nyuuzyou/wikis huggingface dataset:
Rule of Assumption/Proof Rule/Tableau Form

== Proof Rule ==
Let $\phi$ be a propositional formula.
The Rule of Assumption is invoked for $\phi$ in a tableau proof in the following manner:
Note that the Description: field of the Rule of Assumption is populated with either:
:$(1): \quad$ Premise if the statement being assumed is one of the premises of the argument
or:
:$(2): \quad$ Assumption if it is to be discharged later in the proof.
On completion of the proof, only those lines annotated Premise are to remain in the pool of assumptions of the conclusion.
== Also defined as ==
Some sources introduce the formula line as:
:Show $\phi$
== Sources ==
* : Chapter $1$: The Propositional Calculus $1$: $2$ Conditionals and Negation
Category:Tableau Proofs
Category:Law of Identity
Rule of Assumption/Proof Rule/Technical Note

== Technical Note on ==
When invoking the in a tableau proof, use the template:
:{{Assumption|line|statement}}
or:
:{{Assumption|line|statement|comment}}
where:
:line is the number of the line on the tableau proof where the assumption is to be invoked
:statement is the statement of logic that is to be displayed in the Formula column, without the $ ... $ delimiters
:comment is the (optional) comment that is to be displayed in the Notes column.
When the assumption being made is a premise, use the template:
:{{Premise|line|statement}}
or:
:{{Premise|line|statement|comment}}
in the same manner as above.
Category:Technical Notes
Rule of Commutation

== Theorem ==
== Conjunction ==
== Disjunction ==
Its abbreviation in a tableau proof is $\text{Comm}$.
== Also known as ==
The rule of commutation is also known as the commutative law.
Note that this term is also used throughout mathematics in the context of addition and multiplication of numbers:
:the Commutative Law of Addition
:the Commutative Law of Multiplication
so it is wise to be aware of the context.
== Also see ==
* Rule of Association
* Principle of Commutation
== Technical Note ==
Category:Conjunction
Category:Disjunction
Category:Examples of Commutative Operations
Rule of Commutation/Conjunction

== Theorem ==
Conjunction is commutative:
=== Formulation 1 ===
=== Formulation 2 ===
Its abbreviation in a tableau proof is $\text{Comm}$.
== Also see ==
* Rule of Association
* Principle of Commutation
== Sources ==
* 
* 
Category:Conjunction
Category:Rule of Commutation
Rule of Commutation/Conjunction/Formulation 1

== Theorem ==
:$p \land q \dashv \vdash q \land p$
== Proof 1 ==
== Proof by Truth Table ==
== Sources ==
* : $\S 1.2.1$: Rules for natural deduction: Exercise $1.2: \ 2$
* : $\S 2.3.3$
Category:Rule of Commutation
Rule of Commutation/Conjunction/Formulation 1/Proof by Truth Table

== Theorem ==
== Proof ==
We apply the Method of Truth Tables.
As can be seen by inspection, the truth values under the main connectives match for all boolean interpretations.
$\begin{array}{|ccc||ccc|} \hline
p & \land & q & q & \land & p \\
\hline
\F & \F & \F & \F & \F & \F \\
\F & \F & \T & \T & \F & \F \\
\T & \F & \F & \F & \F & \T \\
\T & \T & \T & \T & \T & \T \\
\hline
\end{array}$
Category:Rule of Commutation
Category:Truth Table Proofs
Rule of Commutation/Conjunction/Formulation 2

== Theorem ==
:$\vdash \paren {p \land q} \iff \paren {q \land p}$
== Proof ==
== Sources ==
* : $\S \text{II}.13$: Symbolism of sentential calculus
* : $\S 3.6$: Reference Formulae: $RF \, 8$
* : $\S 4.7$: The Derivation of Formulae: $D \, 21 - 22$
* : $\text{II}$: 'AND', 'OR', 'IF AND ONLY IF': $\S 3$: Theorem $\text{T24}$
* : $3$: The Method of Deduction: $3.2$: The Rule of Replacement: $11.$
* : $\S \text{II}$: The Logic of Statements $(2): \ 6$: Using logical equivalences: $10$
* : Chapter $1$: Sets and mappings: $\S 1.1$: The need for logic: Exercise $(1) \ \text{(iii)}$
* : $\S 1.14$: Exercise $12 \ (4)$
Category:Rule of Commutation
Rule of Commutation/Disjunction

