hcju/proofwikiqa huggingface dataset:
Consider Instance 1 of a constructed semantics, denoted $\mathscr C_1$. Note that $\neg p$ is not a tautology for $\mathscr C_1$. We will establish that every $\mathscr H_2$-theorem is a $\mathscr C_1$-tautology. That is, that $\mathscr H_2$ is sound for $\mathscr C_1$. Starting with the axioms: {{begin-axiom}} {{axiom|n = A1        |lc = Rule of Idempotence        |m = (p \lor p) \implies p        |rc = Proof of Tautology }} {{axiom|n = A2        |lc = Rule of Addition        |m = q \implies (p \lor q)        |rc = Proof of Tautology }} {{axiom|n = A3        |lc = Rule of Commutation        |m = (p \lor q) \implies (q \lor p)        |rc = Proof of Tautology }} {{axiom|n = A4        |lc = Factor Principle        |m = (q \implies r) \implies \left({ (p \lor q) \implies (p \lor r)}\right)        |rc = Proof of Tautology }} {{end-axiom}} Next it needs to be shown that the rules of inference of $\mathscr H_2$ preserve $\mathscr C_1$-tautologies. === Rule $RST \, 1$: Rule of Uniform Substitution === By definition, any WFF is assigned a value $1$ or $2$. Thus, in applying Rule $RST \, 1$, we are introducing $1$ or $2$ in the position of a propositional variable. But all possibilities of assignments of $1$s and $2$s to such propositional variables were shown not to affect the resulting value $2$ of the axioms. Hence Rule $RST \, 1$ preserves $\mathscr C_1$-tautologies. === Rule $RST \, 2$: Rule of Substitution by Definition === Because the definition of $\mathscr C_1$ was given in terms of Rule $RST \, 2$, it cannot affect any of its results. === Rule $RST \, 3$: Rule of Detachment === Suppose $\mathbf A$ and $\mathbf A \implies \mathbf B$ both take value $2$. Then using Rule $RST \, 2$, definition $(2)$, we get: :$\neg \mathbf A \lor \mathbf B$ taking value $2$ by assumption. But $\neg \mathbf A$ takes value $1$ by definition of $\neg$. So from the definition of $\lor$ it must be that $\mathbf B$ takes value $2$. Hence Rule $RST \, 3$ also produces only WFFs of value $2$. === Rule $RST \, 4$: Rule of Adjunction === Suppose $\mathbf A$ and $\mathbf B$ take value $2$. Then: {{begin-eqn}} {{eqn|l = \mathbf A \land \mathbf B      |r = 2 \land 2 }} {{eqn|r = \neg ( \neg 2 \lor \neg 2 )      |c = Rule $RST \, 2 \, (1)$ }} {{eqn|r = \neg ( 1 \lor 1 ) }} {{eqn|r = \neg 1 }} {{eqn|r = 2 }} {{end-eqn}} proving that Rule $RST \, 4$ also produces only $2$s from $2$s. Hence $\mathscr H_2$ is sound for $\mathscr C_1$. In particular: :$\not\vdash_{\mathscr H_2} \neg p$ Hence $\mathscr H_2$ is consistent. {{qed}}
Define: :$Y := \leftset {x \in X: x}$ is local minimum in $\rightset X$ By definition of $Y$ and definition of local minimum in set: :$\forall x \in Y: \exists y \in \R: y < x \land \openint y x \cap X = \O$ By the Axiom of Choice, define a mapping $f: Y \to \powerset \R$ as: :$\forall x \in Y: \exists y \in \R: \map f x = \openint y x \land y < x \land \map f x \cap X = \O$ We will prove that $f$ is an injection by definition: Let $x_1, x_2 \in Y$ such that :$\map f {x_1} = \map f {x_2}$ By definition of $f$: :$\exists y_1 \in \R: \map f {x_1} = \openint {y_1} {x_1} \land y_1 < x_1 \land \map f {x_1} \cap X = \O$ and: :$\exists y_2 \in \R: \map f {x_2} = \openint {y_2} {x_2} \land y_2 < x_2 \land \map f {x_2} \cap X = \O$ Then: :$\openint {y_1} {x_1} = \openint {y_2} {x_2}$ Thus $x_1 = x_2$. This ends the proof of injection. By Cardinality of Image of Injection: :$(1): \quad \card Y = \card {\map {f^\to} Y} = \card {\Img f}$ where :$\card Y$ denotes the cardinality of $Y$, :$\map {f^\to} Y$ denotes the image of $Y$ under $f$, :$\Img f$ denotes the image of $f$. We will prove that $\Img f$ is pairwise disjoint by definition. Let $A, B \in \Img f$ such that :$A \ne B$. Then by definition of image: :$\exists x_1 \in Y: \map f {x_1} = A$ and :$\exists x_2 \in Y: \map f {x_2} = B$. By difference of $A$ and $B$: :$x_1 \ne x_2$ By definition of $f$: :$\exists y_1 \in \R: \map f {x_1} = \openint {y_1} {x_1} \land y_1 < x_1 \land \map f {x_1} \cap X = \O$ and: :$\exists y_2 \in \R: \map f {x_2} = \openint {y_2} {x_2} \land y_2 < x_2 \land \map f {x_2} \cap X = \O$ Aiming at contradiction suppose that :$A \cap B \ne \O$. $x_1 < x_2$ or $x_1 > x_2$. In case when $x_1 < x_2$, $x_1 \in \map f {x_2}$ what contradicts with $\map f {x_2} \cap X = \O$. In case when $x_1 > x_2$, analogically. This ends the proof that $\Img f$ is pairwise disjoint. By Set of Pairwise Disjoint Intervals is Countable: :$\Img f$ is countable. Thus by $(1)$ and Set is Countable if Cardinality equals Cardinality of Countable Set the result: :$Y$ is countable. {{qed}}
By Between two Real Numbers exists Rational Number: :$\forall A \in X: \exists x, y \in \R, q \in \Q: x < y \land q \in \left({x \,.\,.\, y}\right) \subseteq A$ By the Axiom of Choice define a mapping $f: X \to \Q$: :$\forall A \in X: f \left({A}\right) \in A$ First it needs to be shown that $f$ is an injection by definition. Let $A, B \in X$ such that: :$f \left({A}\right) = f \left({B}\right)$ By definition of $f$: :$f \left({A}\right) \in A$ and $f \left({B}\right) \in B$ By definition of intersection: :$f \left({A}\right) \in A \cap B$ Then by definition of empty set: :$A \cap B \ne \varnothing$ Thus by definition of pairwise disjoint: :$A = B$ Hence $f$ is an injection. By Set is Subset of Itself, $X$ is a subset of $X$. Thus by Cardinality of Image of Injection: :$\left\vert{X}\right\vert = \left\vert{f^\to \left({X}\right)}\right\vert$ By definition of image: :$f^\to \left({X}\right) \subseteq \Q$ By Rational Numbers are Countably Infinite: :$\Q$ is countable. Hence by Subset of Countable Set is Countable: :$f^\to \left({X}\right)$ is countable. Thus by Set is Countable if Cardinality equals Cardinality of Countable Set the result: :$X$ is countable. {{qed}}
Assume that $X$ is countable. By definition of countable set there exists an injection:L :$f: X \to \N$ By definition of cardinality the sets $Y$ and $X$ are equivalent: :$Y \sim X$ Then by definition of set equivalence there exists a bijection: :$g: Y \to X$ By definition of bijection: :$g$ is an injection. Hence by Composite of Injections is Injection: :$f \circ g: Y \to \N$ is an injection. Thus by definition: : $Y$ is countable. {{qed}} Category:Countable Sets 5lec0t36mj905rmtdx7evd47l2rwp68
