nyuuzyou/wikis huggingface dataset:
Conditional Expectation is Monotone

== Theorem ==
Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space.
Let $X$ and $Y$ be integrable random variables on $\struct {\Omega, \Sigma, \Pr}$ such that:
:$X \le Y$ almost everywhere. 
Let $\GG \subseteq \Sigma$ be a sub-$\sigma$-algebra of $\Sigma$.
Let $\expect {X \mid \GG}$ be a version of the conditional expectation of $X$ given $\GG$.
Let $\expect {Y \mid \GG}$ be a version of the conditional expectation of $Y$ given $\GG$.
Then we have:
:$\expect {X \mid \GG} \le \expect {Y \mid \GG}$ almost everywhere.
== Proof ==
We have:
:$Y - X \ge 0$ almost everywhere.
So, for each $A \in \GG$ we have:
:$\paren {Y - X} \cdot 1_A \ge 0$ almost everywhere.
So, from Expectation is Monotone:
:$\expect {\paren {Y - X} \cdot 1_A} \ge 0$
for each $A \in \GG$.
So, from Condition for Conditional Expectation to be Almost Surely Non-Negative, we have:
:$\expect {Y - X \mid \GG} \ge 0$ almost everywhere.
Then, from Conditional Expectation is Linear:
:$\expect {Y \mid \GG} - \expect {X \mid \GG} \ge 0$ almost everywhere.
So that:
:$\expect {X \mid \GG} \le \expect {Y \mid \GG}$ almost everywhere.
Category:Conditional Expectation
Conditional Expectation of Constant

== Theorem ==
Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space.
Let $\GG \subseteq \Sigma$ be a sub-$\sigma$-algebra.
Let $c \in \R$.
Define $X : \Omega \to \R$ by $\map X \omega = c$ for each $\omega \in \Omega$.
Let $\expect {X \mid \GG}$ be a version of the conditional expectation of $X$ given $\GG$.
Then:
:$\expect {X \mid \GG} = c$ almost everywhere.
== Proof ==
From Constant Function is Measurable, $X$ is a real-valued random variable.
The result then follows immediately from Conditional Expectation of Measurable Random Variable.
Category:Conditional Expectation
Conditional Expectation of Measurable Random Variable

== Theorem ==
Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space.
Let $\GG \subseteq \Sigma$ be a sub-$\sigma$-algebra.
Let $X$ be a integrable random variable that is $\GG$-measurable.
Let $\expect {X \mid \GG}$ be a version of the conditional expectation of $X$ given $\GG$.
Then:
:$\expect {X \mid \GG} = X$ almost everywhere.
== Proof ==
We show that $X$ is a version of $\expect {X \mid \GG}$.
Then since conditional expectation is essentially unique by Existence and Essential Uniqueness of Conditional Expectation Conditioned on Sigma-Algebra, we will obtain:
:$\expect {X \mid \GG} = X$ almost everywhere.
Note that $X$ is integrable and $\GG$-measurable , with: 
:$\ds \int_A X \rd \Pr = \int_A X \rd \Pr$
So $X$ satisfies the conditions of conditional expectation, and so:
:$\expect {X \mid \GG} = X$ almost everywhere.
== Sources ==
* : $9.7$: Properties of conditional expectation: a list
Category:Conditional Expectation
Conditional Expectation of Non-Negative Random Variable is Non-Negative

== Theorem ==
Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space.
Let $\GG \subseteq \Sigma$ be a sub-$\sigma$-algebra.
Let $X$ be an integrable random variable such that:
:$X \ge 0$ almost everywhere.
Let $\expect {X \mid \GG}$ be a version of the conditional expectation of $X$ given $\GG$.
Then:
:$\expect {X \mid \GG} \ge 0$ almost everywhere.
== Proof ==
From Conditional Expectation is Monotone, we have:
:$\expect {X \mid \GG} \ge \expect {0 \mid \GG}$ almost everywhere.
From Conditional Expectation of Constant, we have:
:$\expect {0 \mid \GG} = 0$ almost everywhere.
So:
:$\expect {X \mid \GG} \ge 0$ almost everywhere.
== Sources ==
* : $9.7$: Properties of conditional expectation: a list
Category:Conditional Expectation
Conditional Expectation of Sum of Squared Increments of Square-Integrable Martingale