== Theorem ==
Disjunction is commutative:
=== Formulation 1 ===
=== Formulation 2 ===
Its abbreviation in a tableau proof is $\text{Comm}$.
== Also see ==
* Rule of Association
* Principle of Commutation
== Sources ==
* 
* 
Category:Disjunction
Category:Rule of Commutation
Rule of Commutation/Disjunction/Formulation 1/Proof 2

== Theorem ==
== Proof ==
We apply the Method of Truth Tables.
As can be seen by inspection, in both cases, the truth values under the main connectives match for all boolean interpretations.
:$\begin{array}{|ccc||ccc|} \hline
p & \lor & q & q & \lor & p \\
\hline
\F & \F & \F & \F & \F & \F \\
\F & \T & \T & \T & \T & \F \\
\T & \T & \F & \F & \T & \T \\
\T & \T & \T & \T & \T & \T \\
\hline
\end{array}$
== Sources ==
* : Chapter $2$: Propositional Calculus: $\S 2.4$: Logical equivalence and substitution: Theorem $2.4.2$
* : $\S 2.3$: Theorem $2.28$
Category:Truth Table Proofs
Category:Rule of Commutation
Rule of Commutation/Disjunction/Formulation 2

== Theorem ==
:$\vdash \paren {p \lor q} \iff \paren {q \lor p}$
=== Forward Implication ===
== Proof 1 ==
== Proof by Truth Table ==
== Sources ==
* : $\S \text{II}.13$: Symbolism of sentential calculus
* : $\S 3.6$: Reference Formulae: $RF \, 9$
* : $\S 4.7$: The Derivation of Formulae: $D \, 14$
* : $\text{II}$: 'AND', 'OR', 'IF AND ONLY IF': $\S 5$: Theorem $\text{T53}$
* : $3$: The Method of Deduction: $3.2$: The Rule of Replacement: $11.$
* : $\S \text{II}$: The Logic of Statements $(2): \ 6$: Using logical equivalences: $10$
* : Chapter $1$: Sets and mappings: $\S 1.1$: The need for logic: Exercise $(1) \ \text{(iv)}$
* : $\S 1.14$: Exercise $12 \ (5)$
* 
* 
Category:Rule of Commutation
Rule of Commutation/Disjunction/Formulation 2/Forward Implication

== Theorem ==
:$\vdash \left({p \lor q}\right) \implies \left({q \lor p}\right)$
== Proof ==
Using a tableau proof for instance 1 of a Gentzen proof system:
$\vdash \left({p \lor q}\right) \implies \left({q \lor p}\right)$
Line		Pool	Formula 	Rule 	Depends upon	Notes
1 		 	$\neg p, q, p$ 	Axiom 	 	
2 		 	$\neg q, q, p$ 	Axiom 	 	
3 		 	$\neg \left({p \lor q}\right), q, p$ 	$\alpha$-Rule: $\alpha\lor$ 	1, 2 	
4 		 	$\neg \left({p \lor q}\right), q \lor p$ 	$\beta$-Rule: $\beta\lor$ 	3 	
5 		 	$\left({p \lor q}\right) \implies \left({q \lor p}\right)$	$\beta$-Rule: $\beta\implies$	4 	
== Also see ==
* This is an axiom of the following proof system:
** Definition:Hilbert Proof System/Instance 2
== Sources ==
* : $\S 3.2$: Example $3.4$
Category:Rule of Commutation
Rule of Commutation/Disjunction/Formulation 2/Proof by Truth Table

== Theorem ==
== Proof ==
We apply the Method of Truth Tables.
As can be seen by inspection, the truth values under the main connective match for all boolean interpretations.
$\begin{array}{|ccc|c|ccc|} \hline
(p & \lor & q) & \iff & (q & \lor & p) \\
\hline
\F & \F & \F & \T & \F & \F & \F \\
\F & \T & \T & \T & \T & \T & \F \\
\T & \T & \F & \T & \F & \T & \T \\
\T & \T & \T & \T & \T & \T & \T \\
\hline
\end{array}$
Category:Truth Table Proofs
Category:Rule of Commutation
Rule of Commutation/Technical Note

