Take $\alpha =\beta =P(x)$ for a unary predicate symbol. The premise $\forall x(P(x)\to P(x))$ is valid, but the proposed conclusion is $P(x)\to \forall xP(x)$. On a domain $\{0,1\}$ with $P=\{0\}$ and assignment $x\mapsto 0$, its antecedent is true and its consequent false.

Hence dropping the condition that $x$ is not free in $\alpha$ would turn a valid premise into an invalid conclusion. The condition permits the antecedent to remain fixed while the quantified variable ranges over the domain; an antecedent depending on that variable cannot be held true for all its values merely because it is true at the current assignment.

A term $t$ is free for substitution for $x_i$ in $\alpha$ when none of its variables becomes bound at a free occurrence of $x_i$ being replaced. Equivalently, no quantifier surrounding such an occurrence binds a variable of $t$. Bound occurrences of $x_i$ are not replaced. This occurrence-sensitive condition prevents variable capture.

For example take $\alpha (x)=\exists y\neg (x=y)$ and $t=y$. On a two-element domain, $\forall x\exists y\neg (x=y)$ is true. Naive textual substitution produces $\exists y\neg (y=y)$, which is false on every nonempty domain. Thus the corresponding unrestricted A4 implication would be invalid. The problem is that the inserted free $y$ was captured by the existing existential quantifier. Renaming that binder first instead gives the correct instance $\exists z\neg (y=z)$, with $z$ fresh.

The assertion $\Gamma ⊢\varphi$ means that there is a finite formal derivation ending in $\varphi$ with its undischarged assumptions contained in $\Gamma$. In a context-labelled presentation, an axiom instance is available with empty context, a formula $\alpha$ is available from the context $\{\alpha \}$, modus ponens combines derivations of $\alpha$ and $\alpha \to \beta$ to derive $\beta$, and generalization derives $\forall x\alpha$ provided $x$ is not free in its open assumptions. Thinning enlarges a context without changing its conclusion.

Each line or node must specify an axiom instance, an assumption, or an application of a permitted rule to earlier lines or subderivations. The side conditions on substitution, quantification and contexts are part of correctness. In particular a derivation is a finite syntactic object, even when $\Gamma$ is infinite; it uses only finitely many assumptions before a final thinning to $\Gamma$ if necessary. Derivability $⊢$ is thus distinct in definition from semantic entailment $\models$, which quantifies over structures and assignments.

Write $C(\mathbb{R},\mathbb{R})$ for the set of continuous real-valued functions on the real line. Constant functions give an injection of $\mathbb{R}$ into this set. In the other direction, restriction to the rational numbers is injective. Indeed, if continuous functions $f,g$ agree at every rational and $x$ is real, choose rational numbers $q_n\to x$. Continuity givesThus restriction embeds $C(\mathbb{R},\mathbb{R})$ into ${\mathbb{R}}^{\mathbb{Q}}$, which has the same cardinality as the real-sequence space because $\mathbb{Q}$ is countable. The previous part gives the upper bound $2^{{\aleph }_0}$, while constant functions give the matching lower bound. HenceThis computes the cardinality of continuous real functions. All five sets consequently have the same cardinality. The continuity assumption is decisive for restriction to a dense subset to determine a function.

Constant sequences give an injection $\mathbb{R}\to {}^{\omega }\mathbb{R}$. For the opposite bound fix an enumeration $(q_m{)}_{m\in \omega }$ of the rational numbers and encode a sequence $(x_n)$ byDistinct sequences differ at some coordinate $n$; density of the rational numbers then makes their encoded sets differ at a pair $(n,m)$. Thus $E$ is injective. The Cartesian product $\omega \times \omega$ is countable; for example the Cantor pairing functionis a bijection with $\omega$. Consequently $\mathcal{P}(\omega \times \omega )$ is equinumerous with $\mathcal{P}(\omega )$, already equinumerous with $\mathbb{R}$. Apply Cantor-Schröder-Bernstein theorem to obtain

There is an injection from the power set of $\omega$ into the real numbers given byIf $S$ and $T$ first differ at $k$, the contribution at $k$ has magnitude $2/3^{k+1}$, while the entire later tail has magnitude at mostTheir real values are therefore different. This proves injectivity directly and avoids any ambiguity of numeral expansions.

Conversely, map a real number $x$ to the set $\{q\in \mathbb{Q}:q<x\}$. Distinct real numbers give distinct sets because the rational numbers are dense in the real line. Fixing a bijection between $\omega$ and the countable set $\mathbb{Q}$ turns this into an injection $\mathbb{R}\to \mathcal{P}(\omega )$. Cantor-Schröder-Bernstein theorem now gives

Map $S\subseteq \omega$ to its indicator function $1_S:\omega \to \{0,1\}$. Its inverse maps a binary sequence $u$ to $\{n\in \omega :u(n)=1\}$. These maps are mutually inverse, soHere the left superscript denotes the set of functions from $\omega$ into $2$. It is a space of binary sequences, not an ordinal square.

