nyuuzyou/wikis huggingface dataset:
Menelaus's Theorem

== Theorem ==
Let $ABC$ be a triangle.
Let points $D, E, F$ lie on lines $BC, AC, AB$ respectively (produced if necessary).
Then $D, E$ and $F$ are collinear :
: $\dfrac {AF} {FB} \cdot \dfrac {BD} {DC} \cdot \dfrac {CE} {EA} = -1$
In the above, the line segments $AF, BD, EA$ are determined to have negative length if they lie outside the line segments $AB, BC, CA$.
An equivalent definition is that, as directed line segments, $AF$ and $FB$ lie in opposite directions.
Hence the ratio $\dfrac {AF} {FB}$ is negative, accounting for $-1$ in the product above.
== Proof ==
=== Necessary Condition ===
First we check that the equation works out negative.
From Pasch's Axiom, the line $DEF$ must intersect either two sides of $\triangle ABC$:
:
or none of them:
:
That means there is an odd number of negative contributions to the product.
Hence the equation works out to be negative.
Next, the magnitude can be checked.
The method is illustrated using the second of the above options; the first works the same.
:
Construct perpendiculars $AD$, $CH$ and $BI$ from $A$, $B$ and $C$ to $DEF$.
The triangles $\triangle AEG$, $\triangle CEH$ are similar.
The triangles $\triangle BDI$, $\triangle CDH$ are also similar.
The triangles $\triangle BFI$, $\triangle AFG$ are also similar.
Hence:
:$\dfrac {CE} {EA} = \dfrac {CH} {AG}, \dfrac {BD} {DC} = \dfrac {BI} {CH}, \dfrac {AF} {FB} = \dfrac {AG} {BI}$
Thus:
:$\size {\dfrac {AF} {FB} \cdot \dfrac{BD} {DC} \cdot \dfrac{CE} {EA} } = \size {\dfrac {AG} {BI} \cdot \dfrac {BI} {CH} \cdot \dfrac {CH} {AG} } = 1$
Hence the result.
=== Sufficient Condition ===
Suppose $\dfrac {AF} {FB} \cdot \dfrac {BD} {DC} \cdot \dfrac {CE} {EA} = -1$.
Let $F'$ be a point on $AB$ distinct from $F$.
Let us define the measurements of $AF$, $AF'$, $AB$ as $n, n', s$ respectively.
, suppose that $F'$ also satisfies the equation.
Then:
:$\dfrac {AF} {FB} = \dfrac {AF'} {F'B}$
which means:
:$\dfrac n {s - n} = \dfrac {n'} {s - n'}$
and so $n = n'$ and so $F = F'$.
So only one point on line $AB$ can satisfy the equation.
Let us fix $D$ and $E$.
Then if $F$ satisfies the equation, then it must be the point collinear with $D$ and $E$.
By a similar argument, the same applies to $D$ and $E$.
Hence the result.
== Also known as ==
== Also see ==
* Ceva's Theorem, to which this one is related.
== Historical Note ==
== Sources ==
* 
* 
* 
* 
Category:Menelaus's Theorem
Category:Triangles
Mergelyan-Wesler Theorem

== Theorem ==
Let $P = \sequence {D_1, D_2, \dotsc}$ be an infinite sequence of disjoint open disks whose union is the unit disk $D$ except for a set of measure zero.
Let $r_n$ be the radius of $D_n$.
Then:
:$\ds \sum_{k \mathop = 1}^\infty r_k = +\infty$
== Proof ==
== Sources ==
* 
* 
* 
* : $1,30695 1 \ldots$
Category:Complex Analysis
Meromorphic Function is Elliptic iff Doubly Periodic

== Theorem ==
Let $f: \C \to \C$ be a meromorphic function. 
Then $\map f z$ is an elliptic function it is also doubly periodic.
== Proof ==
== Historical Note ==
== Sources ==
* : Chapter $\text {A}.33$: Weierstrass ()
Category:Elliptic Functions
Meromorphic Function is Elliptic iff Doubly Periodic/Historical Note

== Historical Note on Meromorphic Function is Elliptic iff Doubly Periodic ==
Meromorphic Function is Elliptic iff Doubly Periodic was discovered by during his exploration of elliptic functions.
== Sources ==
* : Chapter $\text {A}.33$: Weierstrass ()
Category:Historical Notes
Mersenne Number whose Index is Mersenne Prime