== Technical Note on Rule of Commutation ==
When invoking the Rule of Commutation in a tableau proof, use the template:
:{{Commutation|line|pool|statement|depends|type}}
where:
:line is the number of the line on the tableau proof where Rule of Commutation is to be invoked
:pool is the pool of assumptions (comma-separated list)
:statement is the statement of logic that is to be displayed in the Formula column, without the $ ... $ delimiters
:depends is the line (or lines) of the tableau proof upon which this line directly depends
:type is the type of Rule of Commutation: specifically Disjunction or Conjunction, whose link will be displayed in the Notes column.
Category:Technical Notes
Category:Rule of Commutation
Rule of Conjunction

== Sequent ==
The rule of conjunction is a valid argument in types of logic dealing with conjunctions $\land$.
This includes propositional logic and predicate logic, and in particular natural deduction.
=== Proof Rule ===
=== Sequent Form ===
== Explanation ==
Thus a conjunction is added to a sequent.
== Also known as ==
== Also see ==
* Definition:Conjunction
* Rule of Simplification
Category:Conjunction
Category:Rule of Conjunction
Rule of Conjunction/Proof Rule

== Proof Rule ==
The rule of conjunction is a valid argument in types of logic dealing with conjunctions $\land$.
This includes propositional logic and predicate logic, and in particular natural deduction.
As a proof rule it is expressed in the form:
:If we can conclude both $\phi$ and $\psi$, we may infer the compound statement $\phi \land \psi$.
It can be written:
:$\ds {\phi \qquad \psi \over \phi \land \psi} \land_i$
=== Tableau Form ===
== Explanation ==
== Also known as ==
== Also see ==
* This is a rule of inference of the following proof systems:
** Definition:Natural Deduction
** Definition:Hilbert Proof System/Instance 2
* Rule of Simplification
== Technical Note ==
== Sources ==
* : $\S 4.2$: The Construction of an Axiom System: $RST \, 4$
* : $\text{II}$: 'AND', 'OR', 'IF AND ONLY IF': $\S 3$
* : Chapter $1$: The Propositional Calculus $1$: $3$ Conjunction and Disjunction
* : $\S \text{II}$: The Logic of Statements $(2): \ 1$: Decision procedures and proofs: $9$
* : $\S 1.2.1$: Rules for natural deduction
Category:Rule of Conjunction
Category:Rules of Inference
Rule of Conjunction/Proof Rule/Tableau Form

== Proof Rule ==
Let $\phi$ and $\psi$ be two well-formed formulas in a tableau proof.
The Rule of Conjunction is invoked for $\phi$ and $\psi$ in the following manner:
== Also denoted as ==
Sources which refer to this rule as the rule of adjunction may as a consequence give the abbreviation $\operatorname {Adj}$.
== Sources ==
* : Chapter $1$: The Propositional Calculus $1$: $3$ Conjunction and Disjunction
Category:Rule of Conjunction
Category:Tableau Proofs
Rule of Conjunction/Proof Rule/Technical Note

== Technical Note on ==
When invoking the in a tableau proof, use the template:
:{{Conjunction|line|pool|statement|first|second}}
or:
:{{Conjunction|line|pool|statement|first|second|comment}}
where:
:line is the number of the line on the tableau proof where the is to be invoked
:pool is the pool of assumptions (comma-separated list)
:statement is the statement of logic that is to be displayed in the Formula column, without the $ ... $ delimiters
:first is the first of the two lines of the tableau proof upon which this line directly depends
:second is the second of the two lines of the tableau proof upon which this line directly depends
:comment is the (optional) comment that is to be displayed in the Notes column.
Category:Technical Notes
Rule of Conjunction/Sequent Form/Formulation 1