The Cantor-Schröder-Bernstein theorem states that injections $f:A\to B$ and $g:B\to A$ imply the existence of a bijection $A\to B$. Here is the injection-chain formal proof, so no fixed-point theorem is needed. Defineand setThe second branch is defined because $a\notin C$ implies $a\notin A_0$, hence $a\in g(B)$, and the inverse there is unique by injectivity of $g$. Each branch is injective. A collision between branches would have $f(c)=g^{-1}(a)$ with $c\in C$ and $a\notin C$, implying $a=g(f(c))\in C$, a contradiction. Thus $h$ is injective.

For surjectivity let $b\in B$. If $g(b)\notin C$, then $h(g(b))=b$. If $g(b)\in C$, it is not in $A_0$, so it belongs to some $A_{n+1}=g(f(A_n))$. There is $a\in A_n$ with $g(b)=g(f(a))$; injectivity of $g$ gives $b=f(a)=h(a)$. This proves that $h$ is a bijection and completes the theorem without the axiom of choice.

For the comparison of cardinalities take the power set of $\omega$ as the reference set. Its cardinality is by definitionThe remaining parts give explicit injections or bijections comparing each requested set with this one.

This encoding works, interpreting the outer three-coordinate brackets by the nested ordered pair convention established in item 1. Equality of the outer ordered triples givesFor any fixed set $k$, the map $x\mapsto \{k,x\}$ is injective. If $x=k$, equality $\{k,x\}=\{k,y\}$ makes the right-hand side a singleton and forces $y=k$. If $x\ne k$, membership of $x$ in $\{k,y\}$ forces $x=y$. Applying this argument with $k=0,1,2$ recovers the respective coordinates.

The outer brackets must encode an ordered triple, rather than an unordered set of three doubletons. The source uses three-coordinate angle brackets here; using the successful convention of item 1 makes their meaning explicit and supplies a genuine construction.

This encoding works. Use three distinct tags $0,1,2$ and the ordered pair construction already proved injective. In the setexactly one member has first coordinate $0$, exactly one has first coordinate $1$, and exactly one has first coordinate $2$. Equality of two such sets therefore matches the members carrying each tag. Injectivity of the ordered pair construction then gives equality of $a$, of $b$ and of $c$, respectively.

This is a tagged encoding of a finite tuple. It remains valid when coordinates coincide with one another or with tags: the equality criterion for ordered pairs still distinguishes their first coordinates. Finite set formation using pairing and union makes the encoding a set.

This encoding fails. Choose two distinct sets $0$ and $1$, for example the first two finite von Neumann ordinals. The distinct ordered triples $(0,0,1)$ and $(0,1,1)$ both encode asIndeed the repeated singleton in the first encoding and the repeated two-element set in the second disappear when forming an ordinary set. Thus the middle coordinate cannot always be recovered. The issue is loss of multiplicity, not a failure to construct a set.

Use the Kuratowski ordered pairThe axiom of pairing first produces $\{a\}$ by pairing $a$ with itself, then $\{a,b\}$, and then the pair of these two sets. The axiom of extensionality supplies the usual uniqueness of each specified set. In particular, no use of Axiom schema of replacement is needed to construct a single ordered pair from two given sets.

To recover the coordinates from $P=⟨a,b⟩$, observe thatThe first equality recovers $a$. Once $a$ is known, the second recovers $b$: if its union is the singleton $\{a\}$ then $b=a$, and otherwise $b$ is its unique element different from $a$. Hence equality of two encoded pairs forces equality of both coordinates, including the degenerate case $a=b$.

The first ordered triple encoding works. Ifinjectivity of the ordered pair construction first gives $a=a^{\prime }$ and $⟨b,c⟩=⟨b^{\prime },c^{\prime }⟩$, and a second application gives $b=b^{\prime }$ and $c=c^{\prime }$. The construction uses only two successive ordered pair constructions, so is a set as well.

False. Take $\alpha =\beta =1$ and $\gamma =\omega$. ThenThus ordinal multiplication on the right does not in general distribute over ordinal addition. The side on which ordinal multiplication occurs matters; the left-distributive identity in the previous part does not imply this one.

True: ordinal multiplication on the left distributes over ordinal addition. If $\alpha =0$ both sides are zero. Otherwise fix $\alpha >0$ and $\beta$, and induct on $\gamma$. At zero the identity follows from $\alpha 0=0$. At a successor, use the recursive ordinal multiplication rule and the associativity just proved:For a nonzero limit ordinal $\lambda$, the ordinals $\beta +\xi$ are cofinal in $\beta +\lambda$. Continuity of ordinal multiplication in its right argument givesFor the last equality, $\alpha \xi$ is increasing and cofinal in $\alpha \lambda$, which is a limit ordinal because $\alpha >0$, and ordinal addition is continuous in its right argument. This establishes the limit step and the identity for all ordinals.

