\&$}$\;$}\mathsf{a} \mbox {\boldmath $\:\subseteq \:$}\mbox {\boldmath $\bigcup $}(\mathsf{h}_{0} \mbox {\boldmath $\setminus $}\left \{\mathsf{a}\right \})$ ensues from the definition of hank, $\mbox {\boldmath $\bigcup $}(\mathsf{h}_{0} \mbox {\boldmath $\setminus $}\left \{\mathsf{a}\right \})$ must be non-null; hence $\mathsf{h}_{0} \mbox {\bf $\:$=$\:$}\left \{\mathsf{a} ,\mathsf{b}\right \}$, where $\mathsf{b} \mathbf {\mbox {\bf $\:\not $=$\:$}}\mathsf{a}$. But then $\mbox {\boldmath $\bigcup $}(\mathsf{h}_{0} \mbox {\boldmath $\setminus $}\left \{\mathsf{a}\right \})\mbox {\bf $\:$=$\:$}\mbox {\boldmath $\bigcup $}\left \{\mathsf{b}\right \}$ and $\mbox {\boldmath $\bigcup $}(\mathsf{h}_{0} \mbox {\boldmath $\setminus $}\left \{\mathsf{b}\right \})\mbox {\bf $\:$=$\:$}\mbox {\boldmath $\bigcup $}\left \{\mathsf{a}\right \}$, i.e., $\mbox {\boldmath $\bigcup $}(\mathsf{h}_{0} \mbox {\boldmath $\setminus $}\left \{\mathsf{a}\right \})\mbox {\bf $\:$=$\:$}\mathsf{b}$ and $\mbox {\boldmath $\bigcup $}(\mathsf{h}_{0} \mbox {\boldmath $\setminus $}\left \{\mathsf{b}\right \})\mbox {\bf $\:$=$\:$}\mathsf{a}$; therefore $\mathsf{a} \mbox {\boldmath $\:\subseteq \:$}\mathsf{b}$ and $\mathsf{b} \mbox {\boldmath $\:\subseteq \:$}\mathsf{a}$ ensue from the definition of hank, leading us to the identity $\mathsf{a} \mbox {\bf $\:$=$\:$}\mathsf{b}$, which contradicts an earlier inequality.

The following is the basic case of a general theorem scheme where the length of the chain can be any number >2.

The following is the basic case of a general theorem scheme where the length of the path can be any number >1: Replacing an edge by a path with the same endpoints does not disrupt a hank.

Suppose that the premisses are met by h ₀,w ₀,y ₀,x ₁, and h ₁. In order to prove Hank(h ₁), we assume it to be false, so that the definition of hank implies the existence of an e ₁∈h ₁ and a z ₁∈e ₁ such that $\mathsf{z}_{1} \notin \mbox {\boldmath $\bigcup $}(\mathsf{h}_{1} \mbox {\boldmath $\setminus $}\left \{\mathsf{e}_{1}\right \})$.

Since x ₁ belongs to the distinct edges $\left \{\mathsf{w}_{0} ,\mathsf{x}_{1}\right \} ,\left \{\mathsf{x}_{1} ,\mathsf{y}_{0}\right \}$ of h ₁, clearly $\mathsf{z}_{1} \mathbf {\mbox {\bf $\:\not $=$\:$}}\mathsf{x}_{1}$.

Moreover, e ₁ cannot be one of the edges of $\mathsf{h}_{1} \mbox {\boldmath $\setminus $}\mathsf{h}_{0}$.

We know, at this point, that $\mathsf{e}_{1} \in \mathsf{h}_{0} \mbox {\boldmath $\setminus $}\left \{\left \{\mathsf{w}_{0} ,\mathsf{y}_{0}\right \}\right \}$. Since h ₀ is a hank, z ₁ has in h ₀ at least one incident edge different from e ₁; since the latter is no longer available in $\mathsf{h}_{1} \mbox {\boldmath $\setminus $}\left \{\mathsf{e}_{1}\right \}$, it must be $\left \{\mathsf{w}_{0} ,\mathsf{y}_{0}\right \}$, and either $\mathsf{z}_{1} \mbox {\bf $\:$=$\:$}\mathsf{w}_{0}$ or $\mathsf{z}_{1} \mbox {\bf $\:$=$\:$}\mathsf{y}_{0}$ hence holds. Both cases lead to a contradiction, though; in fact $\left \{\mathsf{w}_{0} ,\mathsf{x}_{1}\right \} ,\left \{\mathsf{x}_{1} ,\mathsf{y}_{0}\right \}$ differ from e ₁ and these edges of h ₁ are, respectively, incident to w ₀ and to y ₀.

The replacement of an edge by a 2-path with the same endpoints does not disrupt a cycle.