== Theorem ==
The Rule of Conjunction can be symbolised by the sequent:
== Proof 1 ==
== Proof by Truth Table ==
== Sources ==
* : $2$ Arguments Containing Compound Statements: $2.3$: Argument Forms and Truth Tables: Exercise $\text{I} \ \mathbf d.$
* : $3$: The Method of Deduction: $3.1$: Formal Proof of Validity: Rules of Inference: $8.$
Category:Rule of Conjunction
Rule of Conjunction/Sequent Form/Formulation 1/Proof by Truth Table

== Theorem ==
== Proof ==
We apply the Method of Truth Tables.
$\begin{array}{|c|c||ccc|} \hline
p & q & p & \land & q\\
\hline
\F & \F & \F & \F & \F \\
\F & \T & \F & \F & \T \\
\T & \F & \T & \F & \F \\
\T & \T & \T & \T & \T \\
\hline
\end{array}$
As can be seen, only when both $p$ and $q$ are true, then so is $p \land q$.
Category:Truth Table Proofs
Category:Rule of Conjunction
Rule of Conjunction/Sequent Form/Formulation 2

== Theorem ==
The Rule of Conjunction can be symbolised by the sequent:
:$\vdash p \implies \paren {q \implies \paren {p \land q} }$
== Proof ==
== Sources ==
* : $\S 4.7$: The Derivation of Formulae: $D \, 11, 12$
Category:Rule of Conjunction
Category:Hilbert Proof System Instance 2
Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive/Formulation 1

== Definition ==
The conjunction operator is left distributive over the disjunction operator:
:$p \land \paren {q \lor r} \dashv \vdash \paren {p \land q} \lor \paren {p \land r}$
This can be expressed as two separate theorems:
=== Forward Implication ===
=== Reverse Implication ===
== Proof by Truth Table ==
== Sources ==
* : Chapter $1$: The Propositional Calculus $1$: $5$ Further Proofs: Résumé of Rules: Exercise $1 \ \text{(c)}$
* : $\S 2.3.3$
* : $\S 2.10$: Exercise $2.3$
Category:Rule of Distribution
Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive/Formulation 1/Forward Implication

== Definition ==
:$p \land \paren {q \lor r} \vdash \paren {p \land q} \lor \paren {p \land r}$
== Proof ==
== Sources ==
* : $\S 1.2.1$: Rules for natural deduction: Example $1.18$
Category:Rule of Distribution
Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive/Formulation 1/Proof by Truth Table

== Theorem ==
== Proof ==
We apply the Method of Truth Tables to the proposition.
As can be seen by inspection, the truth values under the main connectives match for all boolean interpretations.
$\begin{array}{|ccccc||ccccccc|} \hline
p & \land & (q & \lor & r) & (p & \land & q) & \lor & (p & \land & r) \\
\hline
\F & \F & \F & \F & \F & \F & \F & \F & \F & \F & \F & \F \\
\F & \F & \F & \T & \T & \F & \F & \F & \F & \F & \F & \T \\
\F & \F & \T & \T & \F & \F & \F & \T & \F & \F & \F & \F \\
\F & \F & \T & \T & \T & \F & \F & \T & \F & \F & \F & \T \\
\T & \F & \F & \F & \F & \T & \F & \F & \F & \T & \F & \F \\
\T & \T & \F & \T & \T & \T & \F & \F & \T & \T & \T & \T \\
\T & \T & \T & \T & \F & \T & \T & \T & \T & \T & \F & \F \\
\T & \T & \T & \T & \T & \T & \T & \T & \T & \T & \T & \T \\
\hline
\end{array}$
Category:Rule of Distribution
Category:Truth Table Proofs
Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive/Formulation 1/Reverse Implication

== Definition ==
:$\paren {p \land q} \lor \paren {p \land r} \vdash p \land \paren {q \lor r}$
== Proof ==
== Sources ==
* : $\S 1.2.1$: Rules for natural deduction: Exercises $1.4: \ 2 \ \text{(q)}$
Category:Rule of Distribution
Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive/Formulation 2

