The classical proof by contradiction rule, also called reductio ad absurdum, iswith assumption discharge of the displayed $\neg \varphi$. It differs from the negation introduction rule by concluding $\varphi$ from a contradictory assumption of its negation.

To derive double-negation elimination using this rule, start with $\neg \neg \varphi$, assume $\neg \varphi$, and apply the negation elimination rule to those two formulas:No double-negation elimination occurs inside this derivation, so the argument is not circular.

Write $H=(P\wedge Q)\vee (P\wedge R)$ and $K=P\wedge (Q\vee R)$. In the following natural deduction derivations, brackets name temporary assumptions, and cited ranges specify the subproofs discharged. The abbreviations $\wedge E,\wedge I,\vee E,\vee I$ denote the corresponding conjunction elimination rule, conjunction introduction rule, disjunction elimination rule and disjunction introduction rule.

First use the disjunction elimination rule backwards: split the premise into two cases and build the whole conclusion in each case.

Alternatively, use the conjunction introduction rule backwards on the conclusion. First obtain $P$ by cases, then obtain $Q\vee R$ by a second case analysis, and finally combine the results:

The first proof goal transformation starts with case analysis; the second starts by splitting a conjunction goal. Counting only actual applications of the printed inference rules, both use nine steps: the first has four $\wedge E$, two $\vee I$, two $\wedge I$ and one $\vee E$; the second has four $\wedge E$, two $\vee I$, one $\wedge I$ and two $\vee E$. The extra case analysis in the second exactly replaces one conjunction construction. If assumption lines are counted as well, the second display has two more lines because it opens four case subproofs rather than two. Thus the apparent difference in display length need not be a difference in inference count.

The implication introduction rule opens a subproof under a temporary assumption $\varphi$, derives $\psi$, then performs assumption discharge to conclude $\varphi \to \psi$. The negation introduction rule instead derives logical falsity $\perp$ under a temporary assumption $\varphi$, discharges it, and concludes $\neg \varphi$:The rectangular boxes delimit these subproofs and the scope of their temporary assumptions. A line depending on the boxed temporary assumption cannot simply be used outside the box. The relevant introduction rule exports the indicated logical implication or negation; other surrounding assumptions $\Gamma$ stay in force.

The second, third and fourth linear equations giveSetting $x_4=t$ yields $x_3=-1-t$, $x_2=1+t$ and $x_1=1-t$. The first linear equation then gives $(1-t)+(1+t)+(-1-t)+t=1$, and the fifth gives $(1-t)+t=1$, so both remaining conditions hold for every real $t$. ConsequentlyThis is a particular solution plus the whole null space, an affine line in a vector space of solutions. The derivation both produces solutions and shows that none have been omitted.

The reduced row echelon form gives $x_1+x_4=0$, $x_2-x_4=0$, and $x_3+x_4=0$. With free parameter $x_4=t$,The displayed nonzero vector therefore forms a basis for the null space:This agrees with the rank-nullity theorem: there are four columns and matrix rank three.

Let $c_1,\dots ,c_4$ be the original columns. The first three are linearly independent: restricting them to rows 2, 3 and 4 gives the triangular matrixwhose determinant is one. Also $c_4=c_1-c_2+c_3$, so a basis for the column space isThese are columns of the original matrix. Taking columns of its reduced row echelon form would generally give the wrong column space.

Denote the five rows by $r_1,\dots ,r_5$. They satisfyThus $r_2,r_3,r_4$ span the row space. They are linearly independent: in a linear combination $\alpha r_2+\beta r_3+\gamma r_4=0$, coordinate 1 gives $\alpha =0$, coordinate 2 then gives $\beta =0$, and coordinate 4 gives $\gamma =0$. ThereforeAs an alternative basis, the nonzero rows of the reduced row echelon form are

For a real $m\times n$ matrix $A$, the row space is the linear span of its rows in ${\mathbb{R}}^n$. The column space is the linear span of its columns in ${\mathbb{R}}^m$, equivalently $\{A\mathbf{x}:\mathbf{x}\in {\mathbb{R}}^n\}$. The null space isThe matrix rank is the dimension of the column space, which equals the dimension of the row space. In particular,The last equality is the rank-nullity theorem.