Supposing that h ₀,w ₀,y ₀,x ₁,h ₁ constitute a counter-example to the claim, observe that Hank(h ₁) must hold; hence we can consider a strictly smaller hank d ₁ than h ₁. It readily turns out that either $\left \{\mathsf{w}_{0} ,\mathsf{x}_{1}\right \} \in \mathsf{d}_{1}$ or $\left \{\mathsf{x}_{1} ,\mathsf{y}_{0}\right \} \in \mathsf{d}_{1}$; for otherwise h ₀ would strictly include d ₁, since $\mathsf{d}_{1} \mathbf {\mbox {\bf $\:\not $=$\:$}}\mathsf{h}_{0}$ follows from $\left \{\mathsf{w}_{0} ,\mathsf{y}_{0}\right \} \in \mathsf{h}_{0} \mbox {\boldmath $\setminus $}\mathsf{h}_{1}$.

Should one of $\left \{\mathsf{w}_{0} ,\mathsf{x}_{1}\right \} ,\left \{\mathsf{x}_{1} ,\mathsf{y}_{0}\right \}$, but not the other, belong to d ₁, we would easily get a contradiction: the two cases are treated symmetrically. At this point we have derived that both $\left \{\mathsf{w}_{0} ,\mathsf{x}_{1}\right \}$ and $\left \{\mathsf{x}_{1} ,\mathsf{y}_{0}\right \}$ belong to d ₁.

We will show that the set d ₀ obtained by replacing $\left \{\mathsf{w}_{0} ,\mathsf{x}_{1}\right \}$ and $\left \{\mathsf{x}_{1} ,\mathsf{y}_{0}\right \}$ by $\left \{\mathsf{w}_{0} ,\mathsf{y}_{0}\right \}$ in d ₁ is non-null and is a cycle strictly included in h ₀. Obviously $\left \{\mathsf{w}_{0} ,\mathsf{x}_{1}\right \} \mathbf {\mbox {\bf $\:\not $=$\:$}}\left \{\mathsf{w}_{0} ,\mathsf{y}_{0}\right \} \mbox {$\;${\footnotesize $\&$}$\;$}\left \{\mathsf{x}_{1} ,\mathsf{y}_{0}\right \} \mathbf {\mbox {\bf $\:\not $=$\:$}}\left \{\mathsf{w}_{0} ,\mathsf{y}_{0}\right \}$, because x ₁, y ₀, and w ₀ are distinct.

Despite us having assumed at the beginning that h ₀ contains no proper cycle, so that in particular Hank(d ₀) cannot hold, due to an edge e ₀ of d ₀ and to an endpoint x ₀ of e ₀ which is not properly covered in d ₀, …

… we now aim at showing that this offending edge e ₀ of d ₀ will also offend d ₁, which contradicts a fact noted at the beginning. Here we shortly digress to prove that $\mathsf{e}_{0} \mbox {\bf $\:$=$\:$}\left \{\mathsf{w}_{0} ,\mathsf{y}_{0}\right \}$ must hold, else e ₀ would offend d ₁.

Indeed, assuming $\mathsf{e}_{0} \mathbf {\mbox {\bf $\:\not $=$\:$}}\left \{\mathsf{w}_{0} ,\mathsf{y}_{0}\right \}$, e ₀ would also belong to d ₁, and each one of its endpoints must have edges distinct from e ₀ incident to it in d ₁. However, it will turn out that this cannot be the case for the endpoint x ₀, which hence is not properly covered in d ₁.

To see this, let $\mathsf{q}_{2} \mathbf {\mbox {\bf $\:\not $=$\:$}}\mathsf{e}_{0}$ be the edge that covers x ₀ in d ₁.

If this edge q ₂ were one of the two which have been removed, the edge $\left \{\mathsf{w}_{0} ,\mathsf{y}_{0}\right \}$ would satisfactorily cover x ₀ in d ₀.

q ₂ must hence belong to d ₀; but again, this implies that q ₂ would satisfactorily cover x ₀ in d ₀. Therefore, $\mathsf{e}_{0} \mbox {\bf $\:$=$\:$}\left \{\mathsf{w}_{0} ,\mathsf{y}_{0}\right \}$ must hold.

The only remaining possibility, $\mathsf{e}_{0} \mbox {\bf $\:$=$\:$}\left \{\mathsf{w}_{0} ,\mathsf{y}_{0}\right \}$, is also untenable. Indeed, d ₁ has two edges incident to w ₀, one of which is $\left \{\mathsf{w}_{0} ,\mathsf{x}_{1}\right \}$; likewise, d ₁ has two edges incident to y ₀, one of which is $\left \{\mathsf{x}_{1} ,\mathsf{y}_{0}\right \}$. For either one of the endpoints w ₀,y ₀ of e ₀, the second incident edge belongs to d ₁ and differs from $\left \{\mathsf{w}_{0} ,\mathsf{y}_{0}\right \}$, so it must belong to d ₀ as well; since d ₀ also owns the edge $\left \{\mathsf{w}_{0} ,\mathsf{y}_{0}\right \}$ incident to either endpoint, it is not true that e ₀ is an offending edge for d ₀, a fact that contradicts one of the assumptions made.