== Theorem ==
:$\vdash \paren {p \land \paren {q \lor r} } \iff \paren {\paren {p \land q} \lor \paren {p \land r} }$
== Proof 1 ==
== Proof 2 ==
== Proof by Truth Table ==
== Sources ==
* : $\S 3.6$: Reference Formulae: $RF \, 12$
* : $\S 4.7$: The Derivation of Formulae: $D \, 34$
* : $\text{II}$: 'AND', 'OR', 'IF AND ONLY IF': $\S 5$: Theorem $\text{T61}$
* : $3$: The Method of Deduction: $3.2$: The Rule of Replacement: $13.$
* : $\S \text{II}$: The Logic of Statements $(2): \ 6$: Using logical equivalences: $12$
* : Chapter $1$: Sets and mappings: $\S 1.1$: The need for logic: Exercise $(1) \ \text{(vii)}$
* : $\S 1.14$: Exercise $12 \ (6)$
* : Chapter $1$: Elementary, my dear Watson: $\S 1.1$: You have a logical mind if... $\text{(a)}$
::(erroneously referring to it as one of De Morgan's Laws)
Category:Rule of Distribution
Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive/Formulation 2/Proof by Truth Table

== Theorem ==
== Proof ==
We apply the Method of Truth Tables to the proposition.
As can be seen by inspection, the truth values under the main connective is true for all boolean interpretations.
$\begin{array}{|ccccc|c|ccccccc|} \hline
p & \land & (q & \lor & r) & \iff & (p & \land & q) & \lor & (p & \land & r) \\
\hline
\F & \F & \F & \F & \F & \T & \F & \F & \F & \F & \F & \F & \F \\
\F & \F & \F & \T & \T & \T & \F & \F & \F & \F & \F & \F & \T \\
\F & \F & \T & \T & \F & \T & \F & \F & \T & \F & \F & \F & \F \\
\F & \F & \T & \T & \T & \T & \F & \F & \T & \F & \F & \F & \T \\
\T & \F & \F & \F & \F & \T & \T & \F & \F & \F & \T & \F & \F \\
\T & \T & \F & \T & \T & \T & \T & \F & \F & \T & \T & \T & \T \\
\T & \T & \T & \T & \F & \T & \T & \T & \T & \T & \T & \F & \F \\
\T & \T & \T & \T & \T & \T & \T & \T & \T & \T & \T & \T & \T \\
\hline
\end{array}$
== Sources ==
* : $\S \text{II}$: Exercise $14 \ \text{(c)}$
* : $2$ Arguments Containing Compound Statements: $2.4$: Statement Forms: Exercise $\text{II}. \ 5$
Category:Rule of Distribution
Rule of Distribution/Conjunction Distributes over Disjunction/Right Distributive/Formulation 1

== Theorem ==
The conjunction operator is right distributive over the disjunction operator:
:$\paren {q \lor r} \land p \dashv \vdash \paren {q \land p} \lor \paren {r \land p}$
This can be expressed as two separate theorems:
=== Forward Implication ===
=== Reverse Implication ===
== Proof by Truth Table ==
== Sources ==
* : $\S 1$: Informal statement calculus: $\S 1.2$: Truth functions and truth tables: Exercise $6 \ \text{(b)}$
Category:Rule of Distribution
Rule of Distribution/Conjunction Distributes over Disjunction/Right Distributive/Formulation 1/Proof by Truth Table

== Theorem ==
== Proof ==
We apply the Method of Truth Tables to the proposition.
As can be seen by inspection, the truth values under the main connectives match for all boolean interpretations.
:$\begin{array}{|ccccc||ccccccc|} \hline
(q & \lor & r) & \land & p & (q & \land & p) & \lor & (r & \land & p) \\
\hline
\F & \F & \F & \F & \F & \F & \F & \F & \F & \F & \F & \F \\
\F & \F & \F & \F & \T & \F & \F & \T & \F & \F & \F & \T \\
\F & \T & \T & \F & \F & \F & \F & \F & \F & \T & \F & \F \\
\F & \T & \T & \T & \T & \F & \F & \T & \T & \T & \T & \T \\
\T & \T & \F & \F & \F & \T & \F & \F & \F & \F & \F & \F \\
\T & \T & \F & \T & \T & \T & \T & \T & \T & \F & \F & \T \\
\T & \T & \T & \F & \F & \T & \F & \F & \F & \T & \F & \F \\
\T & \T & \T & \T & \T & \T & \T & \T & \T & \T & \T & \T \\
\hline
\end{array}$
Category:Rule of Distribution
Category:Truth Table Proofs
Rule of Distribution/Conjunction Distributes over Disjunction/Right Distributive/Formulation 2