== Theorem ==
Let $\struct {\Omega, \Sigma, \sequence {\FF_t}_{t \ge 0}, \Pr}$ be a continuous-time filtered probability space.
Let $\sequence {X_t}_{t \ge 0}$ be a continuous-time martingale such that $\size {X_t}^2$ is integrable for each $t \in \hointr 0 \infty$.
Let $s, t \in \hointr 0 \infty$ be such that $0 \le s < t$. 
Let:
:$s = t_0 < t_1 < \ldots < t_n = t$
be a finite subdivision of $\closedint s t$. 
Then: 
:$\ds \expect {\sum_{i \mathop = 1}^n \paren {X_{t_i} - X_{t_{i - 1} } }^2 \mid \FF_s} = \expect {X_t^2 - X_s^2 \mid \FF_s} = \expect {\paren {X_t - X_s}^2 \mid \FF_s}$ almost surely.
=== Corollary ===
== Proof ==
We have, for $i < j$: 
Since $\sequence {X_t}_{t \ge 0}$ is a $\sequence {\FF_t}_{t \ge 0}$-martingale, we have: 
:$X_{t_i}$ is $\FF_{t_i}$-measurable
and so: 
:$2 \expect {X_{t_j} X_{t_i} \mid \FF_{t_i} } = 2 X_{t_i} \expect {X_{t_j} \mid \FF_{t_i} }$
from Rule for Extracting Random Variable from Conditional Expectation of Product.
Using the martingale property, we have: 
:$\expect {X_{t_j} \mid \FF_{t_i} } = X_{t_i}$ 
From Conditional Expectation of Measurable Random Variable, we have: 
:$\expect {X_{t_i}^2 \mid \FF_{t_i} } = X_{t_i}^2$
Putting this together we have: 
:$\expect {\paren {X_{t_j} - X_{t_i} }^2 \mid \FF_{t_i} } = \expect {X_{t_j}^2 \mid \FF_{t_i} } - 2 X_{t_i}^2 + X_{t_i}^2 = \expect {X_{t_j}^2 \mid \FF_{t_i} } - X_{t_i}^2$ 
Since $\expect {X_{t_i}^2 \mid \FF_{t_i} } = X_{t_i}^2$, we have: 
:$\expect {\paren {X_{t_j} - X_{t_i} }^2 \mid \FF_{t_i} } = \expect {X_{t_j}^2 - X_{t_i}^2 \mid \FF_{t_i} }$ 
from Conditional Expectation is Linear.
Setting $i = 0$, $j = n$, we have: 
:$\expect {X_t^2 - X_s^2 \mid \FF_s} = \expect {\paren {X_t - X_s}^2 \mid \FF_s}$
We can also compute:
completing the proof.
== Sources ==
* : Proposition $3.14$
Category:Conditional Expectation
Category:Martingales
Category:Conditional Expectation of Sum of Squared Increments of Square-Integrable Martingale
Conditional Expectation of Sum of Squared Increments of Square-Integrable Martingale/Corollary

== Theorem ==
Let $\struct {\Omega, \Sigma, \sequence {\FF_t}_{t \ge 0}, \Pr}$ be a continuous-time filtered probability space.
Let $\sequence {X_t}_{t \ge 0}$ be a continuous-time martingale such that $\size {X_t}^2$ is integrable for each $t \in \hointr 0 \infty$.
Let $s, t \in \hointr 0 \infty$ be such that $0 \le s < t$. 
Let:
:$s = t_0 < t_1 < \ldots < t_n = t$
be a finite subdivision of $\closedint s t$. 
Then: 
:$\ds \expect {\sum_{i \mathop = 1}^n \paren {X_{t_i} - X_{t_{i - 1} } }^2} = \expect {X_t^2 - X_s^2} = \expect {\paren {X_t - X_s}^2}$
== Proof ==
From Conditional Expectation of Sum of Squared Increments of Square-Integrable Martingale:
:$\ds \expect {\sum_{i \mathop = 1}^n \paren {X_{t_i} - X_{t_{i - 1} } }^2 \mid \FF_s} = \expect {X_t^2 - X_s^2 \mid \FF_s} = \expect {\paren {X_t - X_s}^2 \mid \FF_s}$ almost surely.
From Expectation of Conditional Expectation, we then obtain: 
:$\ds \expect {\sum_{i \mathop = 1}^n \paren {X_{t_i} - X_{t_{i - 1} } }^2} = \expect {X_t^2 - X_s^2} = \expect {\paren {X_t - X_s}^2}$
== Sources ==
* : Proposition $3.14$
Category:Conditional Expectation of Sum of Squared Increments of Square-Integrable Martingale
Conditional Expectations of Integrable Random Variable with respect to Filtration forms Martingale/Continuous Time