In our specialized context, where edges are 2-sets whose elements satisfy a peculiar membership constraint, we do not simply require that a Hamiltonian cycle H touches every vertex, but also that every source has an incident membership edge in H.

Suppose that s ₀,t ₀,x ₁,y ₀,h ₀,w ₀,k ₀ make a counterexample to the claim. One reason why

$$\mathsf{SqHamiltonian}(\mathsf{h}_{0} \mbox {\boldmath $\setminus $}\left \{\left \{\mathsf{w}_{0} ,\mathsf{y}_{0}\right \}\right \} \mbox {$\;${\footnotesize $\cup $}$\;$}\left \{\left \{\mathsf{w}_{0} ,\mathsf{x}_{1}\right \} ,\left \{\mathsf{x}_{1} ,\mathsf{y}_{0}\right \}\right \},\mathsf{s}_{0})$$

is violated might be

$$\mbox {\boldmath {$\neg $}}\mbox {\large $\langle $}\mbox {\boldmath {$\forall $}}\mathsf{x} \in \mathsf{s}_{0} \mbox {\boldmath $\setminus $}\mbox {\boldmath $\bigcup $}\mathsf{s}_{0},\mbox {\boldmath {$\exists $}}\mathsf{y} \in \mathsf{x}\,|\:\left \{\mathsf{x} ,\mathsf{y}\right \} \in \mathsf{h}_{0} \mbox {\boldmath $\setminus $}\left \{\left \{\mathsf{w}_{0} ,\mathsf{y}_{0}\right \}\right \} \mbox {$\;${\footnotesize $\cup $}$\;$}\left \{\left \{\mathsf{w}_{0} ,\mathsf{x}_{1}\right \} ,\left \{\mathsf{x}_{1} ,\mathsf{y}_{0}\right \}\right \}\mbox {\large $\rangle $}$$

; if this is the case, we can choose an x′ witnessing this fact.

To see that $\mathsf {x}' \in \mathsf{t}_{0} \mbox {\boldmath $\setminus $}\mbox {\boldmath $\bigcup $}\mathsf{t}_{0}$ follows from the constraint $\mathsf {x}' \in \mathsf{s}_{0} \mbox {\boldmath $\setminus $}\mbox {\boldmath $\bigcup $}\mathsf{s}_{0}$, we assume the contrary and argue as follows: (1) Unless x′ belongs to t ₀, we must have $\mathsf {x}' \mbox {\bf $\:$=$\:$}\mathsf{x}_{1}$, which however has an incident membership edge, namely $\left \{\mathsf{x}_{1} ,\mathsf{y}_{0}\right \}$, in $\mathsf{h}_{0} \mbox {\boldmath $\setminus $}\left \{\left \{\mathsf{w}_{0} ,\mathsf{y}_{0}\right \}\right \} \mbox {$\;${\footnotesize $\cup $}$\;$}\left \{\left \{\mathsf{w}_{0} ,\mathsf{x}_{1}\right \} ,\left \{\mathsf{x}_{1} ,\mathsf{y}_{0}\right \}\right \}$. (2) Thus, since x′∈t ₀, we have $\mathsf {x}' \in \mathsf{t}_{0} \mbox {$\;${\footnotesize $\cap $}$\;$}\mbox {\boldmath $\bigcup $}\mathsf{t}_{0}$ and hence $\mathsf {x}' \in \mathsf{s}_{0} \mbox {$\;${\footnotesize $\cap $}$\;$}\mbox {\boldmath $\bigcup $}\mathsf{s}_{0}$, contradicting the constraint on x′.

Knowing that $\mathsf {x}' \in \mathsf{t}_{0} \mbox {\boldmath $\setminus $}\mbox {\boldmath $\bigcup $}\mathsf{t}_{0}$, we can find a y ₁∈x′ such that $\left \{\mathsf {x}' ,\mathsf{y}_{1}\right \} \in \mathsf{h}_{0}$. Since this membership edge is no longer available after the modification of h ₀, it must be $\left \{\mathsf{w}_{0} ,\mathsf{y}_{0}\right \}$; therefore, $\mathsf {x}' \mbox {\bf $\:$=$\:$}\mathsf{w}_{0}$, for otherwise $\mathsf {x}' \mbox {\bf $\:$=$\:$}\mathsf{y}_{0}$ would (in view of the fact y ₀∈x ₁) contradict the assumption $\mathsf {x}' \in \mathsf{s}_{0} \mbox {\boldmath $\setminus $}\mbox {\boldmath $\bigcup $}\mathsf{s}_{0}$.

