avewright/proofwiki-math huggingface dataset:
Theorem: Plane contains Infinite Number of Lines

A plane contains an infinite number of distinct lines.

Proof:
A plane contains an infinite number of points.
Not all these points are collinear.
Let $A$, $B$ and $C$ be points in a plane $P$.
From Propositions of Incidence: Line in Plane, any two of these points determine a line.
Consider the lines $AB$, $AC$ and $BC$, all of which are distinct.
Let $X$ be one of the infinite number of points on $BC$ which is not $B$ or $C$.
Then $AX$ is a line in $P$ which is distinct from both $AB$ and $AC$.
As there is are infinite number of points on $BC$, there are an infinite number of lines incident to $A$ and $BC$.
All these lines are in $P$.
Hence the result.
{{qed}}
Theorem: Two Planes have Line in Common

Two distinct planes have exactly one (straight) line in common.

Proof:
Take two distinct lines in plane $1$.
From Propositions of Incidence: Plane and Line, they each meet plane $2$ in one point each, say at $A$ and $B$.
Thus $A$ and $B$ both lie in both planes.
Thus the line defined by $A$ and $B$ lies in both planes.
{{qed}}
{{Handwaving}}
Theorem: Three Non-Collinear Planes have One Point in Common

Three planes which are not collinear have exactly one point in all three planes.

Proof:
Let $A$, $B$ and $C$ be the three planes in question.
From Two Planes have Line in Common, $A$ and $B$ share a line, $p$ say.
From Propositions of Incidence: Plane and Line, $p$ meets $C$ in one point.
{{qed}}
Theorem: Desargues' Theorem

Let $\triangle ABC$ and $\triangle A'B'C'$ be triangles lying in the same or different planes.
Let the lines $AA'$, $BB'$ and $CC'$ intersect in the point $O$.
Then:
:$BC$ meets $B'C'$ in $L$
:$CA$ meets $C'A'$ in $M$
:$AB$ meets $A'B'$ in $N$
where $L, M, N$ are collinear.

Proof:
:500px
Let $\triangle ABC$ and $\triangle A'B'C'$ be in different planes $\pi$ and $\pi'$ respectively.
Since $BB'$ and $CC'$ intersect in $O$, it follows that $B$, $B'$, $C$ and $C'$ lie in a plane.
Thus $BC$ must meet $B'C'$ in a point $L$.
By the same argument, $CA$ meets $C'A'$ in a point $M$ and $AB$ meets $A'B'$ in a point $N$.
These points $L, M, N$ are in each of the planes $\pi$ and $\pi'$.
By Two Planes have Line in Common they are therefore collinear on the line where $\pi$ and $\pi'$ meet.
Now let $\triangle ABC$ and $\triangle A'B'C'$ be in the same plane $\pi$.
Let $OPP'$ be any line through $O$ which does not lie in $\pi$.
Then since $PP'$ meets $AA'$ in $O$, the four points $P, P', A, A'$ are coplanar.
Thus $PA$ meets $P'A'$ at a point $A' '$.
Similarly $PB$ meets $P'B'$ at a point $B' '$, and $PC$ meets $P'C'$ at a point $C' '$.
The lines $BC, B'C'$ and $B' 'C' '$ are the three lines of intersection of the three planes $PBC$, $P'B'C'$ and $\pi$ taken in pairs.
So $BC$, $B'C'$ and $B' 'C' '$ meet in a point $L$.
Similarly $CA$, $C'A'$ and $C' 'A' '$ meet in a point $M$ and $AB$, $A'B'$ and $A' 'B' '$ meet in a point $N$.
The two triangles $\triangle ABC$ and $\triangle A' 'B' 'C' '$ are in different planes, and $AA' '$, $BB' '$ and $CC' '$ meet in $P$.
Thus $L$, $M$ and $N$ are collinear by the first part of this proof.
{{qed}}
Theorem: Union of Mappings which Agree is Mapping

Let $A, B, Y$ be sets.
Let $f: A \to Y$ and $g: B \to Y$ be mappings.
Let $X = A \cup B$.
Let $f$ and $g$ agree on $A \cap B$.
Then $f \cup g: X \to Y$ is a mapping.