== Theorem ==
Let $\struct {\Omega, \Sigma, \sequence {\FF_t}_{t \ge 0}, \Pr}$ be a continuous-time filtered probability space.
Let $Z$ be an integrable random variable.
For each $t \in \hointr 0 \infty$, let $\expect {Z \mid \FF_t}$ be a version of the conditional expectation of $Z$ given $\FF_t$. 
For each $t \in \hointr 0 \infty$ set:
:$X_t = \expect {Z \mid \FF_t}$
Then $\sequence {X_t}_{t \ge 0}$ is a $\sequence {\FF_t}_{t \ge 0}$-martingale. 
== Proof ==
From the definition of the conditional expectation of $Z$ given $\FF_t$, we have that: 
:$X_t$ is $\FF_t$-measurable for each $t \in \hointr 0 \infty$. 
So $\sequence {X_t}_{t \ge 0}$ is $\sequence {\FF_t}_{t \ge 0}$-adapted.
Now let $s, t \in \hointr 0 \infty$ have $s \le t$. 
Then we have: 
So $\sequence {X_t}_{t \ge 0}$ is a $\sequence {\FF_t}_{t \ge 0}$-martingale.
== Sources ==
* : $3.3$: Continuous Time Martingales and Supermartingales
Category:Conditional Expectations of Integrable Random Variable with respect to Filtration forms Martingale
Conditional Fatou's Lemma

== Theorem ==
Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space.
Let $X$ be an non-negative integrable random variable.
Let $\sequence {X_n}_{n \in \N}$ be an sequence of non-negative integrable random variables.
Let $\GG \subseteq \Sigma$ be a sub-$\sigma$-algebra.
For each $n \in \N$, let $\expect {X_n \mid \GG}$ be a version of the conditional expectation of $X_n$ conditioned on $\GG$.
Let $\ds \expect {\liminf_{n \mathop \to \infty} X_n \mid \GG}$ be a version of the conditional expectation of $X$ conditioned on $\GG$.
Then we have: 
:$\ds \expect {\liminf_{n \mathop \to \infty} X_n \mid \GG} \le \liminf_{n \mathop \to \infty} \expect {X_n \mid \GG}$ almost everywhere.
== Proof ==
From Conditional Expectation of Measurable Random Variable, we have:
:$\expect {X_n \mid \GG} = \expect {\expect {X_n \mid \GG} \mid \GG}$
It therefore suffices to show, from Conditional Expectation is Linear:
:$\ds \expect {\liminf_{n \mathop \to \infty} \expect {X_n \mid \GG} - \liminf_{n \mathop \to \infty} X_n \mid \GG} \ge 0$ almost everywhere.
From Condition for Conditional Expectation to be Almost Surely Non-Negative, we can show that:
:$\ds \expect {\paren {\liminf_{n \mathop \to \infty} \expect {X_n \mid \GG} - \liminf_{n \mathop \to \infty} X_n} \cdot 1_A} \ge 0$
for $A \in \GG$.
We indeed have, for each $A \in \GG$: 
== Sources ==
* : $9.7$: Properties of conditional expectation: a list
Category:Conditional Expectation
Conditional Jensen's Inequality