If x′∈y ₀, the assumption $\mathsf {x}' \in \mathsf{s}_{0} \mbox {\boldmath $\setminus $}\mbox {\boldmath $\bigcup $}\mathsf{s}_{0}$ would be contradicted similarly: but then, by the initial assumption, we must have $\left \{\mathsf {x}' ,\mathsf{k}_{0}\right \} \in \mathsf{h}_{0} \mbox {$\;${\footnotesize $\&$}$\;$}\mathsf{k}_{0} \mathbf {\mbox {\bf $\:\not $=$\:$}}\mathsf{y}_{0} \mbox {$\;${\footnotesize $\&$}$\;$}\mathsf{k}_{0} \in \mathsf {x}'$, conflicting with Stat2, because $\left \{\mathsf {x}' ,\mathsf{k}_{0}\right \} \mbox {\bf $\:$=$\:$}\left \{\mathsf{w}_{0} ,\mathsf{y}_{0}\right \}$ would imply $\mathsf{k}_{0} \mbox {\bf $\:$=$\:$}\mathsf{y}_{0}$.

At this point the reason why

$$\mathsf{SqHamiltonian}(\mathsf{h}_{0} \mbox {\boldmath $\setminus $}\left \{\left \{\mathsf{w}_{0} ,\mathsf{y}_{0}\right \}\right \} \mbox {$\;${\footnotesize $\cup $}$\;$}\left \{\left \{\mathsf{w}_{0} ,\mathsf{x}_{1}\right \} ,\left \{\mathsf{x}_{1} ,\mathsf{y}_{0}\right \}\right \},\mathsf{s}_{0})$$

is false must be that

$$\mathsf{Hamiltonian}\big (\mathsf{h}_{0} \mbox {\boldmath $\setminus $}\left \{\left \{\mathsf{w}_{0} ,\mathsf{y}_{0}\right \}\right \} \mbox {$\;${\footnotesize $\cup $}$\;$}\left \{\left \{\mathsf{w}_{0} ,\mathsf{x}_{1}\right \} ,\left \{\mathsf{x}_{1} ,\mathsf{y}_{0}\right \}\right \},\mathsf{s}_{0},\mathsf{sqEdges}(\mathsf{s}_{0})\big )$$

is false; however, we will derive a contradiction also in this case.

In fact, after observing that $\left \{\mathsf{w}_{0} ,\mathsf{y}_{0}\right \} \mbox {\boldmath $\:\subseteq \:$}\mbox {\boldmath $\bigcup $}\mathsf{h}_{0}$ must hold, we will be able to discard one by one each potential reason why $\mathsf{Hamiltonian}\big (\mathsf{h}_{0} \mbox {\boldmath $\setminus $}\left \{\left \{\mathsf{w}_{0} ,\mathsf{y}_{0}\right \}\right \} \mbox {$\;${\footnotesize $\cup $}$\;$} \left \{\left \{\mathsf{w}_{0} ,\mathsf{x}_{1}\right \} ,\left \{\mathsf{x}_{1} ,\mathsf{y}_{0}\right \}\right \},\mathsf{s}_{0},\mathsf{sqEdges}(\mathsf{s}_{0})\big )$ should be false.

We prove first that $\mbox {\boldmath $\bigcup $}\mathsf{h}_{1} \mbox {\boldmath $\:\subseteq \:$}\mathsf{s}_{0}$.

The remaining case is $\mathsf{s}_{0} \mbox {\boldmath $\:\not \subseteq \:$}\mbox {\boldmath $\bigcup $}\mathsf{h}_{1}$, which entails that we can find an $\mathsf{a} \in \mathsf{t}_{0} \mbox {\boldmath $\setminus $}\mbox {\boldmath $\bigcup $}\mathsf{h}_{1}$.

Since a∈t ₀ and $\mathsf{t}_{0} \mbox {\bf $\:$=$\:$}\mbox {\boldmath $\bigcup $}\mathsf{h}_{0}$, we can find an e′∈h ₀ such that a∈e′.

Since $\mathsf{h}_{1} \mbox {\bf $\:$=$\:$}\mathsf{h}_{0} \mbox {\boldmath $\setminus $}\left \{\left \{\mathsf{w}_{0} ,\mathsf{y}_{0}\right \}\right \} \mbox {$\;${\footnotesize $\cup $}$\;$}\left \{\left \{\mathsf{w}_{0} ,\mathsf{x}_{1}\right \} ,\left \{\mathsf{x}_{1} ,\mathsf{y}_{0}\right \}\right \}$, we conclude that $\mathsf {e}' \mbox {\bf $\:$=$\:$}\left \{\mathsf{w}_{0} ,\mathsf{y}_{0}\right \}$ must hold. Hence, either $\mathsf{a} \mbox {\bf $\:$=$\:$}\mathsf{w}_{0}$ or $\mathsf{a} \mbox {\bf $\:$=$\:$}\mathsf{y}_{0}$ must hold, both of which yield a contradiction.

Notice that since h ₀ is a Hamiltonian path in t ₀, its unionset must equal t ₀; since x ₀ does not belong to t ₀, but it belongs to $\left \{\mathsf{x}_{0} ,\mathsf{y}_{0}\right \}$, it follows that $\left \{\mathsf{x}_{0} ,\mathsf{y}_{0}\right \}$ cannot belong to h ₀.