== Theorem ==
The conjunction operator is right distributive over the disjunction operator:
:$\vdash \paren {\paren {q \lor r} \land p} \iff \paren {\paren {q \land p} \lor \paren {r \land p} }$
== Proof ==
{{qed}
Category:Rule of Distribution
Rule of Distribution/Disjunction Distributes over Conjunction/Left Distributive/Formulation 1

== Definition ==
The disjunction operator is left distributive over the conjunction operator:
:$p \lor \paren {q \land r} \dashv \vdash \paren {p \lor q} \land \paren {p \lor r}$
This can be expressed as two separate theorems:
=== Forward Implication ===
=== Reverse Implication ===
== Proof by Truth Table ==
== Sources ==
* : Chapter $1$: The Propositional Calculus $1$: $5$ Further Proofs: Résumé of Rules: Exercise $1 \ \text{(d)}$
* : $\S 2.3.3$
Category:Rule of Distribution
Rule of Distribution/Disjunction Distributes over Conjunction/Left Distributive/Formulation 1/Proof by Truth Table

== Theorem ==
== Proof ==
We apply the Method of Truth Tables to the proposition.
As can be seen by inspection, the truth values under the main connectives match for all boolean interpretations.
$\begin{array}{|ccccc||ccccccc|} \hline
p & \lor & (q & \land & r) & (p & \lor & q) & \land & (p & \lor & r) \\
\hline
\F & \F & \F & \F & \F & \F & \F & \F & \F & \F & \F & \F \\
\F & \F & \F & \F & \T & \F & \F & \F & \F & \F & \T & \T \\
\F & \F & \T & \F & \F & \F & \T & \T & \F & \F & \F & \F \\
\F & \T & \T & \T & \T & \F & \T & \T & \T & \F & \T & \T \\
\T & \T & \F & \F & \F & \T & \T & \F & \T & \T & \T & \F \\
\T & \T & \F & \F & \T & \T & \T & \F & \T & \T & \T & \T \\
\T & \T & \T & \F & \F & \T & \T & \T & \T & \T & \T & \F \\
\T & \T & \T & \T & \T & \T & \T & \T & \T & \T & \T & \T \\
\hline
\end{array}$
Category:Rule of Distribution
Category:Truth Table Proofs
Rule of Distribution/Disjunction Distributes over Conjunction/Left Distributive/Formulation 2

== Theorem ==
:$\vdash \paren {p \lor \paren {q \land r} } \iff \paren {\paren {p \lor q} \land \paren {p \lor r} }$
=== Forward Implication ===
=== Reverse Implication ===
== Proof 1 ==
== Proof by Truth Table ==
== Sources ==
* : $\S 3.6$: Reference Formulae: $RF \, 13$
* : $\S 4.7$: The Derivation of Formulae: $D \, 33$
* : $\text{II}$: 'AND', 'OR', 'IF AND ONLY IF': $\S 5$: Theorem $\text{T62}$
* : $3$: The Method of Deduction: $3.2$: The Rule of Replacement: $13.$
* : $\S \text{II}$: The Logic of Statements $(2): \ 6$: Using logical equivalences: $12$
* : Chapter $1$: Sets and mappings: $\S 1.1$: The need for logic: Exercise $(1) \ \text{(viii)}$
* : $\S 1.14$: Exercise $12 \ (7)$
* : Chapter $1$: Elementary, my dear Watson: $\S 1.1$: You have a logical mind if... $\text{(b)}$
::(erroneously referring to it as one of De Morgan's Laws)
Category:Rule of Distribution
Rule of Distribution/Disjunction Distributes over Conjunction/Left Distributive/Formulation 2/Forward Implication