== Theorem ==
Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space.
Let $\GG$ be a sub-$\sigma$-algebra of $\Sigma$. 
Let $X$ be an integrable random variable.
Let $\expect {X \mid \GG}$ be a version of the conditional expectation of $X$ given $\GG$.
Let $f : \R \to \R$ be a convex function such that $\map f X$ is integrable.
Let $\expect {\map f X \mid \GG}$ be a version of the conditional expectation of $\map f X$ given $\GG$.
Then: 
:$\map f {\expect {X \mid \GG} } \le \expect {\map f X \mid \GG}$ 
== Proof ==
Note that from Convex Real Function is Measurable and Composition of Measurable Mappings is Measurable, $\map f X$ is $\Sigma$-measurable, so the hypotheses make sense. 
By Convex Real Function is Pointwise Supremum of Affine Functions: Corollary, there exists a countable set $\SS \subseteq \R^2$ such that: 
:$\ds \map f x = \sup_{\tuple {a, b} \mathop \in \SS} \paren {a x + b}$
so: 
:$\ds \map f X = \sup_{\tuple {a, b} \mathop \in \SS} \paren {a X + b}$
So for each $\tuple {a, b} \in \SS$, we have: 
:$a X + b \le \map f X$ 
From Conditional Expectation is Monotone and Conditional Expectation is Linear, we have: 
:$a \expect {X \mid \GG} + b \le \expect {\map f X \mid \GG}$ almost surely
for each $\tuple {a, b} \in \SS$. 
For each $\tuple {a, b} \in \SS$, let $A_{\tuple {a, b} }$ be the set of sample points $\omega \in \Omega$ such that this inequality fails.
Then for:
:$\ds \omega \in \Omega \setminus \bigcup_{\tuple {a, b} \mathop \in \SS} A_{\tuple {a, b} }$
we have: 
:$a \map {\expect {X \mid \GG} } \omega + b \le \map {\expect {\map f X \mid \GG} } \omega$ for all $\tuple {a, b} \in \SS$
From Null Sets Closed under Countable Union, we have: 
:$\ds \map \Pr {\bigcup_{\tuple {a, b} \mathop \in \SS} A_{\tuple {a, b} } } = 0$
So we have: 
:$a \expect {X \mid \GG} + b \le \expect {\map f X \mid \GG}$ for all $\tuple {a, b} \in \SS$ almost surely.
We therefore have: 
:$\ds \sup_{\tuple {a, b} \mathop \in \SS} \paren {a \expect {X \mid \GG} + b} \le \expect {\map f X \mid \GG}$ almost surely.
that is: 
:$\map f {\expect {X \mid \GG} } \le \expect {\map f X \mid \GG}$ almost surely.
== Sources ==
* : $9.7$: Properties of conditional expectation: a list
Category:Conditional Expectation
Category:Convex Real Functions
Conditional Monotone Convergence Theorem

== Theorem ==
Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space.
Let $X$ be an non-negative integrable random variable.
Let $\sequence {X_n}_{n \in \N}$ be an sequence of non-negative integrable random variables converging almost surely to $X$, such that: 
:$X_n \le X_{n + 1}$ almost everywhere for each $n \in \N$.
Let $\GG \subseteq \Sigma$ be a sub-$\sigma$-algebra. 
For each $n \in \N$, let $\expect {X_n \mid \GG}$ be a version of the conditional expectation of $X_n$ conditioned on $\GG$.
Let $\expect {X \mid \GG}$ be a version of the conditional expectation of $X$ conditioned on $\GG$.
Then:
:$\ds \lim_{n \mathop \to \infty} \expect {X_n \mid \GG} = \expect {X \mid \GG}$ almost everywhere.
== Proof ==
Note that for almost all $\omega \in \Omega$, $\sequence {\map {X_n} \omega}_{n \mathop \in \N}$ is an increasing real sequence with $\map {X_n} \omega \to \map X \omega$. 
From Monotone Convergence Theorem (Real Analysis), we then have that: 
:$\map {X_n} \omega \le \map X \omega$
for almost all $\omega \in \Omega$.
From Conditional Expectation is Monotone, we then have: 
:$\expect {X_n \mid \GG} \le \expect {X \mid \GG}$
Also: 
:$X_n \le X_{n + 1}$ almost everywhere
by hypothesis, so again applying Conditional Expectation is Monotone we have: 
:$\expect {X_n \mid \GG} \le \expect {X_{n + 1} \mid \GG}$
So for each $\omega \in \Omega$, $\sequence {\map {\paren {\expect {X_n \mid \GG} } } \omega}_{n \mathop \in \N}$ is an increasing real sequence bounded above by $\map {\paren {\expect {X \mid \GG} } } \omega$.
So by Monotone Convergence Theorem (Real Analysis):
:$\ds \lim_{n \mathop \to \infty} \map {\paren {\expect {X_n \mid \GG} } } \omega$ exists for all $\omega \in \Omega$.
So define: 
:$\ds Y = \lim_{n \mathop \to \infty} \expect {X_n \mid \GG}$
Then from Pointwise Limit of Measurable Functions is Measurable, $Y$ is a real-valued random variable. 
We want to show that $Y$ is a version of the conditional expectation of $X$ conditioned on $\GG$.
We just need to check that: 
:$\ds \int_A Y \rd \Pr = \int_A X \rd \Pr$
for each $A \in \GG$.
We have: 
and hence $Y$ is a version of the conditional expectation of $X$ conditioned on $\GG$ and hence we have: 
:$\ds Y = \lim_{n \mathop \to \infty} \expect {X_n \mid \GG} = \expect {X \mid \GG}$ almost everywhere.
== Sources ==
* : $9.7$: Properties of conditional expectation: a list
Category:Conditional Expectation
Conditional Probability Defines Probability Space