Proof:
By definition, $f \cup g$ is a relation whose domain is $X = A \cup B$.
Let $\tuple {x, y_1} \in f \cup g$ and $\tuple {x, y_2} \in f \cup g$.
At least one of the following must be true:
:$(1): \quad \tuple {x, y_1} \in f, \tuple {x, y_2} \in f$
:$(2): \quad \tuple {x, y_1} \in g, \tuple {x, y_2} \in g$
:$(3): \quad \tuple {x, y_1} \in f, \tuple {x, y_2} \in g$
:$(4): \quad \tuple {x, y_1} \in g, \tuple {x, y_2} \in f$
Because $f$ and $g$ are mappings, $(1)$ and $(2)$ imply that $y_1 = y_2$.
If $(3)$ holds, then $y_1 = \map f x$ and $y_2 = \map g x$.
But then $x \in A \cap B$.
So by hypothesis:
:$y_1 = \map f x = \map g x = y_2$
Similarly if $(4)$ holds.
Thus in all cases:
:$\tuple {x, y_1}, \tuple {x, y_2} \in f \cup g \implies y_1 = y_2$
and so by definition $f \cup g: X \to Y$ is a mapping.
{{qed}}
Theorem: Domain of Composite Mapping

Let $S_1, S_2, S_3$ be sets.
Let $f_1: S_1 \to S_2$ and $f_2: S_2 \to S_3$ be mappings.
Let $f_2 \circ f_1: S_1 \to S_3$ be the composite mapping of $f_1$ and $f_2$.
Then:
:$\Dom {f_1} = \Dom {f_2 \circ f_1}$
where $\Dom {f_1}$ denotes the domain of $f_1$.

Proof:
By definition of composition of mappings:
:$f_2 \circ f_1 := \set {\tuple {x, z} \in S_1 \times S_3: \exists y \in S_2: \tuple {x, y} \in f_1 \land \tuple {y, z} \in f_2}$
{{explain|Not quite what that definition says any more}}
Let $x \in \Dom {f_2 \circ f_1}$.
Then:
:$\exists z \in S_3: \tuple {x, z} \in S_1 \times S_3$
and:
:$\exists y \in S_2: \tuple {x, y} \in f_1$
That is:
:$x \in \Dom {f_1}$
Thus by definition of subset:
:$\Dom {f_2 \circ f_1} \subseteq \Dom {f_1}$
Now suppose $x \in \Dom {f_1}$.
By the definition of mapping:
:$\exists y \in S_2: \tuple {x, y} \in f_1$
As $f_2$ is likewise a mapping:
:$\exists z \in S_3: \tuple {y, z} \in f_2$
and so by definition of composition of mappings:
:$\tuple {x, z} \in S_1 \times S_3$
and so:
:$x \in \Dom {f_2 \circ f_1}$
Thus by definition of subset:
:$\Dom {f_1} \subseteq \Dom {f_2 \circ f_1}$
By definition of set equality:
:$\Dom {f_1} = \Dom {f_2 \circ f_1}$
{{qed}}
Theorem: Identity Mapping is Idempotent

Let $S$ be a set.
Let $I_S: S \to S$ be the identity mapping on $S$.
Then $I_S$ is idempotent:
:$I_S \circ I_S = I_S$

Proof:
From Identity Mapping is Left Identity:
:$I_S \circ f = f$
for all mappings $f: S \to S$.
From Identity Mapping is Right Identity:
:$f \circ I_S = f$
for all mappings $f: S \to S$.
Substituting $I_S$ for $f$ in either one and the result follows.
{{qed}}
Theorem: Union of Functions Theorem

Let $X$ be a set.
Let $\sequence {X_i: i \in \N}$ be an exhausting sequence of sets on $X$.
For each $i \in \N$, let $g_i: X_i \to Y$ be a mapping such that:
:$g_{i + 1} \restriction X_i = g_i$
where $g_{i + 1} \restriction X_i$ denotes the restriction of $g_{i + 1}$ to $g_i$.
Then:
:$\ds \bigcup \set {g_i: i \in \N}$
is a mapping from $X$ to $Y$.

