Introduce a fresh existential witness $v$ with $R(v)$. The universal premise gives $v=0\vee v=1$. In the first case, equality elimination rule gives $R(0)$. In the second, it gives $R(1)$, contradicting $\neg R(1)$; explosion then supplies the common case conclusion $R(0)$. Close both cases and then the witness scope:

HenceFor both uses of equality elimination rule, the formula template is $R(z)$. At the final existential elimination, $v$ is fresh in all three original premises and absent from $R(0)$, so the eigenvariable condition holds. No additional assumption that the constants $0$ and $1$ denote distinct objects is needed.

Use reflexivity to obtain $T_1=T_1$. In equality elimination rule, take $\varphi (z)$ to be $z=T_1$, and replace the designated first occurrence of $T_1$ using the premise $T_1=T_2$:

ThereforeThe second occurrence of $T_1$ in $T_1=T_1$ is part of the fixed parameter in $\varphi (z)$ and is not replaced.

The equality introduction rule is reflexivity: any well-formed first-order term is equal to itself, with no premises. The equality elimination rule permits replacement of equals in a first-order formula:In the second rule, write $\varphi (S)$ and $\varphi (T)$ as the results of substituting $S$ and $T$ into the same designated free occurrences of a variable in $\varphi$. Both substitutions must avoid variable capture. The rule preserves the surrounding assumptions. This expresses first-order equality rather than treating equality as an arbitrary relation symbol.

The original scan repeats the premise and conclusion of part (i) exactly; it is not a missing or unreadable alternative exercise.^[ref] Here is a second formal ordering of the same argument. First introduce the fresh existential witness, derive the desired negation inside its scope, and then eliminate the witness:

Thus, again,At the last existential elimination, $v$ occurs in neither the exported conclusion nor the remaining premise. The eigenvariable condition is therefore satisfied. This proof retains a different nesting of the subproofs from part (i).

Choose an arbitrary fresh eigenvariable $v$. If $\neg R(v)$ held, existential quantifier introduction rule would contradict the premise. Derive $R(v)$ by double-negation elimination, then generalize:

HenceAt the final universal introduction, $v$ is free in no open assumption: the only remaining premise is the closed first-order formula $\neg \exists x\neg R(x)$. Thus the eigenvariable condition is satisfied. The passage from double negation to the positive relation uses the paper's classical rule.

Assume that every object fails $R$. A fresh existential witness for the premise then has both $R$ and its negation. Eliminate the witness before closing the assumption about all objects:

The conclusion isThe witness $v$ occurs neither in the conclusion of existential elimination nor in its other open assumptions, so the eigenvariable condition holds.

Temporarily deny the desired logical disjunction. If $\neg P$ held, it would introduce that very logical disjunction, producing logical falsity; double-negation elimination therefore gives $P$. The same argument gives $Q$, contradicting the premise. The complete natural deduction derivation is:

ConsequentlyThe uses of double-negation elimination are justified by the supplied rules. This distinguishes the proof from the preceding constructive direction of De Morgan's laws.

Under the premise, assuming either disjunct would introduce the forbidden logical disjunction. Discharging those two assumptions gives the two separate negations, which can then be conjoined.

ThusThe redundant closing parenthesis printed after the conclusion on p. 8 does not alter the readable propositional formula.^[ref]

Let $L=\{0^m{1}^n:m,n\ge 0,m\ne n\}$, and suppose it were a regular language. The formal language $B=0^{\ast }{1}^{\ast }$ is a regular language. By closure of regular languages under set difference,would be a regular language. But the pumping lemma for regular languages rules this out: for pumping length $p$, every permitted pumped block in $0^p{1}^p$ consists of $k>0$ zeros, and deleting it produces $0^{p-k}{1}^p$, with unequal counts. This contradiction provesThe closure step can also be realized by a product automaton accepting when the automaton for $B$ accepts and the automaton for $L$ rejects. If the exponents are instead required to be positive, use $B=0^{+}{1}^{+}$; the same argument applies.