== Conjecture ==
A Mersenne number whose index is a Mersenne prime is itself a Mersenne prime.
== Refutation ==
The smallest counterexample is $M_{8191}$.
We have that $M_{13} = 2^{13} - 1 = 8191$ is a Mersenne prime.
But $M_{8191} = 2^{8191} - 1$ is composite.
== Historical Note ==
== Sources ==
* 
* : $8191$
* : $2^{8191} - 1$
* : $8191$
* : $2^{8191} - 1$
Category:Famous False Conjectures
Category:Mersenne Numbers
Category:Mersenne Primes
Mersenne Number whose Index is Mersenne Prime/Historical Note

== Historical Note on Mersenne Number whose Index is Mersenne Prime ==
According to , in his of $1997$, this conjecture was made by .
The composite nature of $2^{8191} - 1$ was reported on by in $1954$, in an article in the :
:The corresponding Mersenne number was actually tested in $1953$ by on the Illiac, at the University of Illinois, one hundred hours of machine time being required.
== Sources ==
* 
* : $2^{8191} - 1$
* : $2^{8191} - 1$
Category:Historical Notes
Mersenne Number whose Index is Mersenne Prime/Mistake

== Source Work ==
:
:The Dictionary
::$8191$
== Mistake ==
:It had been conjectured that although most Mersenne numbers appear to be composite, a Mersenne number whose index was prime would itself be prime.
== Correction ==
The correct wording of the end of this sentence is:
:... a Mersenne number whose index was a Mersenne prime would itself be prime.
This has been corrected in of $1997$.
== Sources ==
* : $8191$
Category:Mistakes/Curious and Interesting Numbers
Mersenne Prime/Current Status

== Currently known Mersenne Primes ==
The list of all known Mersenne primes is as follows:
Note that the index numbers of Mersenne primes after no. $48$ are uncertain, as there may still be undiscovered Mersenne primes between the $48$th and $51$st.
Not all numbers in that range have been explored yet.
== Also see ==
== Sources ==
* : $28$
* : Chapter $\text {B}.2$: More about Numbers: Irrationals, Perfect Numbers and Mersenne Primes
* : $28$
* 
* 
* 
* 
* : Chapter $7$: Patterns in Numbers: Euclid
* 
Category:Mersenne Primes
Mertens' Convergence Theorem

== Theorem ==
Let $\sequence {a_n}_{n \mathop \ge 0}$ and $\sequence {b_n}_{n \mathop \ge 0}$ be real or complex sequences.
Let:
:$\ds \sum_{n \mathop = 0}^\infty a_n$ converge to $A$
:$\ds \sum_{n \mathop = 0}^\infty b_n$ converge to $B$.
Let either $\ds \sum_{n \mathop = 0}^\infty a_n$ or $\ds \sum_{n \mathop = 0}^\infty b_n$ converge absolutely.
Then:
:$\ds \paren {\sum_{n \mathop = 0}^\infty a_n} \paren {\sum_{n \mathop = 0}^\infty b_n} = \sum_n \paren {\sum_{j \mathop + k \mathop = n} a_j b_k}$
If both $\ds \sum_{n \mathop = 0}^\infty a_n$ and $\ds \sum_{n \mathop = 0}^\infty b_n$ converge absolutely, then so also does the Cauchy product.
== Proof ==
== Sources ==
* 
Category:Convergence
Mertens' Convergence Theorem/Mistake

== Source Work ==
:
:Merten's theorem
== Mistake ==
:Merten's theorem
== Correction ==
This is a grammatical error.
This should be expressed as Mertens' theorem.
Also note that there are a number of other theorems also referred to as Mertens' theorem, relating to the distribution of prime numbers.
== Sources ==
* 
Category:Mistakes/Dictionary of Mathematics
Mertens' Third Theorem

== Theorem ==
:$\ds \lim_{x \mathop \to \infty} \ln x \prod_{\substack {p \mathop \le x \\ \text {$p$ prime} } } \paren {1 - \dfrac 1 p} = e^{-\gamma}$
where $\gamma$ denotes the Euler-Mascheroni constant.
== Proof ==
== Sources ==
* : $0,56145 94835 66885 \ldots$
Category:Euler-Mascheroni Constant
Metacategory Induces Morphisms-Only Metacategory