Proof:
By definition, $\ds g = \bigcup \set {g_i: i \in \N}$ is a relation whose domain is $X$.
{{AimForCont}} $g$ is not a mapping.
Then for some $x \in X$ and $i, h \in \N$:
:$(1): \quad x \in X_i, \map {g_i} x \ne \map {g_{i + h} } x$
Let $k \in \N$ be the smallest such that:
:$\map {g_i} x \ne g_{i + k}$
where $x$ and $i$ are the same as in $(1)$.
Then:
:$\map {g_{i + k - 1} } x = g_i$
But then:
:$g_{i + k} \restriction X_{i + k - 1} = g_{i + k - 1}$
From this contradiction it follows that our supposition that $g$ is not a mapping must be false.
That is: 
:$\ds \bigcup \set {g_i: i \in \N}$
is a mapping from $X$ to $Y$.
{{qed}}
Theorem: Inductive Definition of Sequence

Let $X$ be a set.
Let $h \in \N$.
Let $a_i \in X$ for all $i \in \set {1, 2, \ldots, h}$.
Let $S$ be the set of all finite sequences whose codomains are in $X$.
Let $G: S \to X$ be a mapping.
Then there is a unique sequence $f$ whose codomain is in $X$ such that:
:$f_i = \begin{cases} a_i & : i \in \set {1, 2, \ldots, h} \\ \map G {f_1, f_2, \ldots, f_{i - 1} } & : i \ge h + 1 \end{cases}$

Proof:
{{finish|tedious}}
Theorem: Antireflexive and Transitive Relation is Antisymmetric

Let $\RR \subseteq S \times S$ be a relation which is not null.
Let $\RR$ be antireflexive and transitive.
Then $\RR$ is also antisymmetric.

Proof:
Let $\RR \subseteq S \times S$ be antireflexive and transitive.
From Antireflexive and Transitive Relation is Asymmetric it follows that $\RR$ is asymmetric.
The result follows from Asymmetric Relation is Antisymmetric.
{{qed}}
Theorem: Tableau Confutation implies Unsatisfiable

Let $\mathbf H$ be a collection of WFFs of propositional logic.
Suppose there exists a tableau confutation of $\mathbf H$.
Then $\mathbf H$ is unsatisfiable for boolean interpretations.

Proof:
Let $\struct {T, \mathbf H, \Phi}$ be a tableau confutation of $\mathbf H$.
Suppose that $v$ were a boolean interpretation model for $\mathbf H$, that is:
:$v \models_{\mathrm{BI}} \mathbf H$
By Model of Root of Propositional Tableau is Model of Branch, it follows that:
:$v \models_{\mathrm{BI}} \Phi \sqbrk \Gamma$
for some branch $\Gamma$ of $T$.
Since $T$ is a tableau confutation, there is some WFF $\mathbf A$ such that:
:$\mathbf A, \neg\mathbf A \in \Phi \sqbrk \Gamma$
Hence $v \models_{\mathrm{BI}} \mathbf A$, that is:
:$\map v {\mathbf A} = T$
But by the truth table for $\neg$, this means:
:$\map v {\neg \mathbf A} = F$
which contradicts that $v \models_{\mathrm{BI}} \neg \mathbf A$.
Hence, no such boolean interpretation can exist.
That is, $\mathbf H$ is unsatisfiable for boolean interpretations.
{{qed}}
Theorem: Diagonal Relation is Serial

Let $S$ be a set.
Let $\Delta_S$ be the diagonal relation on $S$.
Then $\Delta_S$ is a serial relation.

Proof:
By Diagonal Relation is Equivalence it follows {{afortiori}} that $\Delta_S$ is reflexive.
The result follows from Reflexive Relation is Serial.
{{qed}}
Theorem: Serial Relation is not Null

Let $S$ be a set such that $S \ne \O$.
Let $\RR$ be a serial relation on $S$.
Then $\RR$ is not a null relation.

Proof:
As $S$ is non-empty set:
:$\exists x: x \in S$
As $\RR$ be a serial relation on $S$:
:$\exists y \in S: \tuple {x, y} \in \RR$
That is:
:$\RR \ne \O$
Hence the result by definition of null relation.
{{qed}}
Theorem: Transitive Relation is Antireflexive iff Asymmetric

Let $\RR \subseteq S \times S$ be a relation which is not null.
Let $\RR$ be transitive.
Then $\RR$ is antireflexive {{iff}} $\RR$ is asymmetric.