Arguing by contradiction, assume that $\left \{\left \{\mathsf{x}_{0} ,\mathsf{y}_{0}\right \} ,\left \{\mathsf{y}_{0} ,\mathsf{z}_{0}\right \} ,\left \{\mathsf{z}_{0} ,\mathsf{x}_{0}\right \}\right \}$ is not a ‘square Hamiltonian’ cycle for $\mathsf{s} \mbox {\bf $\:$=$\:$}\left \{\mathsf{x}_{0} ,\mathsf{y}_{0} ,\mathsf{z}_{0}\right \}$, where x ₀∈y ₀ and y ₀∈z ₀ holds. We will first exclude the possibility that $\left \{\left \{\mathsf{x}_{0} ,\mathsf{y}_{0}\right \} ,\left \{\mathsf{y}_{0} ,\mathsf{z}_{0}\right \} ,\left \{\mathsf{z}_{0} ,\mathsf{x}_{0}\right \}\right \}$ is not a Hamiltonian cycle in the ‘square edges’ of s; after discarding this, we will also exclude that this cycle may fail to satisfy the restraining condition that it has a genuine membership edge incident into each source of s.

We conclude by checking that $\left \{\left \{\mathsf{x}_{0} ,\mathsf{y}_{0}\right \} ,\left \{\mathsf{y}_{0} ,\mathsf{z}_{0}\right \} ,\left \{\mathsf{z}_{0} ,\mathsf{x}_{0}\right \}\right \}$ owns a genuine membership edge incident into each source of s; as a matter of fact, z ₀ is the only source of s and $\left \{\mathsf{y}_{0} ,\mathsf{z}_{0}\right \}$ is a membership edge.

Any non-trivial transitive set whose square is devoid of Hamiltonian cycles must strictly comprise certain sets.

Indeed, any set satisfying the premises of our present claim must, due to its transitivity and non-triviality, include either one of the Hamiltonian cycles endowed with the vertices $\emptyset ,\left \{\emptyset \right \} ,\left \{\left \{\emptyset \right \}\right \}$ and $\emptyset ,\left \{\emptyset \right \} ,\left \{\emptyset ,\left \{\emptyset \right \}\right \}$, respectively; but it must also own additional elements, else the last premise would be falsified. Moreover, it cannot have exactly the elements $\emptyset ,\left \{\emptyset \right \} ,\left \{\left \{\emptyset \right \}\right \} ,\left \{\emptyset ,\left \{\emptyset \right \}\right \}$, as these form a Hamiltonian cycle.

The cases $\mathsf{t} \mbox {\bf $\:$=$\:$}\left \{\emptyset ,\left \{\emptyset \right \} ,\left \{\left \{\emptyset \right \}\right \}\right \}$ and $\mathsf{t} \mbox {\bf $\:$=$\:$}\left \{\emptyset ,\left \{\emptyset \right \} ,\left \{\emptyset ,\left \{\emptyset \right \}\right \}\right \}$ must be excluded, because we have the respective Hamiltonian cycles

$$\begin{array}{c} \left \{\left \{\emptyset ,\left \{\emptyset \right \}\right \} ,\left \{\left \{\emptyset \right \} ,\left \{\left \{\emptyset \right \}\right \}\right \} ,\left \{\left \{\left \{\emptyset \right \}\right \} ,\emptyset \right \}\right \}\:,\\\left \{\left \{\emptyset ,\left \{\emptyset \right \}\right \} ,\left \{\left \{\emptyset \right \} ,\left \{\emptyset ,\left \{\emptyset \right \}\right \}\right \} ,\left \{\left \{\emptyset ,\left \{\emptyset \right \}\right \} ,\emptyset \right \}\right \}\:. \end{array}$$

Having thus established that $\mathsf{t} \mbox {\bf $\:$=$\:$}\left \{\emptyset ,\left \{\emptyset \right \} ,\left \{\left \{\emptyset \right \}\right \} ,\left \{\emptyset ,\left \{\emptyset \right \}\right \}\right \}$, we can now exploit Theorem hamiltonian₂ to enrich the Hamiltonian cycle for $\left \{\emptyset ,\left \{\emptyset \right \} ,\left \{\left \{\emptyset \right \}\right \}\right \}$ into one which does to our case.

1.7 A.7 Hamiltonicity of Squared Claw-Free Sets

Non-trivial, claw-free, finite transitive sets have Hamiltonian squares.

For, assuming the opposite, there would exist an inclusion-minimal, finite transitive non-trivial claw-free set whose square lacks a Hamiltonian cycle.

Thanks to the finiteness of such an s ₀, the ${\textsc {Theory}\;}$ pivotsForClawFreeness can be applied to s ₀. We thereby pick an element x from the frontier of s ₀, and an element y of x which is pivotal relative to s ₀. This y will have at most two in-neighbors (one of the two being x) in s ₀. We denote by z an in-neighbor of y in s ₀, such that z differs from x if possible. Observe, among others, that neither one of x,z can belong to the other.