Proof by Counterexample: Consider the number: :$m = 693 = 3^2 \times 7 \times 11$ Thus: :$m = 9 \times 77 = 21 \times 33$ We have that: {{begin-eqn}} {{eqn | l = 9       | r = 4 \times 2 + 1       | rr= \in S }} {{eqn | l = 77       | r = 4 \times 19 + 1       | rr= \in S }} {{eqn | l = 21       | r = 4 \times 5 + 1       | rr= \in S }} {{eqn | l = 33       | r = 4 \times 8 + 1       | rr= \in S }} {{end-eqn}} The divisors of these numbers are as follows: {{begin-eqn}} {{eqn | l = 9       | r = 3^2       | c = where $3 \notin S$ }} {{eqn | l = 77       | r = 7 \times 11       | c = where $7 \notin S$ and $11 \notin S$ }} {{eqn | l = 21       | r = 3 \times 7       | c = where $3 \notin S$ and $7 \notin S$ }} {{eqn | l = 33       | r = 3 \times 11       | c = where $3 \notin S$ and $11 \notin S$ }} {{end-eqn}} Thus $693$ has two different complete factorizations into elements of $S$. Hence the result. {{qed}}
Let $\tuple {x, y, z}$ be non-primitive solution to the Pythagorean equation. Let: :$\exists k \in \Z: k \ge 2, k \divides x, k \divides y$ such that $x \perp y$. Then we can express $x$ and $y$ as $x = k x', y = k y'$. Thus: :$z^2 = k^2 x'^2 + k^2 y'^2 = k^2 z'^2$ for some $z' \in \Z$. Let: :$\exists k \in \Z: k \ge 2, k \divides x, k \divides z$ such that $x \perp z$ Then we can express $x$ and $z$ as $x = k x', z = k z'$. Thus: :$y^2 = k^2 z'^2 - k^2 x'^2 = k^2 y'^2$ for some $y' \in \Z$. Similarly for any common divisor of $y$ and $z$. Thus any common divisor of any pair of $x, y, z$ has to be a common divisor of Integers of the other. Hence any non-primitive solution to the Pythagorean equation is a constant multiple of some primitive solution. {{qed}}
First we show that $\tuple {2 m n, m^2 - n^2, m^2 + n^2}$ is a Pythagorean triple: {{begin-eqn}} {{eqn | l = \paren {2 m n}^2 + \paren {m^2 - n^2}^2       | r = 4 m^2 n^2 + m^4 - 2 m^2 n^2 + n^4       | c = }} {{eqn | r = m^4 + 2 m^2 n^2 + n^4       | c = }} {{eqn | r = \paren {m^2 + n^2}^2       | c = }} {{end-eqn}} So $\tuple {2 m n, m^2 - n^2, m^2 + n^2}$ is indeed a Pythagorean triple. Now we establish that $\tuple {2 m n, m^2 - n^2, m^2 + n^2}$ is primitive: {{AimForCont}} to the contrary, that $\tuple {2 m n, m^2 - n^2, m^2 + n^2}$ is not primitive. So there is a prime divisor $p$ of both $2 m n$ and $m^2 - n^2$. That is: :$p \in \mathbb P: p \divides \paren {2 m n}, p \divides \paren {m^2 - n^2}$ Then from Prime Divides Power: :$p \divides \paren {2 m n}^2$ and $p \divides \paren {m^2 - n^2}^2$ Hence from Common Divisor Divides Integer Combination: :$p \divides \paren {m^2 + n^2}^2$ and from Prime Divides Power again: :$p \divides \paren {m^2 + n^2}$ So from Common Divisor Divides Integer Combination: :$p \divides \paren {m^2 + n^2} + \paren {m^2 - n^2} = 2 m^2$ :$p \divides \paren {m^2 + n^2} - \paren {m^2 - n^2} = 2 n^2$ But $p \ne 2$ as, because $m$ and $n$ are of opposite parity, $m^2 - n^2$ must be odd. So $p \divides n^2$ and $p \divides m^2$ and so from Prime Divides Power, $p \divides n$ and $p \divides m$. But as we specified that $m \perp n$, this is a contradiction. Therefore $\paren {2 m n, m^2 - n^2, m^2 + n^2}$ is primitive. {{qed|lemma}} Now we need to show that every primitive Pythagorean triple is of this form: So, suppose that $\tuple {x, y, z}$ is ''any'' primitive Pythagorean triple given in canonical form. From Parity of Smaller Elements of Primitive Pythagorean Triple, $x$ and $y$ are of opposite parity. By definition of canonical form $x$ is even and $y$ and $z$ are both odd. As $y$ and $z$ are both odd, their sum and difference are both even. Hence we can define: :$s, t \in Z: s = \dfrac {z + y} 2, t = \dfrac {z - y} 2$. Note that $s \perp t$ as any common divisor would also divide $s + t = z$ and $s - t = y$, and we know that $z \perp