== Theorem ==
Let $\struct {\Omega, \Sigma, \Pr}$ be a measure space.
Let $B \in \Sigma$ such that $\map \Pr B > 0$.
Let $Q: \Sigma \to \R$ be the real-valued function defined as:
:$\map Q A = \condprob A B$
where:
:$\condprob A B = \dfrac {\map \Pr {A \cap B} }{\map \Pr B}$
is the conditional probability of $A$ given $B$.
Then $\struct {\Omega, \Sigma, Q}$ is a probability space.
== Proof ==
It is to be shown that $Q$ is a probability measure on $\struct {\Omega, \Sigma}$.
As $\Pr$ is a measure, we have that:
:$\forall A \in \Omega: \map Q A \ge 0$
Also, we have that:
Now, suppose that $A_1, A_2, \ldots$ are disjoint events in $\Sigma$.
Then:
== Sources ==
* : $\S 1.6$: Theorem $1 \text {A}$
Category:Conditional Probabilities
Category:Probability Theory
Conditional Reverse Fatou's Lemma

== Theorem ==
Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space.
Let $X$ be an integrable random variable.
Let $\sequence {X_n}_{n \mathop \in \N}$ be an sequence of non-negative integrable random variables such that:
:there exists an integrable random variable $Y$ such that:
::$\size {X_n} \le Y$ almost surely. 
Let $\GG \subseteq \Sigma$ be a sub-$\sigma$-algebra.
For each $n \in \N$, let $\expect {X_n \mid \GG}$ be a version of the conditional expectation of $X_n$ conditioned on $\GG$.
Let $\expect {X_n \mid \GG}$ be a version of the conditional expectation of $X_n$ conditioned on $\GG$.
Let $\ds \expect {\limsup_{n \mathop \to \infty} X_n \mid \GG}$ be a version of the conditional expectation of $\ds \limsup_{n \mathop \to \infty} X_n$ conditioned on $\GG$.
Then: 
:$\ds \limsup_{n \mathop \to \infty} \expect {X_n \mid \GG} \le \expect {\limsup_{n \mathop \to \infty} X_n \mid \GG}$ almost surely.
== Proof ==
We should first verify that a version of the conditional expectation of $\ds \limsup_{n \mathop \to \infty} X_n$ conditioned on $\GG$ exists.
We have:
:$-Y \le X_n \le Y$
and so:
:$\ds -Y \le \limsup_{n \mathop \to \infty} X_n \le Y$ almost surely
so that: 
:$\ds -\infty < -\expect Y \le \expect {\limsup_{n \mathop \to \infty} X_n} \le \expect Y < \infty$
from Expectation is Monotone.
So:
:$\ds \limsup_{n \mathop \to \infty} X_n$ is integrable
and we have that conditional expectation of $\ds \limsup_{n \mathop \to \infty} X_n$ conditioned on $\GG$ exists.
Now, we have that:
:$Y - X_n \ge 0$
So, we can apply Conditional Fatou's Lemma to obtain:
:$\ds \expect {\liminf_{n \mathop \to \infty} \paren {Y - X_n} \mid \GG} \le \liminf_{n \mathop \to \infty} \expect {Y - X_n \mid \GG}$ almost surely. 
We have, by Conditional Expectation is Linear:
:$\ds \expect {Y - X_n \mid \GG} = \expect {Y \mid \GG} - \expect {X_n \mid \GG}$
So we have:
Similarly:
So we have:
:$\ds \expect {Y \mid \GG} - \expect {\limsup_{n \mathop \to \infty} X_n \mid \GG} \le \expect {Y \mid \GG} - \limsup_{n \mathop \to \infty} \expect {X_n \mid \GG}$ almost surely.
So:
:$\ds \limsup_{n \mathop \to \infty} \expect {X_n \mid \GG} \le \expect {\limsup_{n \mathop \to \infty} X_n \mid \GG}$ almost surely.
Category:Conditional Expectation
Conditional and Converse are not Equivalent