This is a regular language. A deterministic finite automaton needs only to remember the number of consecutive trailing ones, up to three, and whether a forbidden run of four has already occurred. Use states $q_0,q_1,q_2,q_3,q_{\times }$, start at $q_0$, and accept in $q_0,q_1,q_2,q_3$. Its transition function of a finite automaton is:

Reading a zero resets the trailing count unless rejection has already occurred. A fourth successive one enters the permanent rejecting state $q_{\times }$. Thus the accepted strings are exactly those with no substring $1111$, and

These strings are the even-length palindromes $u{u}^R$. Suppose this were a regular language, and let $p$ be its length from the pumping lemma for regular languages. The stringbelongs to the formal language, with $u=0^{p}1$. In any allowed decomposition $w=abc$, where $|ab|\le p$ and $|b|>0$, the pumped block is $b=0^k$ with $1\le k\le p$. Pumping it away givesThis is not a palindrome: its initial and final runs of zeros have different lengths. Hence it cannot have the form $u{u}^R$, contradicting the pumping lemma for regular languages. Therefore

Every finite formal language is a regular language. A single string $a_1\cdots a_k$ is described by the regular expression formed by concatenating its symbols; the empty string is described by $\epsilon$. A finite union of these regular expressions describes any specified finite collection of strings. The empty collection is described by $\varnothing$. ThusEquivalently, a deterministic finite automaton can remember which prefix of one of the finitely many accepted strings has been read, moving to a rejecting dead state when no such prefix remains possible.

The partial fraction decompositionmakes the sum a telescoping sum. All interior terms cancel:

For $s>0$, the denominator $s(s+1)$ is positive, so every value of $f$ is positive. Conversely, for every $r>0$, the positive number $s=(\sqrt{1+4/r}-1)/2$ satisfies $f(s)=r$. Thus its image of a function, or range, isOne can also see the full range from continuity and monotonicity: $f$ is a continuous function with $f^{\prime }(s)=-(2s+1)/(s^2+s{)}^2<0$, and its endpoint limits are $+\infty$ as $s\to 0^{+}$ and $0$ as $s\to \infty$. The endpoints themselves are not attained.

For $r>0$, solve $f(s)=r$ with $s>0$. Multiplying by the positive denominator gives $s^2+s=1/r$, so the quadratic formula givesThe minus sign gives a negative number and is inadmissible. The plus sign gives a positive number for every $r>0$, and it is the unique positive solution. Hence the inverse function isIndeed $g(r)(g(r)+1)=1/r$, so $f(g(r))=r$. Conversely, for $s>0$, $\sqrt{1+4/f(s)}=\sqrt{(2s+1{)}^2}=2s+1$, giving $g(f(s))=s$. These verify both inverse function identities.

The domain of a function is fixed by its declared type, soAlthough the rational expression also has values at some negative arguments, they are not in the specified domain of a function. Part (a) of the printed question only supplies the definition used in these four subparts.^[ref]

Here the velocity contribution in D'Alembert's formula isThus, with arbitrary twice continuously differentiable $h$,For the added term, $u_{tt}=e^x\sinh t=u_{xx}$ and $u_t(x,0)=e^x\cosh 0=e^x$. Its displacement at $t=0$ is zero, leaving $h$ free as required.

The velocity contribution in D'Alembert's formula is $\frac{1}{2}{\int }_{x-t}^{x+t}1ds=t$. Hence, with arbitrary twice continuously differentiable $h$,The added term has both second partial derivatives zero and has time derivative one, so it supplies precisely the required initial condition while preserving the wave equation.

For the wave equation with wave speed one, D'Alembert's formula with arbitrary twice continuously differentiable initial displacement $h(x)=u(x,0)$ isThe arbitrary function $h$ is essential: only the initial velocity is prescribed. For $g=0$, this givesBoth travelling terms solve the wave equation, and their time derivatives cancel at $t=0$. Conversely every classical solution has this form for its initial displacement $h$.

Use the product rule for the Laplace operator:Thus Laplace's equation, away from the origin, reduces to $g^{\prime \prime }+3g^{\prime }/r=0$. On the line $x=0$ the displayed expression vanishes automatically; requiring the equation on an open region gives the same radial equation by continuity from $x\ne 0$. Multiplying by $r^3$ gives $(r^3{g}^{\prime }{)}^{\prime }=0$, so $g^{\prime }=C{r}^{-3}$ and $g=A+B{r}^{-2}$. ThereforeIf extension through the origin is required, $B=0$: along the horizontal axis the second term is $B/x$. The remaining harmonic function is $Ax$.