y$ from Elements of Primitive Pythagorean Triple are Pairwise Coprime. Then from the Pythagorean equation: :$x^2 = z^2 - y^2 = \paren {z + y} \paren {z - y} = 4 s t$ Hence: :$\paren {\dfrac x 2}^2 = s t$ As $x$ is even, $\dfrac x 2$ is an integer and so $s t$ is a square. So each of $s$ and $t$ must be square as they are coprime. Now, we write $s = m^2$ and $t = n^2$ and substitute back: :$x^2 = 4 s t = 4 m^2 n^2$ and so $x = 2 m n$ :$y = s - t = m^2 - n^2$ :$z = m^2 + n^2$ Finally, note that: :$m \perp n$ from $s \perp t$ and Prime Divides Power :$m$ and $n$ have opposite parity otherwise $y$ and $z$ would be even. Thus, our primitive Pythagorean triple is of the form $\tuple {2 m n, m^2 - n^2, m^2 + n^2}$. {{qed}}
Suppose the Goldbach Conjecture holds. Let $n \in \Z$ such that $n > 5$. Let $n$ be an odd integer. Then $n - 3$ is an even integer greater than $2$. By the Goldbach Conjecture: :$n - 3 = p_1 + p_2$ where $p_1$ and $p_2$ are both primes. Then: :$n = p_1 + p_2 + 3$ As $3$ is prime, the result follows. Let $n$ be an even integer. Then $n - 2$ is an even integer greater than $3$ and so greater than $2$. By the Goldbach Conjecture: :$n - 2 = p_1 + p_2$ where $p_1$ and $p_2$ are both primes. Then: :$n = p_1 + p_2 + 2$ As $2$ is prime, the result follows. {{qed}} Category:Prime Numbers Category:Goldbach Conjecture ru92lcffn8n43wrt1rqdv6uzsj1evo7
Let $a \equiv b \pmod p$. Then by Congruence of Powers: :$a^2 \equiv b^2 \pmod p$ Hence: :$x^2 \equiv a \pmod p$ has a solution {{iff}} $x^2 \equiv b \pmod p$. Hence the result. {{qed}} Category:Quadratic Residues rqg7tzzvmszbgwmoy21zvr5o7qdyw94
By definition of Sorgenfrey line, the set: :$\BB = \set {\hointr x y: x, y \in \R \land x < y}$ is a basis of $T$. By definition of weight: :$\map w T \le \card \BB$ where $\card \BB$ denotes the cardinality of $\mathcal B$. By Cardinality of Basis of Sorgenfrey Line not greater than Continuum: :$\card \BB \le \mathfrak c$ Thus :$\map w T \le \mathfrak c$ It remains to show that: :$\mathfrak c \le \map w T$ Aiming for a contradiction, suppose that :$\mathfrak c \not \le \map w T$ Then: :$\map w T < \mathfrak c$ By definition of weight, there exists a basis $\BB_0$ of $T$: :$\map w T = \card {\BB_0}$ Then by Set of Subset of Reals with Cardinality less than Continuum has not Interval in Union Closure: :$\exists x, y \in \R: x < y \land \hointr x y \notin \set {\bigcup A: A \subseteq \BB_0} = \tau$ By definition of $\BB$: :$\hointr x y \in \BB \subseteq \tau$ Thus this contradicts by definition of subset with: :$\hointr x y \notin \tau$ {{qed}}
It remains to be demonstrated that the line segment $NM$ is the side of a regular hepadecagon inscribed in circle $ACB$. This will be done by demonstrating that $\angle LOM$ is equal to $\dfrac {2 \pi} {17}$ radians, that is, $\dfrac 1 {17}$ of the full circle $ACB$. For convenience, let the radius $OA$ be equal to $4 a$. By Pythagoras's Theorem, $AD = a \sqrt {17}$. By definition of tangent, $OE = a \arctan \left({\dfrac {\angle ODA} 4}\right)$. By construction, $\angle EDF = \dfrac \pi 4$ radians. Thus: {{begin-eqn}} {{eqn | l = \frac {\tan \angle ODE + \tan \angle ODF} {1 - \tan \angle ODE \tan \angle ODF}       | r = \tan \angle EDF       | c = Tangent of Sum }} {{eqn | r = 1       | c = Tangent of $\dfrac \pi 4$ }} {{end-eqn}} {{ProofWanted|finish off -- the algebra gets complicated from here on in.}}