Proof:
=== Necessary Condition ===
Let $\RR \subseteq S \times S$ be antireflexive.
Then by Antireflexive and Transitive Relation is Asymmetric it follows that $\RR$ is asymmetric.
{{qed|lemma}}
Theorem: Symmetric and Antisymmetric Relation is Transitive

Let $S$ be a set.
Let $\RR \subseteq S \times S$ be a relation in $S$ which is both symmetric and antisymmetric.
Then $\RR$ is transitive.

Proof:
Let $\tuple {x, y}, \tuple {y, z} \in \RR$.
By Relation is Symmetric and Antisymmetric iff Coreflexive:
:$x = y, y = z$
and so trivially:
:$\tuple {x, z} = \tuple {x, x} \in \RR$
Thus $\RR$ is transitive.
{{qed}}
Theorem: Relation is Reflexive Symmetric and Antisymmetric iff Diagonal Relation

Let $S$ be a set.
Let $\RR \subseteq S \times S$ be a relation in $S$.
Then:
:$\RR$ is reflexive, symmetric and antisymmetric
{{iff}}:
:$\RR$ is the diagonal relation $\Delta_S$.

Proof:
=== Necessary Condition ===
Let $\RR$ is reflexive, symmetric and antisymmetric.
By definition of reflexive:
:$\Delta_S \subseteq \RR$
From Relation is Symmetric and Antisymmetric iff Coreflexive:
:$\RR \subseteq \Delta_S$
By definition of set equality:
:$\RR = \Delta_S$
{{qed}}
Theorem: Relation is Reflexive and Coreflexive iff Diagonal

Let $S$ be a set.
Let $\RR \subseteq S \times S$ be a relation on $S$.
Then $\RR$ is reflexive and coreflexive {{iff}}:
: $\RR = \Delta_S$
where $\Delta_S$ is the diagonal relation.

Proof:
=== Necessary Condition ===
Let $\RR \subseteq S \times S$ be reflexive and coreflexive.
Then:
{{begin-eqn}}
{{eqn | l = \RR
      | o = \supseteq
      | r = \Delta_S
      | c = {{Defof|Reflexive Relation|index = 2}}
}}
{{eqn | l = \RR
      | o = \subseteq
      | r = \Delta_S
      | c = {{Defof|Coreflexive Relation|index = 2}}
}}
{{eqn | ll= \leadsto
      | l = \RR
      | r = \Delta_S
      | c = {{Defof|Set Equality|index = 2}}
}}
{{end-eqn}}
{{qed|lemma}}
Theorem: Symmetric Preordering is Equivalence Relation

Let $\RR \subseteq S \times S$ be a preordering on a set $S$.
Let $\RR$ also be symmetric.
Then $\RR$ is an equivalence relation on $S$.

Proof:
By definition, a preordering on $S$ is a relation on $S$ which is:
:$(1): \quad$ reflexive
and:
:$(2): \quad$ transitive.
Thus $\RR$ is a relation on $S$ which is reflexive, transitive and symmetric.
Thus by definition $\RR$ is an equivalence relation on $S$.
{{qed}}
Theorem: Antisymmetric Preordering is Ordering

Let $\RR \subseteq S \times S$ be a preordering on a set $S$.
Let $\RR$ also be antisymmetric.
Then $\RR$ is an ordering on $S$.

Proof:
By definition, a preordering on $S$ is a relation on $S$ which is:
:$(1): \quad$ reflexive
and:
:$(2): \quad$ transitive.
Thus $\RR$ is a relation on $S$ which is reflexive, transitive and antisymmetric.
Thus by definition $\RR$ is an ordering on $S$.
{{qed}}
Theorem: Relation is Antisymmetric and Reflexive iff Intersection with Inverse equals Diagonal Relation

Let $\RR \subseteq S \times S$ be a relation on a set $S$.
Then $\RR$ is both antisymmetric and reflexive {{iff}}:
:$\RR \cap \RR^{-1} = \Delta_S$
where $\Delta_S$ denotes the diagonal relation.

