Assume the two implications in turn, then derive a contradiction from $\neg P$. The natural deduction proof isHere RAA is the derived proof by contradiction rule. The last two assumption discharge steps establish the requested right-associated implication as a theorem.

Apply the biconditional introduction rule from part (iv) to the two already established derivations:The left numerator is part (iii), the right is part (i). Introduction discharges both displayed assumptions, so the result has no premises.

Use a biconditional introduction rule that discharges the two displayed temporary assumptions:Suitable elimination rules areThese are equivalent to using the biconditional elimination rule to extract the two implications and then applying modus ponens rule. Together they express both directions, with no extra arbitrary assumption retained after introduction.

Put $H=\neg (P\to \neg Q)$. First derive $Q$, then $P$, using the derived proof by contradiction rule. Indentation records the subproofs:Line 11 uses explosion; lines 5 and 12 discharge only their inner assumptions. Each RAA step can be expanded into negation introduction followed by double-negation elimination, as in part (ii).

Proof by contradiction is the derived ruleGiven the subproof under $\neg A$, first discharge it by negation introduction to get $\Gamma ⊢\neg \neg A$. Apply double-negation elimination to obtain $A$. This justification uses the stated classical rules; unrestricted RAA would not follow from intuitionistic rules alone.

Use conjunction elimination to obtain both premises needed for a contradiction under a temporary implication assumption. The following natural deduction derivation discharges precisely assumption 4:Here $\to E$ is the implication elimination rule and $\neg I$ is negation introduction. The original conjunction remains the only undischarged premise.

Yes. Part (a) diagonalizes $G=B$, whose spectral radius is $3/8<1$. Thusand every initial error converges to zero. For the specified initial vector, solving $Ax=b$ gives $x=(1,-1{)}^T$ and $e_0=(-1,1{)}^T$. In fact,The special formula is not asserted for $n=0$, because the zero-eigenvalue component disappears only after the first step.

The exact solution obeys $(D-L)x=Ux+b$. Subtract this from the iteration to get $(D-L)e_n=U{e}_{n-1}$. Hence the stationary iteration matrix isThis is exactly the matrix $B$ from part (a); the sign follows the given $A=D-L-U$ convention.

As $x_0=0$, solve $(D-L)x_1=b$. Its first row gives $4(x_1{)}_1=1$; its second gives $(x_1{)}_1+2(x_1{)}_2=-1$. ThereforeThe second component uses the new first component $1/4$, distinguishing Gauss-Seidel iteration from Jacobi iteration.

The Gauss-Seidel method solves each component using the already updated components from the same iteration. In the splitting $A=D-L-U$, those updated diagonal and lower-triangular terms stay on the left, while upper-triangular terms use the previous iterate. ThusHere $D-L$ is lower triangular with nonzero diagonal, so each new iterate is obtained by forward substitution, starting from the specified zero vector.

For eigenvalue zero take $s_1=(1,0{)}^T$. For eigenvalue $3/8$, the first row of $(B-3I/8)s_2=0$ gives $x=-2y$, so take $s_2=(-2,1{)}^T$. These are linearly independent eigenvectors, givingThe identities $B{s}_i={\lambda }_i{s}_i$ give $BS=SD$, hence the required diagonalization $S^{-1}BS=D$.

Since $B$ is upper triangular, its eigenvalues are its diagonal entries. Equivalently, its characteristic polynomial is $\lambda (\lambda -3/8)$. Hence

Translate the mirror line to zero, reflect there, then translate back. Under the column-vector convention the rightmost operation acts first:This composition of affine transformations maps $(x,y)$ to $(4-x,y)$, fixing exactly the points with $x=2$.

Reflection in $x=0$ sends $(x,y)$ to $(-x,y)$. Therefore its homogeneous-coordinate matrix isIts square is the identity, as required for applying the same reflection twice.

With column-vector homogeneous coordinates, the translation matrix isIt maps $(x,y,1{)}^T$ to $(x+a,y+b,1{)}^T$. The final coordinate incorporates the constant displacement into matrix multiplication.

Introduce $z=Ux$. Solve $Lz=b$ by forward substitution and then $Ux=z$ by back substitution. For these factors, the recurrences arefollowed byThe nonzero diagonal entries make both triangular solves well-defined. No matrix inverse is calculated.

First subtract row one from row two, using $E_1$; then subtract four times the updated row two from row three, using $E_2$:The updated second row is $(0,1,2)$ and the final third row is $(0,0,-4)$. Hence $E_2{E}_{1}A=U$ and $L=E_1^{-1}{E}_2^{-1}$. The resulting LU decomposition isReversing the order when undoing the elementary operations is essential.

The elementary row operation and its reverse areThus $FA$ replaces row three by row three minus three times row two. The inverse adds those three copies back.

The required elementary matrix isLeft multiplication exchanges the two rows. Applying the same exchange twice restores every row, so $E^2=I_3$.

For add, select rd and the arithmetic logic unit result. For sw, select destination zero while retaining the memory-store controls and source-register reads. For jal, select register 31 and the return-address data. These are the same actual register writes as before, with sw using destination zero in place of disabling RegWrite.

The link value is determined by the particular MIPS architecture datapath convention: $\mathrm{PC}+4$ for the usual teaching datapath without a delay slot, or $\mathrm{PC}+8$ for classic delayed-branch MIPS. This write-enable change does not alter whichever link convention the original machine implements.