False. Use $\alpha =2$ and $\beta =\omega$. The recursive definition of ordinal multiplication givesThe first product concatenates countably many finite two-point blocks. The second concatenates two whole copies of the order type $\omega$, producing a strictly larger ordinal. Hence ordinal multiplication is not commutative.

False. Take $\alpha =1$ and $\beta =\omega$. By the limit clause for ordinal addition,The first order type is one point followed by a countably infinite sequence, which is still of type $\omega$. The second has a greatest element after that infinite sequence, so its order type is the successor ordinal $\omega +1$. Thus ordinal addition is not commutative, even though the two resulting sets have equal cardinality.

First give the recursive definitions, keeping the right argument as the recursion variable. For ordinals $\alpha ,\beta$ and nonzero limit ordinals $\lambda$,Here $\xi +1$ is the successor ordinal, and the supremum of a set of ordinals is its union. These are ordinal arithmetic, defined by transfinite recursion, and differ from cardinal sum and ordinal multiplication.

Associativity of ordinal addition is true. Fix $\alpha ,\beta$ and use transfinite induction on $\gamma$. The case $\gamma =0$ is immediate. In the successor step the recursive definitions and the induction hypothesis giveFor a nonzero limit ordinal $\lambda$, the ordinals $\beta +\xi$ for $\xi <\lambda$ form an increasing cofinal subset of the limit ordinal $\beta +\lambda$. The limit clause for ordinal addition, restricted to this cofinal subset, givesCofinal restriction is valid because ordinal addition by a fixed ordinal on the left is increasing and continuous in its right argument, directly from its recursive definition. This completes the induction.

Let $T=\mathrm{Th}(\mathbb{C})$ in the finite language of rings: it is the set of all first-order sentences true in the field of complex numbers. It is satisfiable, with first-order model $\mathbb{C}$, so the countable-language Löwenheim-Skolem theorem gives an at most countable first-order model $K$ of $T$.

All field axioms belong to $T$, hence $K$ is a field. For every positive integer $n$, the first-order sentence saying that there are at least $n$ distinct elements is also true in $\mathbb{C}$ and belongs to $T$. Thus $K$ cannot be finite, and is countably infinite.

Finally, for each first-order sentence $\sigma$, either $\sigma \in T$ or $\neg \sigma \in T$. Since $K$ satisfies all of $T$, it satisfies $\sigma$ exactly when $\mathbb{C}$ does. ThereforeElementary equivalence concerns first-order first-order sentences and does not imply an isomorphism; the two fields have different cardinalities. The stronger downward form may even choose $K$ as a countable elementary subfield of $\mathbb{C}$.

For a countable first-order language, the Henkin construction introduces only countably many constant symbols: each round needs at most one for each logical formula, there are countably many finite strings in a countable alphabet, and countably many rounds suffice. There are therefore at most countably many closed terms. The term model, and also its quotient set by equality, has at most countably many elements.

It follows that every satisfiable theory in a countable first-order language has an at most countable first-order model. If the theory has no finite first-order model, the constructed first-order model is countably infinite. More generally the same counting argument gives a first-order model of cardinality at most $\max (|L|,{\aleph }_0)$ for any satisfiable theory in language $L$.

For the elementary-substructure version of the downward Löwenheim-Skolem theorem, begin with a first-order structure $M$ and a set $S$ of designated elements. Close $S$, the named constant symbols and one initial element under the language's functions and under chosen witnesses in $M$ for existential formulas with parameters already chosen. Iterate countably often. The resulting set has cardinality at most $\max (|L|,|S|,{\aleph }_0)$, and its induced substructure $N$ is elementary: induction on logical formulas proves the truth equivalence, with the existential case supplied by the chosen witnesses. This is the same witness mechanism as the Henkin formal proof, now keeping witnesses inside a given first-order model.

In the presence of equality, two syntactically different closed terms can be asserted equal by the Henkin theory, so the raw first-order term domain from part (b) must be replaced by a quotient set. DefineThe equality axioms give reflexivity, symmetry and transitivity, so this is an equivalence relation. Their substitution consequences make it a congruence relation on a first-order structure: replacing an input by an equivalent first-order term preserves equivalence of function values and preserves membership of atomic formulas in $H$.

Take domain $\{[t]:t\text{ is a closed term}\}$. Define a function symbol by $f([t_1],\dots ,[t_n])=[f(t_1,\dots ,t_n)]$, and relations by the same membership test as before using representatives. The congruence relation on a first-order structure makes these definitions independent of representatives. Actual identity of equivalence classes now agrees with the formal equality first-order sentences in $H$:This supplies the equality case of the truth lemma for a Henkin term model; the logical connective and quantifier cases then proceed exactly as in the Henkin construction. Thus the term model built from the quotient set is a genuine first-order model with equality interpreted as identity.