Proof:
=== Necessary Condition ===
Let $\RR$ be both antisymmetric and reflexive.
Then:
{{begin-eqn}}
{{eqn | l = \RR \cap \RR^{-1}
      | o = \subseteq
      | r = \Delta_S
      | c = Relation is Antisymmetric iff Intersection with Inverse is Coreflexive
}}
{{eqn | l = \RR
      | o = \supseteq
      | r = \Delta_S
      | c = {{Defof|Reflexive Relation|index = 2}}
}}
{{eqn | ll= \leadsto
      | l = \RR^{-1}
      | o = \supseteq
      | r = \Delta_S
      | c = Inverse of Reflexive Relation is Reflexive
}}
{{eqn | ll= \leadsto
      | l = \RR \cap \RR^{-1}
      | o = \supseteq
      | r = \Delta_S
      | c = Intersection is Largest Subset
}}
{{eqn | ll= \leadsto
      | l = \RR \cap \RR^{-1}
      | r = \Delta_S
      | c = {{Defof|Set Equality|index = 2}}
}}
{{end-eqn}}
{{qed|lemma}}
Theorem: Reflexive and Transitive Relation is Idempotent

Let $\RR \subseteq S \times S$ be a relation on a set $S$.
Let $\RR$ be both reflexive and  transitive.
Then $\RR$ is idempotent, in the sense that:
:$\RR \circ \RR = \RR$
where $\circ$ denotes composition of relations.

Proof:
Let $\RR$ be both reflexive and  transitive.
By definition of transitive relation:
:$\RR \circ \RR \subseteq \RR$
Let $\tuple {x, y} \in \RR$.
By definition of reflexive relation:
:$\tuple {y, y} \in \RR$
By definition of composition of relations:
:$\tuple {x, y} \in \RR \circ \RR$
Hence:
:$\RR \subseteq \RR \circ \RR$
Thus by definition of set equality:
:$\RR \circ \RR = \RR$
{{qed}}
Theorem: Equivalence of Definitions of Ordering/Proof 2

The following definitions of ordering are equivalent:

Proof:
=== Definition 1 implies Definition 2 ===
Let $\RR$ be a relation on $S$ satisfying:
{{begin-axiom}}
{{axiom | n = 1
        | lc= $\RR$ is reflexive
        | q = \forall a \in S
        | m = a \mathrel \RR a
}}
{{axiom | n = 2
        | lc= $\RR$ is transitive
        | q = \forall a, b, c \in S
        | m = a \mathrel \RR b \land b \mathrel \RR c \implies a \mathrel \RR c
}}
{{axiom | n = 3
        | lc= $\RR$ is antisymmetric
        | q = \forall a \in S
        | m = a \mathrel \RR b \land b \mathrel \RR a \implies a = b
}}
{{end-axiom}}
==== Condition $(1)$ ====
Let $\tuple {x, y} \in \RR \circ \RR$.
Then there exists a $z \in \RR$ such that:
:$\tuple {x, z}, \tuple {z, y} \in \RR$
By $\RR$ being transitive:
:$\tuple {x, y} \in \RR$
Hence:
:$\RR \circ \RR \subseteq \RR$
Now let $\tuple {x, y} \in \RR$.
By $\RR$ being reflexive:
:$\tuple {y, y} \in \RR$
Hence by the definition of relation composition:
:$\tuple {x, y} \in \RR \circ \RR$
Hence:
:$\RR \subseteq \RR \circ \RR$
==== Condition $(2)$ ====
Follows immediately from Relation is Antisymmetric iff Intersection with Inverse is Coreflexive and $\RR$ being reflexive.
Thus $\RR$ is an ordering by definition 2.
{{qed|lemma}}
=== Definition 2 implies Definition 1 ===
Let $\RR$ be a relation which fulfils the conditions:
:$(1): \quad \RR \circ \RR = \RR$
:$(2): \quad \RR \cap \RR^{-1} = \Delta_S$
==== Reflexivity ====
By Intersection is Subset the condition:
:$\RR \cap \RR^{-1} = \Delta_S$
implies:
:$\Delta_S \subseteq \RR$
Thus $\RR$ is reflexive by definition.
==== Antisymmetry ====
By Relation is Antisymmetric iff Intersection with Inverse is Coreflexive the condition:
:$\RR \cap \RR^{-1} = \Delta_S$
implies that $\RR$ is antisymmetric.
==== Transitivity ====
Let $\tuple {x, y}, \tuple {y, z} \in \RR$.
Then by the definition of relation composition:
:$\tuple {x, z} \in \RR \circ \RR$
But by the condition:
:$\RR \circ \RR = \RR$
It follows that:
:$\tuple {x, z} \in \RR$
Hence $\RR$ is transitive.
Thus $\RR$ is an ordering by definition 1.
{{qed}}
Theorem: Image of Domain of Relation is Image Set