Define: :$\mathcal F = \left\{{\bigcup \mathcal G: \mathcal G \subseteq \mathcal B}\right\}$ Define: :$ Z = \left\{{x \in \R: \exists U \in \mathcal F: x}\right.$ is local minimum in $\left.U\right\}$ By Set of Subsets of Reals with Cardinality less than Continuum Cardinality of Local Minimums of Union Closure less than Continuum: :$\left\vert{Z}\right\vert < \mathfrak c$ Then by Cardinalities form Inequality implies Difference is Nonempty: :$\R \setminus Z \ne \varnothing$ Hence by definition of empty set: :$\exists z: z \in \R \setminus Z$ By definition of difference: :$z \in \R \land z \notin Z$ Thus $z < z+1$. We will show that $z$ is local minimum in $\left[{z \,.\,.\, z+1}\right)$. Thus: :$z \in \left[{z \,.\,.\, z+1}\right)$ Hence: : $z-1 < z$ Thus: : $\left({z-1 \,.\,.\, z}\right) \cap \left[{z \,.\,.\, z+1}\right) = \varnothing$ Then by definition $z$ is a local minimum in $\left[{z \,.\,.\, z+1}\right)$. Because $z \notin Z$: :$\left[{z \,.\,.\, z+1}\right) \notin \mathcal F$ {{qed}}
Define a mapping $f: \BB \to \R \times \R$: :$\forall I \in \BB: \map f I = \tuple {\min I, \sup I}$ That is: :$\map f {\hointr x y} = \tuple {x, y} \forall x, y \in \R: x < y$ We will show that $f$ is an injection by definition. Let $I_1, I_2 \in \BB$ such that: :$\map f {I_1} = \map f {I_2}$ {{begin-eqn}} {{eqn | l = I_1       | r = \hointr {\min I_1} {\sup I_1}       | c = {{Defof|Half-Open Real Interval}} }} {{eqn | r = \hointr {\min I_2} {\sup I_2}       | c = by $\map f {I_1} = \map f {I_2}$ }} {{eqn | r = I_2       | c = {{Defof|Half-Open Real Interval}} }} {{end-eqn}} So: :$I_1 = I_2$ Thus $f$ is an injection. By Injection implies Cardinal Inequality: :$\card \BB \le \card {\R \times \R}$ By Cardinal Product Equal to Maximum: :$\card {\R \times \R} = \map \max {\mathfrak c, \mathfrak c}$ Thus: :$\card \BB \le \mathfrak c$ {{qed}}
Aiming for a contradiction suppose that : $Y \setminus X = \varnothing$ Then by Set Difference with Superset is Empty Set: :$Y \subseteq X$ Hence by Subset implies Cardinal Inequality: :$\left\vert{Y}\right\vert \leq \left\vert{X}\right\vert$ This contradicts: :$\left\vert{X}\right\vert < \left\vert{Y}\right\vert$ Hence the result. {{qed}}
We will prove that: :$(1): \quad \size \BB \aleph_0 < \mathfrak c$ where $\aleph_0 = \size \N$ by Aleph Zero equals Cardinality of Naturals. In the case when $\size \BB = \mathbf 0$ we have by Zero of Cardinal Product is Zero: :$\size \BB \aleph_0 = \mathbf 0 < \mathfrak c$ In the case when $\mathbf 0 < \size \BB < \aleph_0$: {{begin-eqn}} {{eqn | l = \size \BB \aleph_0       | r = \aleph_0 \size \BB       | c = Product of Cardinals is Commutative }} {{eqn | r = \size {\N \times \BB}       | c = {{Defof|Product of Cardinals}} }} {{eqn | r = \map \max {\size \N, \size \BB}       | c = Cardinal Product Equal to Maximum }} {{eqn | r = \aleph_0       | c = because $\size \BB < \aleph_0$ }} {{eqn | o = <       | r = \mathfrak c       | c = Aleph Zero is less than Continuum }} {{end-eqn}} In the case when $\size \BB \ge \aleph_0$ we have: {{begin-eqn}} {{eqn | l = \size \BB \aleph_0       | r = \size {\BB \times \N}       | c = definition of Definition:Product of Cardinals }} {{eqn | r = \map \max {\size \BB, \size \N}       | c = Cardinal Product Equal to Maximum }} {{eqn | r = \size \BB       | c = because $\size \BB \ge \aleph_0$ }} {{eqn | o = <       | r = \mathfrak c       | c = assumption }} {{end-eqn}} Define: :$Y = \leftset {x \in \R: \exists U \in \BB: x}$ is local minimum in $\rightset U$ We will show that $X \subseteq Y$ by definition of subset. Let $x \in X$. By definition of $X$: :$\exists U \in \leftset {\bigcup \GG: \GG \subseteq \BB}: x$ is local minimum in $\rightset U$ :$\exists \GG \subseteq \BB: U = \bigcup \GG$ By definition of local minimum: :$x \in U$ By definition of union: :$\exists V \in \GG: x \in V$ By definition of subset :$V \in \BB$ By definition of local minimum :$\exists y \in \R: y < x \land \openint y x \cap U = \O$ By Set is Subset of Union: :$V \subseteq U$ Then: :$\exists y \in \R: y < x \land \openint y x \cap V = \O$ By definition: :$x$ is local minimum in $V$ Thus by definition of $Y$ :$x \in Y$ So :$(2): \quad X \subseteq Y$ Define $\family {Z_A}_{A \mathop \in \BB}$ as: :$Z_A = \leftset {x \in \R: x}$ is local minimum in $\rightset A$ We will prove that: :$(3): \quad Y \subseteq \ds \bigcup_{A \mathop \in \BB} Z_A$ Let $x \in Y$. By definition of $Y$: :$\exists U \in \BB: x$ is local minimum in $U$ By definition of $Z_U$: :$x \in Z_U$ Thus by definition of union: :$x \in \ds \bigcup_{A \mathop \in \BB} Z_A$ This ends the proof of inclusion. {{qed|lemma}} By Set of Local Minimum is Countable: :$\forall A \in \BB: Z_A$ is countable By Countable iff Cardinality not greater Aleph Zero: :$\forall A \in \BB: \size {Z_A} \le \aleph_0$ By Cardinality of Union not greater than Product: :$(4): \quad \ds \size {\bigcup_{A \mathop \in \BB} Z_A} \le \size \BB \aleph_0$ Thus: {{begin-eqn}} {{eqn | l = \size X       | o = \le       | r = \size Y       | c = $(2)$ and Subset implies Cardinal Inequality }} {{eqn | o = \le       | r = \size {\bigcup_{A \mathop \in \BB} Z_A}       | c = $(3)$ and Subset implies Cardinal Inequality }} {{eqn | o = \le       | r = \size \BB \aleph_0       | c = $(4)$ }} {{eqn | o = <       | r = \mathfrak c       | c = $(1)$ }} {{end-eqn}} {{qed}}
The slope of $AB$ is defined as the change in $y$ divided by the change in $x$. Between $A$ and $B$: :the change in $x$ is $\paren {x + h} - x = h$ :the change in $y$ is $\map f {x + h} - \map f x$. Hence the result. {{qed}}
Let $f: \R \to \R$ be a real function. :400px Let the graph $G$ of $f$ be depicted on a Cartesian plane. Let $A = \tuple {\xi, \map f \xi}$ be a point on $G$. Consider the secant $AB$ to $G$ where $B = \tuple {\xi + h, \map f {\xi + h} }$. From Slope of Secant, the slope of $AB$ is given by: :$\dfrac {\map f {x + h} - \map f x} h$ By taking $h$ smaller and smaller, the secant approaches more and more closely the tangent to $G$ at $A$. {{qed}}
:$X$ is countable  {{iff}}: :there exists an injection $f: X \to \N$ by definition of countable set {{iff}}: :$\left\vert{X}\right\vert \leq \left\vert{\N}\right\vert$ by Injection iff Cardinal Inequality {{iff}}: :$\left\vert{X}\right\vert \leq \aleph_0$ {{qed}}
{{begin-eqn}} {{eqn | l = \aleph_0       | r = \left\vert{\aleph_0}\right\vert       | c = Cardinal of Cardinal Equal to Cardinal }} {{eqn | r = \left\vert{\omega}\right\vert       | c = definition of aleph mapping }} {{eqn | r = \left\vert{\N}\right\vert       | c = construction of natural numbers }} {{end-eqn}} {{qed}}