== Theorem ==
Let $\mathbf C$ be a metacategory.
Then the collection of morphisms $\mathbf C_1$ of $\mathbf C$ is a morphisms-only metacategory.
== Proof ==
In order to check that $\mathbf C_1$ is a morphisms-only metacategory, we need to interpret the symbols $\operatorname{dom}$, $\operatorname{cod}$ and $R_\circ$.
Set $\operatorname{dom} x := \operatorname{id}_{\operatorname{dom} x}$, and $\operatorname{cod} x := \operatorname{id}_{\operatorname{cod} x}$ with the symbols on the right as defined in $\mathbf C$.
Further, put $\map {R_\circ} {x, y, z} \iff x \circ y = z$.
Now we can proceed to verify the axioms for a morphisms-only metacategory.
For $(MOCT0)$, that is (omitting universal quantifiers):
:$\paren {\map {R_\circ} {x, y, z} \land \map {R_\circ} {x, y, z'} } \implies z = z'$
we conclude that the antecedent holds iff $z = x \circ y = z'$, hence $(MOCT0)$ holds.
The content of $(MOCT1)$ is:
:$\operatorname{dom} x = \operatorname{cod} y \iff \exists z: \map {R_\circ} {x, y, z}$
This says that a composite morphism $x \circ y$ is defined $x$ and $y$ are composable morphisms.
Axiom $(MOCT2)$ is the statement:
:$\map {R_\circ} {x, y, z} \implies \paren {\operatorname{dom} z = \operatorname{dom} y \land \operatorname{cod} z = \operatorname{cod} x}$
which prescribes the domain and codomain of $x \circ y$.
It follows directly from axiom $(C1)$ for metacategories that this is satisfied by $\mathbf C_1$.
Axiom $(MOCT3)$ is:
:$\map {R_\circ} {x, y, a} \land \map {R_\circ} {y, z, b} \implies \paren {\exists w: \map {R_\circ} {x, b, w} \land \map{R_\circ} {a, z, w} }$
which expresses that $\circ$ is an associative operation, exactly as axiom $(C3)$ for metacategories specifies.
Finally, axiom $(MOCT4)$ is the statement:
:$\map {R_\circ} {x, \operatorname{dom} x, x} \land \map {R_\circ} {\operatorname{cod} x, x, x}$
which will hold :
:$x \circ \operatorname{id}_{\operatorname{dom} x} = x$
and:
:$\operatorname{id}_{\operatorname{cod} x} \circ x = x$
This is the content of axiom $(C2)$ for metacategories.
Having verified the axioms, we conclude $\mathbf C_1$ to be a morphisms-only metacategory.
== Also see ==
* Morphisms-Only Metacategory Induces Metacategory
* Morphisms-Only Metacategory
* Characterization of Metacategory via Equations
Category:Category Theory
Metacompact Countably Compact Space is Compact

== Theorem ==
Let $T = \struct {S, \tau}$ be a countably compact space which is also metacompact.
Then $T$ is compact.
== Proof ==
From the definition of countably compact space, every countable open cover of $S$ has a finite subcover.
Let $T = \struct {S, \tau}$ be a countably compact space which is also metacompact.
Let $\UU_\alpha$ be an open cover of $S$.
Then let $\VV_\beta$ be the open refinement which is point finite, guaranteed by its metacompactness.
Let $x \in S$.
We have that $x$ is an element of only a finite number of the elements of $\VV_\beta$, as $\VV_\beta$ is point finite.
Consider all the subcovers of $\VV_\beta$, and order them by subset.
Consider any chain $\CC$ of such subcovers.
 $x$ is not in the intersection of $\CC$.
Then it would fail to be covered by one of the elements of $\CC$.
This would be a contradiction of the fact that every element of $\CC$ is a cover of $\VV_\beta$. 
So the intersection of a chain of such subcovers is itself a subcover.
Hence it must be that $\VV$ has a minimal subcover, $\VV_\gamma$, say.
Now each element $V_\gamma$ of $\VV_\gamma$ must itself contain a unique element $x_\gamma \in V_\gamma$ which belongs to no element of $\VV_\gamma$.
This is because $\VV_\gamma$ is minimal.
If $\VV_\gamma$ were infinite, the set of all $\set {x_\gamma}$ would be an infinite set without an $\omega$-accumulation point.
This can not be the case.
Thus $\VV_\gamma$ is a finite cover.
So we have constructed a finite subcover for $\UU_\alpha$, demonstrating that $T$ is compact.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $3$: Compactness: Paracompactness
Category:Compact Spaces
Category:Countably Compact Spaces
Category:Metacompact Spaces
Metacompact Space is Countably Metacompact