Let $S$ and $T$ be sets.
Let $\RR \subseteq S \times T$ be a relation.
The image of the domain of $\RR$ is the image set of $\RR$:
:$\RR \sqbrk {\Dom \RR} = \Img \RR$
where $\Img \RR$ is the image of $\RR$.

Proof:
Let $y \in \RR \sqbrk {\Dom \RR}$.
{{begin-eqn}}
{{eqn | l = y
      | o = \in
      | r = \RR \sqbrk {\Dom \RR}
      | c = 
}}
{{eqn | ll= \leadsto
      | q = \exists x \in \Dom \RR
      | l = \tuple {x, y}
      | o = \in
      | r = \RR
      | c = {{Defof|Image of Subset under Relation}}
}}
{{eqn | ll= \leadsto
      | l = y
      | o = \in
      | r = \Img \RR
      | c = {{Defof|Image of Relation}}
}}
{{eqn | ll= \leadsto
      | l = \RR \sqbrk {\Dom \RR}
      | o = \subseteq
      | r = \Img \RR
      | c = {{Defof|Subset}}
}}
{{end-eqn}}
Let $y \in \Img \RR$.
{{begin-eqn}}
{{eqn | l = y
      | o = \in
      | r = \Img \RR
      | c = 
}}
{{eqn | ll= \leadsto
      | q = \exists x \in S
      | l = \tuple {x, y}
      | o = \in
      | r = \RR
      | c = {{Defof|Image of Relation}}
}}
{{eqn | ll= \leadsto
      | l = y
      | o = \in
      | r = \RR \sqbrk {\Dom \RR}
      | c = {{Defof|Domain of Relation}}
}}
{{eqn | ll= \leadsto
      | l = \Img \RR
      | o = \subseteq
      | r = \RR \sqbrk {\Dom \RR}
      | c = {{Defof|Subset}}
}}
{{end-eqn}}
The result follows by definition of set equality.
{{qed}}
Category:Relation Theory
bicnvpj1sktvr4vl336vn6fm7n8bx8n
Theorem: Condition for Relation to be Transitive and Antitransitive

Let $S$ be a set.
Let $\RR \subseteq S \times S$ be a relation in $S$.
Then:
:$\RR$ is both transitive and antitransitive
{{iff}}:
:$\neg \paren {\exists x, y, z \in S: x \mathrel \RR y \land y \mathrel \RR z}$

Proof:
=== Necessary Condition ===
Suppose $\neg \paren {\exists x, y, z \in S: x \mathrel \RR y \land y \mathrel \RR z}$.
Then $\RR$ is both transitive and antitransitive vacuously.
{{qed|lemma}}
Theorem: Transitive and Antitransitive Relation is Asymmetric

Let $S$ be a set.
Let $\RR \subseteq S \times S$ be a relation in $S$.
Let $\RR$ be both transitive and antitransitive.
Then $\RR$ is asymmetric.

Proof:
Let $\tuple {x, y} \in \RR$ for some $x, y \in S$.
Then as $\RR$ is antitransitive:
:$\tuple {x, x} \notin \RR$
and so as $\RR$ is transitive and $\tuple {x, x} \notin \RR$:
:$\tuple {y, x} \notin \RR$
That is, $\RR$ is asymmetric.
{{qed}}
Theorem: Valid Syllogism in Figure I needs Affirmative Minor Premise and Universal Major Premise

Let $Q$ be a valid categorical syllogism in Figure $\text I$.
Then it is a necessary condition that:
:The major premise of $Q$ be a universal categorical statement
and
:The minor premise of $Q$ be an affirmative categorical statement.