== Theorem ==
A conditional statement:
:$p \implies q$
is not logically equivalent to its converse:
:$q \implies p$
== Proof ==
We apply the Method of Truth Tables to the proposition:
:$\paren {p \implies q} \iff \paren {q \implies p}$
$\begin{array}{|ccc|c|ccc|} \hline
p & \implies & q) & \iff & (q & \implies & p) \\
\hline
\F & \T & \F & \T & \F & \T & \F \\
\F & \T & \T & \F & \T & \F & \F \\
\T & \F & \F & \F & \F & \T & \T \\
\T & \T & \T & \T & \T & \T & \T \\
\hline
\end{array}$
As can be seen by inspection, the truth values under the main connectives do not match for all boolean interpretations.
== Also see ==
* Affirming the Consequent
== Sources ==
* : $\S \text{II}.14$: Application of laws of sentential calculus in inference
* : Appendix $\text{A}.5$: Theorems and Proofs
* : Chapter $1$: Sets and mappings: $\S 1.1$: The need for logic: Exercise $(3)$
Category:Conditional
Conditional and Inverse are not Equivalent

== Theorem ==
A conditional statement:
:$p \implies q$
is not logically equivalent to its inverse:
:$\lnot p \implies \lnot q$
== Proof ==
We apply the Method of Truth Tables to the proposition:
:$\paren {p \implies q} \iff \paren {\lnot p \implies \lnot q}$
$\begin{array}{|ccc|c|ccc|} \hline
p & \implies & q) & \iff & (\lnot & p & \implies & \lnot & q) \\
\hline
\F & \T & \F & \T & \T & \F & \T & \T & \F \\
\F & \T & \T & \F & \T & \F & \F & \F & \T \\
\T & \F & \F & \F & \F & \T & \T & \T & \F \\
\T & \T & \T & \T & \F & \T & \T & \F & \T \\
\hline
\end{array}$
As can be seen by inspection, the truth values under the main connectives do not match for all boolean interpretations.
== Also see ==
* Denying the Antecedent
== Sources ==
* : $\S \text{II}.14$: Application of laws of sentential calculus in inference
Category:Conditional
Conditional iff Biconditional of Antecedent with Conjunction

== Theorem ==
:$p \implies q \dashv \vdash p \iff \paren {p \land q}$
== Proof ==
We apply the Method of Truth Tables.
As can be seen by inspection, the appropriate truth values match for all boolean interpretations.
$\begin{array}{|ccc||ccccc|} \hline
p & \implies & q & p & \iff & (p & \land & q) \\
\hline
\F & \T & \F & \F & \T & \F & \F & \F \\
\F & \T & \T & \F & \T & \F & \F & \T \\
\T & \F & \F & \T & \F & \T & \F & \F \\
\T & \T & \T & \T & \T & \T & \T & \T \\
\hline
\end{array}$
== Sources ==
* : $\S 2.10$: Exercise $2.6$
Category:Conditional
Category:Biconditional
Category:Conjunction
Conditional iff Biconditional of Consequent with Disjunction

== Theorem ==
:$p \implies q \dashv \vdash q \iff \paren {p \lor q}$
== Proof ==
We apply the Method of Truth Tables.
As can be seen by inspection, the appropriate truth values match for all boolean interpretations.
$\begin{array}{|ccc||ccccc|} \hline
p & \implies & q & q & \iff & (p & \lor & q) \\
\hline
\F & \T & \F & \F & \T & \F & \F & \F \\
\F & \T & \T & \T & \T & \F & \T & \T \\
\T & \F & \F & \F & \F & \T & \T & \F \\
\T & \T & \T & \T & \T & \T & \T & \T \\
\hline
\end{array}$
== Sources ==
* : $\S 2.10$: Exercise $2.6$
Category:Conditional
Category:Biconditional
Category:Disjunction
Conditional in terms of NAND