Thus it turns out readily that removal of x,z from s ₀ leads to a set t to which, unless t is ‘trivial’ (i.e. a subset of $\left \{\emptyset ,\left \{\emptyset \right \}\right \}$), the inductive hypothesis applies; hence, by that hypothesis, there is a Hamiltonian cycle h ₀ for t.

In order that t be trivial, we should have $\mathsf{s}_{0} \mbox {\boldmath $\:\subseteq \:$}\left \{\emptyset ,\left \{\emptyset \right \} ,\left \{\left \{\emptyset \right \}\right \} ,\left \{\emptyset ,\left \{\emptyset \right \}\right \}\right \}$; but then, as already shown in the proof of Theorem hamiltonian₄, we have the ability, either directly, or by enrichment of a Hamiltonian cycle for $\left \{\emptyset ,\left \{\emptyset \right \} ,\left \{\left \{\emptyset \right \}\right \}\right \}$, to construct a Hamiltonian cycle for s ₀: thus, if we suppose $\mathsf{t} \mbox {\boldmath $\:\subseteq \:$}\left \{\emptyset ,\left \{\emptyset \right \}\right \}$ then we get a contradiction.

It follows from y being a source of $\mathsf{t} \mbox {\bf $\:$=$\:$}\mbox {\boldmath $\bigcup $}\mathsf{h}_{0}$ that there is an edge $\left \{\mathsf{y} ,\mathsf{w}\right \}$, with w∈y, in h ₀, If $\mathsf{x} \mbox {\bf $\:$=$\:$}\mathsf{z}$, to get a Hamiltonian cycle h ₁ for s ₀ (a fact conflicting with the inductive hypothesis) it will suffice to take $\mathsf{h}_{1} \mbox {\bf $\:$=$\:$}\mathsf{h}_{0} \mbox {\boldmath $\setminus $}\left \{\left \{\mathsf{y} ,\mathsf{w}\right \}\right \} \mbox {$\;${\footnotesize $\cup $}$\;$}\left \{\left \{\mathsf{x} ,\mathsf{y}\right \} ,\left \{\mathsf{x} ,\mathsf{w}\right \}\right \}$, where $\left \{\mathsf{x} ,\mathsf{w}\right \}$ is a square edge because w∈y and y∈x both hold. On the other hand, if $\mathsf{x} \mathbf {\mbox {\bf $\:\not $=$\:$}}\mathsf{z}$, claw-freeness implies that either w∈x or w∈z. Assume the former for definiteness, and put $\mathsf{h}_{2} \mbox {\bf $\:$=$\:$}\mathsf{h}_{0} \mbox {\boldmath $\setminus $}\left \{\left \{\mathsf{y} ,\mathsf{w}\right \}\right \} \mbox {$\;${\footnotesize $\cup $}$\;$}\left \{\left \{\mathsf{y} ,\mathsf{z}\right \} ,\left \{\mathsf{z} ,\mathsf{x}\right \} ,\left \{\mathsf{x} ,\mathsf{w}\right \}\right \}$, where $\left \{\mathsf{x} ,\mathsf{z}\right \}$ is a square edge and $\left \{\mathsf{x} ,\mathsf{w}\right \}$ and $\left \{\mathsf{y} ,\mathsf{z}\right \}$ are genuine edges incident in the sources x,z. We are again facing a contradiction, because h ₂ is a Hamiltonian cycle for s ₀.

1.8 A.8 Perfect Matchings

Next we introduce the notion of perfect matching. This is a partition consisting of doubletons one of whose elements belongs to the other. Special cases of a perfect matching are: the empty set and, more generally, all subsets of a perfect matching.

By adjoining a pair $\left \{\mathsf{x} ,\mathsf{y}\right \}$ with y∈x to a perfect matching none of whose blocks has either x or y as an element, we always obtain a perfect matching.

If, in a perfect matching m, we replace one block $\left \{\mathsf{y} ,\mathsf{w}\right \}$ by pairs $\left \{\mathsf{y} ,\mathsf{z}\right \} ,\left \{\mathsf{x} ,\mathsf{w}\right \}$, then, under suitable conditions ensuring disjointness between blocks and membership within each block, we get a perfect matching again.

For assuming that m,y ₀,w ₀,x ₀,z ₀ are a counterexample to the claim, we could get a contradiction arguing as follows. Begin by observing that neither y ₀ nor w ₀ can belong to the unionset of the perfect submatching $\mathsf{m} \mbox {\boldmath $\setminus $}\left \{\left \{\mathsf{y}_{0} ,\mathsf{w}_{0}\right \}\right \}$ of m.