Proof:
Consider Figure $\text I$:
{{:Definition:First Figure of Categorical Syllogism}}
Let the major premise of $Q$ be denoted $\text{Maj}$.
Let the minor premise of $Q$ be denoted $\text{Min}$.
Let the conclusion of $Q$ be denoted $\text{C}$.
$M$ is:
:the subject of $\text{Maj}$
:the predicate of $\text{Min}$.
So, in order for $M$ to be distributed, either:
:$(1): \quad$ From Universal Categorical Statement Distributes its Subject: $\text{Maj}$ must be universal
or:
:$(2): \quad$ From Negative Categorical Statement Distributes its Predicate: $\text{Min}$ must be negative.
Suppose $\text{Min}$ is a negative categorical statement.
Then by Conclusion of Valid Categorical Syllogism is Negative iff one Premise is Negative:
:$\text{C}$ is a negative categorical statement.
From $(2)$:
:$P$ is distributed in $\text{C}$.
From Distributed Term of Conclusion of Valid Categorical Syllogism is Distributed in Premise:
:$P$ is distributed in $\text{Maj}$.
From Negative Categorical Statement Distributes its Predicate:
:$\text{Maj}$ is a negative categorical statement.
Thus both:
:$\text{Min}$ is a negative categorical statement
:$\text{Maj}$ is a negative categorical statement.
But from No Valid Categorical Syllogism contains two Negative Premises, this means $Q$ is invalid.
Thus $\text{Min}$ is not a negative categorical statement in Figure $\text I$.
As $\text{Min}$ needs to be an affirmative categorical statement, $M$ is not distributed in $\text{Min}$.
From Middle Term of Valid Categorical Syllogism is Distributed at least Once, this means $M$ must be distributed in $\text{Maj}$.
As $M$ is the subject of $\text{Maj}$ in Figure $\text I$, it follows from $(1)$ that:
:$\text{Maj}$ is a universal categorical statement.
Hence, in order for $Q$ to be valid:
:$\text{Maj}$ must be a universal categorical statement
:$\text{Min}$ must be an affirmative categorical statement.
{{qed}}
Theorem: Valid Syllogism in Figure II needs Negative Conclusion and Universal Major Premise

Let $Q$ be a valid categorical syllogism in Figure $\text{II}$.
Then it is a necessary condition that:
:The major premise of $Q$ be a universal categorical statement
and
:The conclusion of $Q$ be a negative categorical statement.

Proof:
Consider Figure $\text{II}$:
{{:Definition:Figure of Categorical Syllogism/II}}
Let the major premise of $Q$ be denoted $\text{Maj}$.
Let the minor premise of $Q$ be denoted $\text{Min}$.
Let the conclusion of $Q$ be denoted $\text{C}$.
$M$ is:
:the predicate of $\text{Maj}$
:the predicate of $\text{Min}$.
So, in order for $M$ to be distributed, either:
:$(1): \quad$ From Negative Categorical Statement Distributes its Predicate: $\text{Maj}$ must be negative
or:
:$(2): \quad$ From Negative Categorical Statement Distributes its Predicate: $\text{Min}$ must be negative.
Note that from No Valid Categorical Syllogism contains two Negative Premises, it is not possible for both $\text{Maj}$ and $\text{Min}$ to be negative.
From Conclusion of Valid Categorical Syllogism is Negative iff one Premise is Negative:
:$\text{C}$ is a negative categorical statement.
From Negative Categorical Statement Distributes its Predicate:
:$P$ is distributed in $\text{C}$.
From Distributed Term of Conclusion of Valid Categorical Syllogism is Distributed in Premise:
:$P$ is distributed in $\text{Maj}$.
From Universal Categorical Statement Distributes its Subject:
:$\text{Maj}$ is a universal categorical statement.
Hence, in order for $Q$ to be valid:
:$\text{Maj}$ must be a universal categorical statement
:Either $\text{Maj}$ or $\text{Min}$, and therefore $\text{C}$, must be a negative categorical statement.
{{qed}}