== Theorem ==
Let $T = \struct {S, \tau}$ be a metacompact space.
Then $T$ is a countably metacompact space.
== Proof ==
From the definition, $T$ is metacompact space iff every open cover of $S$ has an open refinement which is point finite.
This also applies to all countable open covers.
So every countable open cover of $S$ has an open refinement which is point finite.
This is precisely the definition for a countably metacompact space.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $3$: Compactness: Paracompactness
Category:Metacompact Spaces
Category:Countably Metacompact Spaces
Method of Infinite Descent

== Theorem ==
Let $\map P {n_\alpha}$ be a propositional function depending on $n_\alpha \in \N$.
Let $\map P {n_\alpha} \implies \map P {n_\beta}$ such that $0 < n_\beta < n_\alpha$.
Then we may deduce that $\map P n$ is false for all $n \in \N$.
That is, suppose that by assuming the truth of $\map P {n_\alpha}$ for any natural number $n_\alpha$, we may deduce that there always exists some number $n_\beta$ strictly less than $n_\alpha$ for which $\map P {n_\beta}$ is also true, then $\map P {n_\alpha}$ cannot be true after all.
This technique is known as the method of infinite descent.
The process of deducing the truth of $\map P {n_\beta}$ from $\map P {n_\alpha}$ such that $0 < n_\beta < n_\alpha$ is known as the descent step.
== Proof 1 ==
== Proof 2 ==
== Historical Note ==
== Sources ==
* : Chapter $\text{IV}$: The Prince of Amateurs
Category:Method of Infinite Descent
Category:Proof Techniques
Method of Infinite Descent/Historical Note

== Historical Note on Method of Infinite Descent ==
The Method of Infinite Descent was devised by .
He used it to develop his proof of Fermat's Two Squares Theorem, as he describes in a letter to :
:For a long time I was unable to apply my method to affirmative proposition, because the twist and the trick for getting at them is much more troublesome than that which I use for negative propositions. Thus, when I had to prove that every prime number which exceeds a multiple of $4$ by $1$ is composed of two squares, I found myself in a fine torment. But at last a meditation many times repeated gave me the light I lacked, and now affirmative propositions submit to my method, with the aid of certain new principles which necessarily must be adjoined to it. The course of my reasoning in affirmative propositions is such: if an arbitrarily chosen prime of the form $4 n + 1$ is not a sum of two squares, [I prove that] there will be another of the same nature, less than the one chosen, and [therefore] next a third still less, and so on. Making an infinite descent in this way we finally arrive at the number $5$, the least of all the numbers of this kind [$4 n + 1$]. [By the proof mentioned and the previous argument from it], it follows that $5$ is not a sum of two squares. But it is. Therefore we must infer by a reductio ad absurdum that all numbers of the form $4 n + 1$ are sums of two squares.
== Sources ==
* : Chapter $\text{IV}$: The Prince of Amateurs
Category:Historical Notes
Method of Infinite Descent/Proof 1

== Theorem ==
== Proof ==
Let $G$ be the set of all $n$ such that $\map P n$ hold.
That is:
:$G = \set {n \in \N : \map P n}$
 suppose that $G$ is non-empty.
By the Well-Ordering Principle $G$ contains a smallest element $n_\alpha$.
But by the descent step, there exists $n_\beta \in \N$ that strictly precedes $n_\alpha$ for which $\map P {n_\beta}$.
But then $n_\beta \in G$, which contradicts $n_\alpha$ being the smallest element.
Therefore $G$ is empty.
That is $\map P n$ is false for every $n$.
== Historical Note ==
Category:Method of Infinite Descent
Method of Infinite Descent/Proof 2