Thus, taking into account that w ₀∈x ₀ and that $\mathsf{x}_{0} \notin \mbox {\boldmath $\bigcup $}\mathsf{m}$ which is a superset of $\mbox {\boldmath $\bigcup $}(\mathsf{m} \mbox {\boldmath $\setminus $}\left \{\left \{\mathsf{y}_{0} ,\mathsf{w}_{0}\right \}\right \})$, we can extend the perfect matching $\mathsf{m} \mbox {\boldmath $\setminus $}\left \{\left \{\mathsf{y}_{0} ,\mathsf{w}_{0}\right \}\right \}$ with the doubleton $\left \{\mathsf{x}_{0} ,\mathsf{w}_{0}\right \}$.

Observe next that $\mathsf{x}_{0} \mathbf {\mbox {\bf $\:\not $=$\:$}}\mathsf{y}_{0}$ and $\mathsf{z}_{0} \mathbf {\mbox {\bf $\:\not $=$\:$}}\mathsf{w}_{0}$, because $\mathsf{x}_{0} \notin \mbox {\boldmath $\bigcup $}\mathsf{m}$ and $\mathsf{z}_{0} \notin \mbox {\boldmath $\bigcup $}\mathsf{m}$ whereas $\mathsf{y}_{0}\in \mbox {\boldmath $\bigcup $}\mathsf{m}$ and $\mathsf{w}_{0}\in \mbox {\boldmath $\bigcup $}\mathsf{m}$. It then follows from $\mathsf{z}_{0} \mathbf {\mbox {\bf $\:\not $=$\:$}}\mathsf{w}_{0}$, thanks to the assumption $\mathsf{z}_{0} \mathbf {\mbox {\bf $\:\not $=$\:$}}\mathsf{x}_{0}$, that z ₀ does not belong to the unionset of the perfect matching $\mathsf{m} \mbox {\boldmath $\setminus $}\left \{\left \{\mathsf{y}_{0} ,\mathsf{w}_{0}\right \}\right \} \mbox {$\;${\footnotesize $\cup $}$\;$}\left \{\left \{\mathsf{x}_{0} ,\mathsf{w}_{0}\right \}\right \}$.

y ₀ cannot equal w ₀ either, the reason being that the set $\left \{\mathsf{y}_{0} ,\mathsf{w}_{0}\right \}$ is a block of a perfect matching and hence it cannot be a singleton. If follows, thanks to the assumption y ₀∈x ₀ (entailing that $\mathsf{y}_{0} \mathbf {\mbox {\bf $\:\not $=$\:$}}\mathsf{x}_{0}$), that y ₀ does not belong to $\mbox {\boldmath $\bigcup $}(\mathsf{m} \mbox {\boldmath $\setminus $}\left \{\left \{\mathsf{y}_{0} ,\mathsf{w}_{0}\right \}\right \} \mbox {$\;${\footnotesize $\cup $}$\;$}\left \{\left \{\mathsf{x}_{0} ,\mathsf{w}_{0}\right \}\right \})$ either.

We now know that the perfect matching $\mathsf{m} \mbox {\boldmath $\setminus $}\left \{\left \{\mathsf{y}_{0} ,\mathsf{w}_{0}\right \}\right \} \mbox {$\;${\footnotesize $\cup $}$\;$}\left \{\left \{\mathsf{x}_{0} ,\mathsf{w}_{0}\right \}\right \}$ can be extended with the doubleton $\left \{\mathsf{y}_{0} ,\mathsf{z}_{0}\right \}$, which readily leads us to the sought contradiction.

1.9 A.9 Each Claw-Free Set Admits a Near-Perfect Matching

Every claw-free finite, transitive set admits a perfect matching perhaps omitting one of its elements.

For, supposing the contrary, there would be an inclusion-minimal finite set s ₀ which is transitive and claw-free, and such that no perfect matching m partitions the set s ₀ possibly deprived of an element y ₀.

We observe that such an s ₀ cannot equal ∅ or $\left \{\emptyset \right \}$, else the null perfect matching would cover it.

Thanks to the finiteness of s ₀, the ${\textsc {Theory}\;}$ pivotsForClawFreeness can be applied to s ₀. We thereby pick an element x from the frontier of s ₀ and an element y of x which is pivotal relative to s ₀. This y will have at most two ∈-predecessors (one of the two being x) in s ₀. We denote by z a predecessor of y in s ₀, such that z differs from x if possible.

Moreover, we take t′ to be s ₀ deprived of the predecessors of y and, if $\mathsf{x} \mathbf {\mbox {\bf $\:\not $=$\:$}}\mathsf{z}$, we take $\mathsf{t} \mbox {\bf $\:$=$\:$}\mathsf {t}'$ else we take $\mathsf{t} \mbox {\bf $\:$=$\:$}\mathsf {t}' \mbox {\boldmath $\setminus $}\left \{\mathsf{y}\right \}$.

Thus it turns out readily that t is transitive; hence, by the inductive hypothesis, there is a perfect matching m ₀ for t.