Find a stationary distribution $\mathbf{\pi }=(u,v,w{)}^T$ from $P\mathbf{\pi }=\mathbf{\pi }$ and $u+v+w=1$. The first two independent linear equations areThey give $u=9w/8$ and $v=7w/4$. Normalization gives $w=8/31$, hence

To justify the limit using the assumptions, distinct eigenvalues give an eigenbasis, over the complex numbers if necessary. Choose $\mathbf{\pi }$ as the eigenvector for 1 and let ${\mathbf{v}}_2,{\mathbf{v}}_3$ correspond to the other eigenvalues ${\lambda }_2,{\lambda }_3$. Expand the initial probability distribution asSince $1^{T}P=1^T$, each ${\lambda }_j\ne 1$ satisfiesMultiplying the expansion by $1^T$ therefore gives $\alpha =1$. It follows thatbecause $|{\lambda }_2|,|{\lambda }_3|<1$. The components of the limiting probability distribution are approximately $0.2903,0.4516,0.2581$.

For $1=(1,1,1{)}^T$, the entries of $P^{T}1$ are the column sums of $P$. Henceso $1$ is an eigenvalue of $P^T$, with eigenvector $1$.

The determinant is unchanged by matrix transposition, soThus $P-I$ has a nontrivial null space, givingThe same argument with $\lambda I$ shows that a matrix and its transpose have the same characteristic polynomial. It does not say that they have the same eigenvectors: here $P1=(0.9,1.2,0.9{)}^T\ne 1$.

Condition on the intermediate parent's height and use the Markov property. The three mutually exclusive paths through Tall, Medium and Short contributeEquivalently, this is entry $(3,1)$ of the matrix power $P^2$. Hence

Use the state order Tall, Medium, Short and column probability distributions, so that ${\mathbf{p}}_{n+1}=P{\mathbf{p}}_n$. The $j$th column records the child's probability distribution given parent state $j$. Thus the transition matrix isEvery entry is nonnegative and every column sums to one. This convention also gives the requested property $P^T(1,1,1{)}^T=(1,1,1{)}^T$. A row-distribution convention would use the transposed transition matrix consistently instead. The stated Markov chain is the mathematical model to analyze; no additional biological assumptions are needed.

A scalar $\lambda$ is an eigenvalue of an $n\times n$ matrix $P$ if there is a nonzero vector $\mathbf{v}$ such thatAny such vector is an eigenvector corresponding to $\lambda$. The requirement that an eigenvector be nonzero is essential: the zero vector would satisfy the equation for every scalar and would convey no information.

Since $F_1=200+t$ and $t\ge 0$, its minimum is attained at $t=0$:Zero flow on the $B\to C$ arc is allowed; no strictly positive lower bound was specified.

Apply flow conservation at the four junctions, using the directions in the original diagram:The total external inflow and outflow are both 250 cars per hour. Accordingly one of these linear equations is redundant. Put $F_2=t$. Successive substitution yieldsin cars per hour. The nonnegativity restriction expresses clockwise traffic on every arc. Conversely, every $t\ge 0$ gives nonnegative flows satisfying all four conservation equations, so this is the whole physically allowable family. Increasing $t$ adds a common network circulation around the roundabout without changing any external inflow or outflow.

Source gap / TODO: the original printed page 4, confirmed visually, omits the unknown vector. It literally equates a $3\times 3$ matrix with a $3\times 1$ vector, so it does not define a well-formed system of linear equations. The following answer is conditional on the usual intended interpretation $A\mathbf{x}=(1,k^2,k{)}^T$; the missing factor is not present in the source.

Under that interpretation, the rows satisfy $r_1-2r_2+r_3=0$. Applying this linear relation to the right-hand side gives the necessary consistency conditionIt is sufficient too: the upper-left $2\times 2$ minor has determinant $4-9=-5\ne 0$, so the first two rows are linearly independent. The third row is $2r_2-r_1$, and imposes precisely the condition already found. ThereforeFor completeness, setting $x_3=t$ givesand direct substitution satisfies all three linear equations exactly for those two values of $k$. An unconditional answer to the malformed printed equality awaits a source correction.

The register file in this datapath supplies both read ports combinationally for whatever read addresses the instruction fields present. Reading a processor register is harmless, and a result that the current MIPS instruction does not need can simply be ignored. Only writes require an explicit enable because writes change architectural state. There is consequently no functional need for RegRead in this design.

The data-memory read interface has an explicit MemRead enable. Non-load MIPS instructions can present irrelevant or invalid addresses at the memory address input; an external memory transaction should not be initiated merely because those bits happen to be present. An explicit enable also permits interfaces with memory-mapped input-output, where a read can have a side effect. It can reduce unnecessary activity even for ordinary memory.This is a distinction in the supplied datapath, rather than a universal physical prohibition: other register files can have read enables, and an ideal side-effect-free memory can be modeled with continuous reads.