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Aggregate excess demand subtracts total endowment from the sum of consumers' demand correspondences. A market-clearing price satisfies $z(p)=0$. Under Walras law, one equation is redundant after price normalization.

A demand correspondence assigns a nonempty set of chosen consumption bundles to each price and wealth pair. Every chosen bundle satisfies the budget constraint. A utility function rationalizes the demand correspondence when the chosen bundles are exactly its maximizers over the budget set.

Let $B(p,w)=\{z\in {\mathbb{R}}^L:p\cdot z\le w\}$ be the set allowed by the budget constraint. A demand correspondence is understood to choose only feasible bundles: $x(p,w)\subseteq B(p,w)$. Define the direct revealed preference relation byThe indirect revealed preference relation is its reflexive transitive closure. Thus $yRz$ precisely when there is a finite chain $y=y_0,y_1,\dots ,y_m=z$ whose successive comparisons satisfy $y_j{R}_D{y}_{j+1}$. Chains of length zero are included. This makes $R$ a preorder on all of ${\mathbb{R}}^L$, including alternatives that are never chosen. The associated strict preference isIf a convention defines indirect revelation using only positive-length chains, one must add the diagonal before the intersection assertion in part (c) can hold: every complete weak preference relation contains the diagonal.

The congruence axiom requires that an affordable alternative indirectly revealed at least as desirable as a chosen alternative is itself chosen. In the notation of part (a),Equivalently, a chosen bundle $x$ and an affordable but unchosen bundle $y$ satisfy $\neg (yRx)$. Since $x{R}_{D}y$ already holds, this means $xPy$. The axiom therefore distinguishes a genuine strict preference for a chosen bundle from indifference between two chosen bundles; it applies to the entire demand correspondence, including budgets with several chosen alternatives.

First pass from the preorder $R$ to a partial order. Define the indifference relation $y\sim z$ by $yRz$ and $zRy$. Reflexivity and transitivity make this an equivalence relation. On the quotient set $C={\mathbb{R}}^L/\sim$, putTransitivity makes this independent of the representatives, and the quotient relation is antisymmetric, hence a partial order.

We need its order-extension property, including freedom to order incomparable classes. Here is the argument. Order all partial order extensions of $⊵$ by inclusion. The union of a chain of extensions is again a partial order: any finite list of comparisons needed to check an axiom occurs together in one member of the chain. Zorn's lemma therefore supplies a maximal extension. If two classes $A,B$ remain incomparable, adjoin $A⊵B$ and take the transitive closure of a relation. The only possible new comparisons have the form $X⊵A⊵B⊵Y$. A cycle would require a previously existing path $B⊵A$, contradicting incomparability. The enlarged relation is still a partial order, contradicting maximality. Consequently the maximal extension is a total order. This also proves the Szpilrajn extension theorem used here.

Lift such a total order to the original bundles by comparing their equivalence classes. The lifted preference relation $R^{\prime }$ is complete and transitive, contains $R$, and preserves its strict preference: if $yPz$, their distinct classes satisfy $[y]⊵[z]$, and antisymmetry of the extended order prevents the reverse comparison. In particular, the class $\mathcal{P}$ is nonempty and $R\subseteq {\bigcap }_{R^{\prime }\in \mathcal{P}}{R}^{\prime }$.

For the reverse inclusion, take a pair with $\neg (yRz)$. If $zRy$, then $zPy$, so every member of $\mathcal{P}$ strictly ranks $z$ above $y$ and excludes $y{R}^{\prime }z$. If neither comparison holds, their classes are incomparable. First adjoin $[z]⊵[y]$, which creates no cycle by the preceding argument, and then extend to a total order. Its lift belongs to $\mathcal{P}$ and again excludes $y{R}^{\prime }z$. Thus every pair outside $R$ is outside at least one member of $\mathcal{P}$, provingThe intersection proof is an order-theoretic fact about any preorder. The congruence axiom additionally ensures exact rationalization of the observed demand correspondence: every chosen bundle weakly outranks every affordable bundle, while an affordable unchosen bundle is strictly worse by part (b). Hence each completion's budget-maximizing set is exactly the observed chosen set.

Write the first endowment as $(a,t)$. The budget constraint at normalized prices is $px+y\le pa+t$. Because the quasilinear utility is increasing in $y$, the budget binds, and its maximization reduces to maximizing $\log x-px+pa+t$ over $x>0$. Its derivative is $1/x-p$ and its second derivative is $-1/x^2<0$, giving the unique demand correspondenceThe first-good demand has no wealth effect. Since the second consumer's endowment and preference relation are fixed, $Z_t=0$ everywhere. Thus the joint condition of part (a) becomes $Z_p\ne 0$ at equilibrium; it is not automatically satisfied.

Here is a complete smooth counterexample, with the second consumer's consumption set specified explicitly. The question does not impose the first consumer's consumption set on the second consumer. Take endowments $(a,t)=(1/10,t)$ and $(b,c)=(2,1)$. Give the second consumer consumption set $X_2=(-\infty ,A)\times \mathbb{R}$, where $A=43/20$, and put $B=31/20$. Its utility function isThis defines complete, transitive, continuous preference relations. It is in fact smooth on $X_2$: writing $r=A-x>0$ and $s=y-B>0$, the positive branch is $\exp (-r^2/(4s)-r/2-s/4)$; at $s=0$ all derivatives tend to zero locally uniformly where $r>0$. The endowment belongs to $X_2$. Allowing negative holdings is explicit here, just as the first consumer's second coordinate is unrestricted in the printed question.

We now derive, rather than posit, the second consumer's demand correspondence. For $v>0$, the upper contour $U_2\ge e^{-v}$ is described by $(r+s{)}^2\le 4vs$, with $r,s>0$. At a fixed $s$, its cheapest first coordinate uses $r=2\sqrt{vs}-s$. Minimizing total expenditure over $s$ therefore means minimizingThe unique minimum occurs at $\sqrt{s}=p\sqrt{v}/(1+p)$, with $r=vp(p+2)/(1+p{)}^2>0$. The expenditure necessary to reach this contour is consequentlySet $\delta =A-b=3/20$ and $\epsilon =B-c=11/20$. Wealth is $pb+c$, so the best affordable contour hasAll affordable positive-utility bundles have $v\ge v_{\ast }$, and the minimum-expenditure bundle on that contour is unique and spends precisely the budget. Zero-utility bundles cannot be optimal because this positive-utility bundle is available. Substitution yields globally smooth unique demands for every $p>0$:The aggregate excess demand is nowFor every $t>0$, the unique equilibrium price is $p=1$, and $Z_p(1,t)=Z_t(1,t)=0$. At this price the allocations are $(1,t-9/10)$ and $(11/10,19/10)$, whose sum equals the aggregate endowment $(21/10,t+1)$. Second-good excess demand is $-pZ$, so it too is smooth and clears. Hence every economy in this smooth parameterized family is singular. This demonstrates why the first consumer's missing wealth effect alone gives no transversality guarantee; the stronger conclusion in part (c) comes from the second consumer's additional demand restriction.