The possibility that y does not belong to $\mbox {\boldmath $\bigcup $}\mathsf{m}_{0}$ is then discarded; in fact, if this were the case, then by adding the pair $\left \{\mathsf{x} ,\mathsf{y}\right \}$ to m ₀ we would get a perfect matching for s ₀, while we have assumed that such a matching does not exist. From the fact $\mathsf{y} \in \mbox {\boldmath $\bigcup $}\mathsf{m}_{0}$ it follows that y belongs to t, hence that $\mathsf{t} \mbox {\bf $\:$=$\:$}\mathsf {t}'$ and that x,z are distinct.

It also follows that y is the tail of that arc p ₁ of m ₀ to which it belongs. In fact, if y were instead the head of p ₁, then the tail x ₂ of p ₁, which must belong to $\mbox {\boldmath $\bigcup $}\mathsf{m}_{0}$ would belong to $\mathsf{s}_{0} \mbox {\boldmath $\setminus $}\left \{\mathsf{x} ,\mathsf{z}\right \}$, hence would be inside s ₀ but outside the set of predecessors of y, which is absurd.

Call w the head of the arc issuing from y in m ₀. Then y,x,z,w form a potential claw; this implies, since s ₀ is claw-free that either w∈x or w∈z.

Obviously, $\mathsf{w} \in \mbox {\boldmath $\bigcup $}\mathsf{m}_{0}$. Moreover, through elementary reasoning we derive $\mbox {\boldmath $\bigcup $}\mathsf{m}_{0} \mbox {$\;${\footnotesize $\cup $}$\;$}\left \{\mathsf{z} ,\mathsf{x}\right \} \mbox {\bf $\:$=$\:$}\mathsf{s}_{0} \mbox {\boldmath $\setminus $}\left \{\mathsf{y}_{1}\right \}$, where y ₁ lies outside s ₀ if both x and z has been covered by the matching m ₀, otherwise y ₁ equals the one of x,z (which might be the same set) left over by m ₀.

If there is an edge between w and x, then we deviate the perfect matching by replacing $\left \{\mathsf{y} ,\mathsf{w}\right \}$ by $\left \{\mathsf{y} ,\mathsf{z}\right \}$ and $\left \{\mathsf{x} ,\mathsf{w}\right \}$; otherwise we replace $\left \{\mathsf{y} ,\mathsf{w}\right \}$ by $\left \{\mathsf{y} ,\mathsf{x}\right \}$ and $\left \{\mathsf{z} ,\mathsf{w}\right \}$. Plainly we get a perfect matching for s ₀ in either case, which leads us to the desired contradiction.

1.10 A.10 From Membership Digraphs to General Graphs

Let us now place the results presented so far under the more general perspective that motivates this work. We display in this section the interfaces of two Theorys (not developed formally with Ref, as of today), explaining why we can work with membership as a convenient surrogate for the edge relationship of general graphs.

  One of these, ${\textsc {Theory}\;}$ finGraphRepr, will implement the proof that any finite graph (v ₀,e ₀) is ‘isomorphic’, via a suitable orientation of its edges and an injection $\mathsf {\varrho}_{\varTheta }$ of v ₀ onto a set $\mathsf {\nu}_{\varTheta }$, to a digraph $(\mathsf {\nu}_{\varTheta },\{(x,y):\:x\in \mathsf {\nu}_{\varTheta },\:y\in x\cap \mathsf {\nu}_{\varTheta }\})$ that enjoys the weak extensionality property: “distinct non-sink vertices have different out-neighborhoods”.

Although accessory, the weak extensionality condition is the clue for getting the desired isomorphism; in fact, for any weakly extensional digraph, acyclicity always ensures that a variant of Mostowski’s collapse is well-defined: in order to get it, one starts by assigning a distinct set Mt to each sink t and then proceeds by putting recursively

$$\begin{array}{rcl}Mw&=&\{\,Mu:\:(w,u)\;\mbox{is an arc}\,\}\end{array}$$

for all non-sink vertices w; plainly, injectivity of the function u↦Mu can be ensured globally by a suitable choice of the images Mt of the sinks t.

The other Theory, cfGraphRepr, will specialize finGraphRepr to the case of a connected, claw-free (undirected, finite) graph—connectedness and claw-freeness are specified, respectively, by the second and by the third assumption of this Theory. For these graphs, we can insist that the orientation be so imposed as to ensure extensionality in full: “distinct vertices have different out-neighborhoods”. Consequently, the following will hold:

• there is a unique sink, ∅; moreover,

• the set $\mathsf {\nu}_{\varTheta }$ of vertices underlying the image digraph is transitive. And, trivially,

• $\mathsf {\nu}_{\varTheta }$ is a claw-free set, in an even stronger sense than the definition with which we have been working throughout this proof scenario.

(As regards the third of these points, it should be clear that none of the four non-isomorphic membership renderings of a claw are induced by any quadruple of elements of $\mathsf {\nu}_{\varTheta }$; our definition forbade only two of them, though!)

Via the ${\textsc {Theory}\;}$ cfGraphRepr, the above-proved existence results about perfect matchings and Hamiltonian cycles can be transferred from a realm of special membership digraphs to the a priori more general realm of the connected claw-free graphs.