== Theorem ==
== Proof ==
Suppose that $\map P {n_\alpha}$ holds.
Then from the descent step, $\exists n_\beta \in \N_{n_\alpha}: \map P {n_\beta}$.
The descent step then tell us we can deduce a smaller positive solution, $n_\gamma$, such that $\map P {n_\gamma}$ is true and $n_\gamma \in \N_{n_\beta}$.
And again, the descent step tells us we can deduce a still smaller positive solution, $n_\delta$, such that $\map P {n_\delta}$ is true and $n_\delta \in \N_{n_\gamma}$.
Now, consider the unending sequence: $n_\alpha > n_\beta > n_\gamma > n_\delta > \cdots > 0$.
The set $S = \set {n_\alpha, n_\beta, n_\gamma, n_\delta, \ldots}$ is not bounded below, as for any $\forall x \in S: \exists y \in S: y < x$.
By the Well-Ordering Principle, any non-empty subset of $\N$ must have a least element.
As $S$ is not bounded below, it has no least element, so must be empty.
=== Axiom of Choice ===
Care is required to avoid accidentally using the Axiom of Countable Choice for Finite Sets, which is independent of Zermelo-Fraenkel Set Theory.
Before beginning the descent, construct:
:$\set {x : \map P x \land x < m}$
for each $m \leq n_\alpha$.
If it would be empty, define it to be any nonempty set instead.
Applying the Principle of Finite Choice, choose an element from each one.
Now, perform the descent, using the choice function to choose the next element at each step.
Because at each step, there are smaller solutions, it will never be the arbitrary set.
Therefore, the sequence is defined as required.
== Historical Note ==
Category:Method of Infinite Descent
Method of Least Squares (Approximation Theory)/Examples/Arbitrary Example 1

== Examples of Method of Least Squares ==
Let $B$ be a false balance.
$2$ items are weighed on $B$: first individually and then together.
The recorded weights are:
:$17 \, \mathrm g$ and $25 \, \mathrm g$ for the separate items
:$40 \, \mathrm g$ for the combined weight.
The least squares estimates of the true weights are the values of $\hat {w_1}$ and $\hat {w_2}$ that minimize:
:$L = \paren {w_1 - 25}^2 + \paren {w_1 - 17}^2 + \paren {w_1 + w_2 - 40}^2$
Differentiating $w_1$ and $w_2$ and equating the derivatives to zero, gives us:
== Proof ==
== Sources ==
* 
* 
Category:Examples of Method of Least Squares (Approximation Theory)
Method of Least Squares (Statistics)/Examples/Linear Response

== Examples of Use of Method of Least Squares in context of Statistics ==
Let the expected value of an observation be of the form:
:$\expect y = \alpha + \beta x$
Let a set of $n$ paired observations $\tuple {x_i, y_i}$ be given.
The least squares estimators of the unknowns $\alpha$ and $\beta$ are the values $a$ and $b$ which minimise:
:$\ds \sum_i \paren {y_i - a - b x_i}^2$
== Sources ==
* 
* 
Category:Examples of Method of Least Squares (Statistics)
Method of Truth Tables/Indirect Technique

== Proof Technique ==
The conventional technique of proving a tautology by means of a truth table can be unwieldy when there are many variables.
There is an indirect method of proving a tautology which is shorter.
It is recognised that the truth table for a tautology contains all $\T$ down the column containing the main connective.
Hence it follows that if there exist any instances of $\F$ in that column, the proposition under investigation is not a tautology.
Suppose we make the hypothesis that the proposition under investigation is not a tautology.
Then there exists an instance of $\F$ under the main connective.
We construct the heading of the truth table in the same way as for the conventional technique: one column for each variable and one for each connective.
Under the main connective we place $\F$.
Examining the conditions under which the consequence is $\F$ for that connective, we place truth values under the elements of the proposition which are in the scope of the main connective so as to allow $\F$.
Similarly we work backwards through each of the components of the proposition.
If it is impossible to complete the truth table row consistent with the behaviour of the connectives, then no $\F$ can appear under the main connective.
Therefore the proposition under investigation is a tautology.
If it is possible to complete the truth table row consistently, then the proposition under investigation is not a tautology.
=== Example ===
== Sources ==
* : $\S 3.4$: An Indirect Method of Truth-Table Decision
Category:Truth Tables
Method of Truth Tables/Indirect Technique/Example

