Use De Morgan's laws to remove the complement of a set, then distribute the two unions:The first and last intersections in the middle line are the empty set. The remaining union is precisely the collection of elements belonging to exactly one of the original sets. Therefore both expressions define the same symmetric difference:

In proof by contradiction, assume the negation of a desired conclusion $A$, derive logical falsity, and discharge the assumption to infer $A$. A rule expressing this isHere the label $i$ identifies the assumption being discharged; any other open assumptions remain.

To derive this rule using the supplied rules, apply negation introduction to the subproof from $\neg A$ to $\perp$. This discharges $\neg A$ and yields $\neg \neg A$. Apply double-negation elimination to obtain $A$. Thus the extra rule is admissible because Figure 1 explicitly includes double-negation elimination.^[ref]

This is the classical form of proof by contradiction; simply deriving $\neg A$ from an assumption $A$ leading to logical falsity uses negation introduction alone.

The following natural deduction tables record the active assumption lines explicitly. A rule may use an earlier line only while all of that line's assumptions remain available; assumption discharge closes the indicated case or hypothesis. The first proof makes the case split first, constructs the whole required logical conjunction in each case, and then closes the cases.

The second proof instead splits the goal first. It proves each required logical disjunction by a separate use of disjunction elimination, then applies conjunction introduction at the end.

The essential difference is the order of proof construction: the first shares one case split between both goals, whereas the second duplicates that split and combines the results outside the cases. Both use only the stated natural deduction rules, and both conclude

The conjunction introduction rule combines derivations of $\varphi$ and $\psi$ into a derivation of $\varphi \wedge \psi$. Both conclusions must be available under the current assumptions.

The disjunction elimination rule is reasoning by cases: from $\varphi \vee \psi$, a subproof deriving the same conclusion $\kappa$ under $\varphi$, and another subproof deriving $\kappa$ under $\psi$, conclude $\kappa$. It discharges the two case assumptions, not any surrounding premises.

The implication introduction rule derives $\varphi \to \psi$ by temporarily assuming $\varphi$ and deriving $\psi$; it then discharges that occurrence of the assumption $\varphi$.

ThusA conclusion inside a box cannot simply be reused outside it without an appropriate inference rule. Any other undischarged assumptions continue to constrain the result. These are exactly the rules printed in Figure 1.^[ref]

The other eigenvalues are $1/2$ and $0$. Their magnitudes are strictly less than one. All three eigenvalues are distinct, so their eigenvectors form a basis and the matrix is diagonalizable. Every component outside the stationary direction therefore decays under repeated application of $P$.

For an explicit justification, chooseThen $Pw=w/2$, $Pu=0$, andThe zero-eigenvalue component disappears after one step. For every $n\ge 1$,Thus the shares converge to $v$ and the resource vector converges to $10v$, since the conserved total is ten. For any other initial resource vector of total $T$, the same eigenvector argument gives the limit $Tv$.

The factor $(\lambda -1)$ proves that one is an eigenvalue. To find its normalized eigenvector, solve $Pv=v$. The second and third rows giveso $v_2=v_3=v_1/2$. The normalization $v_1+v_2+v_3=1$ now givesIts entries are nonnegative and sum to one, and direct matrix multiplication verifies $Pv=v$. As a probability vector it is the stationary distribution; as resource shares it assigns half to the USA and a quarter to each other region.

Use the characteristic polynomial convention $p(\lambda )=\det (\lambda I-P)$, with $I$ the identity matrix. Put $u=\lambda -1/2$. ThenThe required characteristic polynomial equation for this matrix is thereforeUsing $\det (P-\lambda I)$ instead negates the cubic polynomial and leaves the eigenvalues unchanged.

Apply the transition matrix once and then again:Therefore, in USA, Europe, Rest of World order and in trillions of dollars,Each total remains ten, giving a quick check on the matrix multiplication.

For a square matrix $M$, an eigenvalue–eigenvector pair is a scalar $\lambda$ and a nonzero vector $v$ satisfyingThus the linear map represented by $M$ scales this particular direction by $\lambda$. The nonzero condition excludes the zero vector, which would otherwise satisfy the displayed equation for every scalar.