== Theorem ==
:$p \implies q \dashv \vdash p \uparrow \paren {q \uparrow q}$
where $\implies$ denotes the conditional and $\uparrow$ denotes the logical NAND.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Chapter $1$: Sets and mappings: $\S 1.1$: The need for logic: Exercise $(5)$
Category:Conditional in terms of NAND
Category:Logical NAND
Category:Conditional
Conditional is Equivalent to Negation of Conjunction with Negative

== Theorem ==
==== Formulation 1 ====
==== Formulation 2 ====
== Sources ==
* : $2$ Arguments Containing Compound Statements: $2.2$: Conditional Statements
* : Chapter $1$: Elementary, my dear Watson: $\S 1.1$: You have a logical mind if...: Example $1.1.2$
Category:Conditional is Equivalent to Negation of Conjunction with Negative
Category:Conjunction
Category:Conditional
Category:Logical Negation
Conditional is Equivalent to Negation of Conjunction with Negative/Formulation 1

== Theorem ==
:$p \implies q \dashv \vdash \neg \paren {p \land \neg q}$
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: Theorems $35$
* : $2$ Arguments Containing Compound Statements: $2.4$: Statement Forms
* : $\S 2.3.3$
* : $\S 2.10$: Exercise $2.3$
Category:Conditional is Equivalent to Negation of Conjunction with Negative
Conditional is Equivalent to Negation of Conjunction with Negative/Formulation 1/Forward Implication

== Theorem ==
:$p \implies q \vdash \neg \paren {p \land \neg q}$
== Proofs ==
== Sources ==
* : Chapter $1$: The Propositional Calculus $1$: $5$ Further Proofs: Résumé of Rules: Theorem $35 \ \text{(a)}$
Category:Conditional is Equivalent to Negation of Conjunction with Negative
Conditional is Equivalent to Negation of Conjunction with Negative/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.
$\begin{array}{|ccc||ccccc|} \hline
p & \implies & q & \neg & (p & \land & \neg & q) \\
\hline
\F & \T & \F & \T & \F & \F & \T & \F \\
\F & \T & \T & \T & \F & \F & \F & \T \\
\T & \F & \F & \F & \T & \T & \T & \F \\
\T & \T & \T & \T & \T & \F & \F & \T \\
\hline
\end{array}$
== Sources ==
* : $\S 2.4$: Relations between Truth-Functions
* : Chapter $2$: The Propositional Calculus $2$: $3$ Truth-Tables
* : $2$ Arguments Containing Compound Statements: $2.2$: Conditional Statements
Category:Conditional is Equivalent to Negation of Conjunction with Negative
Category:Truth Table Proofs
Conditional is Equivalent to Negation of Conjunction with Negative/Formulation 1/Reverse Implication

== Theorem ==
:$\neg \paren {p \land \neg q} \vdash p \implies q$
== Proof ==
== Sources ==
* : Chapter $1$: The Propositional Calculus $1$: $5$ Further Proofs: Résumé of Rules: Theorems $35 \ \text{(b)}$
Category:Conditional is Equivalent to Negation of Conjunction with Negative
Conditional is Equivalent to Negation of Conjunction with Negative/Formulation 2

== Theorems ==
:$\vdash \paren {p \implies q} \iff \paren {\neg \paren {p \land \neg q} }$
This can be expressed as two separate theorems:
=== Forward Implication ===
=== Reverse Implication ===
== Proof 1 ==
== Proof by Truth Table ==
== Sources ==
* : $\S 3.6$: Reference Formulae: $RF \, 18$
Category:Conditional is Equivalent to Negation of Conjunction with Negative
Conditional is Equivalent to Negation of Conjunction with Negative/Formulation 2/Proof 1

== Theorem ==
== Proof ==
== Sources ==
* : $\text{II}$: 'AND', 'OR', 'IF AND ONLY IF': $\S 3$: Theorem $\text{T37}$
Category:Conditional is Equivalent to Negation of Conjunction with Negative