== Example of Indirect Truth Table Technique ==
Consider the proposition:
:$\paren {\paren {p \implies q} \land \neg q} \implies \neg p$
First the truth table is set up, with $F$ under the main connective:
$\begin{array}{cccccc|c|cc}
((p & \implies & q) & \land & \neg & q) & \implies & \neg & p \\
\hline
 & & & & & & \F & & \\
\end{array}$
The first step is to assign truth values to the connectives in the scope of the main connective which are consistent with that $\F$.
In this case, the main connective is a conditional, so to make it false we need $\T$ under the antecedent and $\F$ under the consequent:
$\begin{array}{cccccc|c|cc}
((p & \implies & q) & \land & \neg & q) & \implies & \neg & p \\
\hline
 & & & \T & & & \F & \F & \\
 & & & 1 & & & & 1 & \\
\end{array}$
Note that the numbers in the second row are for illustrative purposes only, and are not actually required as part of the proof.
The second step does the same thing with the $\land$:
$\begin{array}{cccccc|c|cc}
((p & \implies & q) & \land & \neg & q) & \implies & \neg & p \\
\hline
 & \T & & \T & \T & & \F & \F & \\
 & 2 & & 1 & 2 & & & 1 & \\
\end{array}$
The third step does the same thing with the $\neg p$ and $\neg q$:
$\begin{array}{cccccc|c|cc}
((p & \implies & q) & \land & \neg & q) & \implies & \neg & p \\
\hline
 & \T & & \T & \T & \F & \F & \F & \T \\
 & 2 & & 1 & 2 & 3 & & 1 & 3 \\
\end{array}$
In the fourth step, we carry across the truth values of $p$ and $q$ which resulted after step $3$:
$\begin{array}{cccccc|c|cc}
((p & \implies & q) & \land & \neg & q) & \implies & \neg & p \\
\hline
\T & \T & \F & \T & \T & \F & \F & \F & \T \\
 4 & 2 & 4 & 1 & 2 & 3 & & 1 & 3 \\
\end{array}$
At this point it can be seen that the truth table contains an inconsistency:
$\begin{array}{ccc}
p & \implies & q \\
\hline
\T & \T & \F \\
\end{array}$
Thus the assumption that the main connective can be assigned the value $\F$ must have been false.
Hence the proposition:
:$\paren {\paren {p \implies q} \land \neg q} \implies \neg p$
is a tautology.
== Sources ==
* : $\S 3.4$: An Indirect Method of Truth-Table Decision
Category:Truth Tables
Method of Truth Tables/Proof of Interderivability

== Proof Technique ==
Suppose we have two propositional formulas $P$ and $Q$ and we wish to see whether $P \dashv \vdash Q$ or not.
:Example: Let $P$ be $p \uparrow q$ and let $Q$ be $\neg \paren {p \land q}$.
There are two things we can do.
=== Express two statements as a single WFF ===
We express $P \dashv \vdash Q$ as a single WFF $\paren {P \iff Q}$ and perform the method of truth tables on that.
Demonstrating this with the example given:
$\begin{array}{cc||cccccccc}
p & q & (p & \uparrow & q) & \iff & \neg & (p & \land & q) \\
\hline
\F & \F & \F & \T & \F & \T & \T & \F & \F & \F \\
\F & \T & \F & \T & \T & \T & \T & \F & \F & \T \\
\T & \F & \T & \T & \F & \T & \T & \T & \F & \F \\
\T & \T & \T & \F & \T & \T & \F & \T & \T & \T \\
\end{array}$
As can be seen, the column under the main connective $\iff$ of $\paren {P \iff Q}$ is all $\T$, so $\paren {\paren {p \uparrow q} \iff \neg \paren {p \land q} }$ is a tautology.
Hence from Equivalences are Interderivable, $\paren {\paren {p \uparrow q} \dashv \vdash \neg \paren {p \land q} }$ and the two formulas are interderivable.
=== Compare two WFFs in the same table ===
Alternatively, we can place the two WFFs side by side in the same truth table:
$\begin{array}{cc||ccc||cccc}
p & q & (p & \uparrow & q) & \neg & (p & \land & q) \\
\hline
\F & \F & \F & \T & \F & \T & \F & \F & \F \\
\F & \T & \F & \T & \T & \T & \F & \F & \T \\
\T & \F & \T & \T & \F & \T & \T & \F & \F \\
\T & \T & \T & \F & \T & \F & \T & \T & \T \\
\end{array}$
This time, we need to make sure that the truth values in the columns under the main connectives of both formulae are the same.
Note that this is exactly the same as putting an instance of $\iff$ between the two and making a WFF out of the pair of them.
Category:Proof Techniques
Category:Truth Tables