From Body under Constant Acceleration: Distance after Time: :$\mathbf s = \mathbf u t + \dfrac {\mathbf a t^2} 2$ where: :$\mathbf s$ is the displacement of $B$ at time $t$ from its initial position at time $t$ :$\mathbf u$ is the velocity at time $t = 0$ :$\mathbf a$ is the constant acceleration $t$ In this scenario, $\mathbf u = \mathbf 0$. Thus: :$\mathbf s = \dfrac {\mathbf a} 2 t^2$ and so by taking the magnitudes of the vector quantities: :$s = \dfrac a 2 t^2$ Hence the result, by definition of proportional. {{qed}}
By definition of acceleration: : $\dfrac {\mathrm d \mathbf v} {\mathrm d t} = \mathbf a$ By Solution to Linear First Order Ordinary Differential Equation: : $\mathbf v = \mathbf c + \mathbf a t$ where $\mathbf c$ is a constant vector. We are given the initial condition: :$\big.{\mathbf v}\big\rvert_{\, t \mathop = 0} = \mathbf u$ from which it follows immediately that: : $\mathbf v = \mathbf u + \mathbf a t$ {{qed}} Category:Mechanics ou4smu10w6yt8nldzxc7v4morh1wbvp
From Body under Constant Acceleration: Velocity after Time: :$\mathbf v = \mathbf u + \mathbf a t$ By definition of velocity, this can be expressed as: :$\dfrac {\d \mathbf s} {\d t} = \mathbf u + \mathbf a t$ where both $\mathbf u$ and $\mathbf a$ are constant. By Solution to Linear First Order Ordinary Differential Equation: :$\mathbf s = \mathbf c + \mathbf u t + \dfrac {\mathbf a t^2} 2$ where $\mathbf c$ is a constant vector. We are (implicitly) given the initial condition: :$\bigvalueat {\mathbf s} {t \mathop = 0} = \mathbf 0$ from which it follows immediately that: :$\mathbf s = \mathbf u t + \dfrac {\mathbf a t^2} 2$ {{qed}}
From Body under Constant Acceleration: Velocity after Time :$\mathbf v = \mathbf u + \mathbf a t$ Then: {{begin-eqn}} {{eqn | l = \mathbf v \cdot \mathbf v       | r = \paren {\mathbf u + \mathbf a t} \cdot \paren {\mathbf u + \mathbf a t}       | c =  }} {{eqn | r = \mathbf u \cdot \mathbf u + \mathbf u \cdot \mathbf a t + \mathbf a t \cdot \mathbf u + \paren {\mathbf a t} \cdot \paren {\mathbf a t}       | c = Dot Product Distributes over Addition }} {{eqn | r = \mathbf u \cdot \mathbf u + 2 \mathbf u \cdot \mathbf a t + \paren {\mathbf a t} \cdot \paren {\mathbf a t}       | c = Dot Product Operator is Commutative }} {{eqn | r = \mathbf u \cdot \mathbf u + 2 \mathbf u \cdot \mathbf a t + \mathbf a \cdot \mathbf a t^2       | c = Dot Product Associates with Scalar Multiplication }} {{eqn | n = 1       | r = \mathbf u \cdot \mathbf u + \mathbf a \cdot \paren {2 \mathbf u t + \mathbf a t^2}       | c = Dot Product Distributes over Addition }} {{end-eqn}} From Body under Constant Acceleration: Distance after Time: :$\mathbf s = \mathbf u t + \dfrac {\mathbf a t^2} 2$ Substituting for $\mathbf s$ in $(1)$ gives: :$\mathbf v \cdot \mathbf v = \mathbf u \cdot \mathbf u + 2 \mathbf a \cdot \mathbf s$ and the proof is complete. {{qed}} Category:Mechanics eslteq9h7qxgpbtjmwg2xkktl3nhdzd
By Continuum equals Cardinality of Power Set of Naturals: :$\mathfrak c = \left\vert{\mathcal P \left({\N}\right)}\right\vert$ where: :$\mathcal P \left({\N}\right)$ denotes the power set of $\N$ :$\left\vert{\mathcal P \left({\N}\right)}\right\vert$ denotes the cardinality of $\mathcal P \left({\N}\right)$. By Cardinality of Set less than Cardinality of Power Set: :$\left\vert{\N}\right\vert < \left\vert{\mathcal P \left({\N}\right)}\right\vert$  Thus by Aleph Zero equals Cardinality of Naturals: :$\aleph_0 < \mathfrak c$ {{qed}}