== Theorem ==
:$\vdash \paren {p \lor \paren {q \land r} } \implies \paren {\paren {p \lor q} \land \paren {p \lor r} }$
== Proof 1 ==
== Proof 2 ==
Category:Rule of Distribution
Rule of Distribution/Disjunction Distributes over Conjunction/Left Distributive/Formulation 2/Proof by Truth Table

== Theorem ==
== Proof ==
We apply the Method of Truth Tables to the proposition.
As can be seen by inspection, the truth values under the main connective is true for all boolean interpretations.
$\begin{array}{|ccccc|c|ccccccc|} \hline
p & \lor & (q & \land & r) & \iff & (p & \lor & q) & \land & (p & \lor & r) \\
\hline
\F & \F & \F & \F & \F & \T & \F & \F & \F & \F & \F & \F & \F \\
\F & \F & \F & \F & \T & \T & \F & \F & \F & \F & \F & \T & \T \\
\F & \F & \T & \F & \F & \T & \F & \T & \T & \F & \F & \F & \F \\
\F & \T & \T & \T & \T & \T & \F & \T & \T & \T & \F & \T & \T \\
\T & \T & \F & \F & \F & \T & \T & \T & \F & \T & \T & \T & \F \\
\T & \T & \F & \F & \T & \T & \T & \T & \F & \T & \T & \T & \T \\
\T & \T & \T & \F & \F & \T & \T & \T & \T & \T & \T & \T & \F \\
\T & \T & \T & \T & \T & \T & \T & \T & \T & \T & \T & \T & \T \\
\hline
\end{array}$
== Sources ==
* : $\S \text{II}$: Exercise $14 \ \text{(c)}$
* : $2$ Arguments Containing Compound Statements: $2.4$: Statement Forms: Exercise $\text{II}. \ 7$
Category:Rule of Distribution
Rule of Distribution/Disjunction Distributes over Conjunction/Left Distributive/Formulation 2/Reverse Implication

== Theorem ==
:$\vdash \paren {\paren {p \lor q} \land \paren {p \lor r} } \implies \paren {p \lor \paren {q \land r} }$
== Proof ==
== Sources ==
* : $\S 4.7$: The Derivation of Formulae: $D \, 32$
Category:Rule of Distribution
Rule of Distribution/Disjunction Distributes over Conjunction/Right Distributive/Formulation 1

== Theorem ==
The disjunction operator is right distributive over the conjunction operator:
:$\paren {q \land r} \lor p \dashv \vdash \paren {q \lor p} \land \paren {r \lor p}$
This can be expressed as two separate theorems:
=== Forward Implication ===
=== Reverse Implication ===
== Proof ==
Category:Rule of Distribution
Rule of Distribution/Disjunction Distributes over Conjunction/Right Distributive/Formulation 1/Proof

== Theorem ==
:$\left({q \land r}\right) \lor p \dashv \vdash \left({q \lor p}\right) \land \left({r \lor p}\right)$
== Proof ==
We apply the Method of Truth Tables to the proposition.
As can be seen by inspection, the truth values under the main connectives match for all boolean interpretations.
$\begin{array}{|ccccc||ccccccc|} \hline
(q & \land & r) & \lor & p & (q & \lor & p) & \land & (r & \lor & p) \\
\hline
F & F & F & F & F & F & F & F & F & F & F & F \\
F & F & F & T & T & F & T & T & T & F & T & T \\
F & F & T & F & F & F & F & F & F & T & T & F \\
F & F & T & T & T & F & T & T & T & T & T & T \\
T & F & F & F & F & T & T & F & F & F & F & F \\
T & F & F & T & T & T & T & T & T & F & T & T \\
T & T & T & T & F & T & T & F & T & T & T & F \\
T & T & T & T & T & T & T & T & T & T & T & T \\
\hline
\end{array}$
Category:Rule of Distribution
Category:Truth Table Proofs
Rule of Distribution/Disjunction Distributes over Conjunction/Right Distributive/Formulation 2

== Theorem ==
The disjunction operator is right distributive over the conjunction operator:
:$\vdash \paren {\paren {q \land r} \lor p} \iff \paren {\paren {q \lor p} \land \paren {r \lor p} }$
== Proof ==
Category:Rule of Distribution