2026-11-03

Chapter 3: First-Order Logic

Satisfaction and negation
- For an -structure and an assignment ,

$Misplaced & \begin{aligned} \mathcal M\models\varphi[b]&\iff\hat b(\varphi)=1,\\ \mathcal M\not\models\varphi[b]&\iff\mathcal M\models\neg\varphi[b]. \end{aligned}         $

- Without an assignment, $\mathcal M\models\varphi$ means that $\varphi$ is true under **every** $M$-assignment. Hence

$Misplaced & \mathcal M\not\models\varphi \iff\exists b:\mathcal M\models\neg\varphi[b].         $

	This is not the same as $\mathcal M\models\neg\varphi$, which requires $\neg\varphi$ to hold under every assignment.
- **Example:** In the usual structure of the natural numbers, neither $x=0$ nor $\neg(x=0)$ is true under every assignment: the first fails at $b(x)=1$, the second at $b(x)=0$.
- A [[Knowledge/Sentence (First-Order Logic)|sentence]] $\sigma$ has no free variables. Its truth value does not depend on the assignment, so $\mathcal M\models\sigma[b]$ for one assignment if and only if $\mathcal M\models\sigma$.

Validity and finite validity
- Fix a language . A formula is valid if it holds in every -structure, and finitely valid if it holds in every finite -structure:

$Misplaced & \begin{aligned} \models_{\mathcal L}\varphi &\iff\forall\mathcal M:\ \mathcal M\models\varphi,\\ \models_{\mathcal L}^{\mathrm{fin}}\varphi &\iff\forall\mathcal M\ \bigl(M\text{ finite}\implies\mathcal M\models\varphi\bigr). \end{aligned}         $

- Every valid formula is finitely valid. Fixing a language means using the same language throughout; it does not by itself mean that its alphabet is finite.
- For a fixed finite signature, finite **non-validity** is [[Knowledge/Semi-decidable Predicate|semi-decidable]]: enumerate finite structures and assignments to the finitely many variables occurring in $\varphi$, and halt if one falsifies it. Consequently, finite validity is co-semi-decidable.
- Whether the corresponding sets are decidable depends on the language. Validity in the equality-only language is decidable; undecidability results for finite validity require a sufficiently expressive signature. Enumerating finite structures alone does not decide finite validity when no counterexample exists.

Semantic entailment
- For a set of sentences,

$Misplaced & \Sigma\models_{\mathcal L}\varphi \iff\forall\mathcal M\ \bigl(\mathcal M\models\Sigma\implies\mathcal M\models\varphi\bigr), \qquad \mathcal M\models\Sigma\iff\forall\sigma\in\Sigma:\mathcal M\models\sigma.         $

- In particular, $\emptyset\models\varphi$ means $\models\varphi$: every structure satisfies the empty set of premises.

First-order tautologies
- Let map propositional formulas to first-order formulas while preserving Boolean connectives and truth constants, for example:

$Misplaced & h(A\land B)=h(A)\land h(B),\qquad h(\neg A)=\neg h(A),\qquad h(\top)=\top.         $

- If $A$ is a propositional tautology, then $h(A)$ is valid. For any fixed structure and assignment, the truth values of the formulas $h(p_i)$ give a propositional assignment; $A$ is true under that assignment.
- In this terminology, a first-order tautology is such an image of a propositional tautology. Every tautology is valid, but not every valid formula is a tautology in this narrower sense.
- **Examples:** $(\forall x\,P(x))\to(\forall x\,P(x))$ has the propositional form $p\to p$. The formula $(\forall x\,P(x))\to(\forall y\,P(y))$ is valid, but its validity depends on the meaning of the quantifiers rather than its propositional form alone. Also $\forall x\,(x=x)$ is valid because equality is interpreted as identity.

Equality and Leibniz axioms
- Equality is reflexive, symmetric and transitive:

$Misplaced & \begin{aligned} x&=x,\\ x=y&\to y=x,\\ x=y&\to(y=z\to x=z). \end{aligned}         $

- Equal objects can be substituted into relations and functions. For unary $P,f$ and binary $R$, examples are

$Misplaced & \begin{aligned} x=y&\to(P(x)\leftrightarrow P(y)),\\ x_1=x_2&\to\bigl(y_1=y_2\to(R(x_1,y_1)\to R(x_2,y_2))\bigr),\\ x=y&\to f(x)=f(y). \end{aligned}         $

	Analogous axioms apply to symbols of other arities. These are valid because equal arguments denote the same objects.

Modus ponens
- The inference rule takes and to :

$Misplaced & \frac{\varphi\qquad\varphi\to\psi}{\psi}\ _{\text{(MP)}}         $

- It preserves truth under the same structure and assignment. If $\hat b(\varphi)=1$ and $\hat b(\varphi\to\psi)=1$, then $\hat b(\psi)=1$: otherwise the implication would have value $1\to0=0$.
- Consequently, if $\Sigma\models\varphi$ and $\Sigma\models\varphi\to\psi$, then $\Sigma\models\psi$.

Formal derivations
- For a set of premises, means that is derivable from . This is a syntactic relation; is a semantic relation about structures.
- The derivable formulas form the smallest set containing all logical axioms and all formulas in and closed under modus ponens.
- Equivalently, a formal derivation from is a finite sequence where each is a logical axiom, belongs to , or follows by modus ponens from two earlier formulas. Then exactly when such a sequence ends in .
- Logical axioms are valid formulas specified by the calculus. Premises in need not be valid: they are assumed for this derivation. The axiom list introduced here is not yet complete.
Soundness and completeness
- Soundness: , provided the logical axioms are valid and the inference rule preserves truth.
- Proof by induction on the derivation: An axiom holds in every structure; a premise holds in every model of . If an entry follows by modus ponens, the induction hypothesis gives the two earlier entries in every model of , and modus ponens gives the new entry there as well.
- Gödel’s completeness theorem: For a complete first-order calculus, . Together with soundness,

$Misplaced & \Phi\vdash\psi\quad\Longleftrightarrow\quad\Phi\models\psi.         $

	This converse requires the complete calculus, not merely the partial axiom list introduced so far.

Quantifier axioms
- Distribution:

$Misplaced & \forall x\,(\varphi\to\psi)\to \bigl(\forall x\,\varphi\to\forall x\,\psi\bigr).         $

	If both universally quantified premises hold, then for each $m\in M$, $\varphi\to\psi$ and $\varphi$ hold under $b_{x\mapsto m}$, so $\psi$ holds there as well.
- **Generalisation axiom:** $\varphi\to\forall x\,\varphi$ when $x$ does not occur free in $\varphi$. Changing the value of $x$ cannot change the truth of $\varphi$.
- Logical axioms include the pure axiom instances and their universal closures: if $\theta$ is a logical axiom, then $\forall x\,\theta$ is also a logical axiom. This is not an unrestricted rule for quantifying arbitrary formulas derived from premises.
- **Existence axioms:** $\exists x\,\varphi\to\neg\forall x\,\neg\varphi$ and $\neg\forall x\,\neg\varphi\to\exists x\,\varphi$.

Example formal proof
- Write , and , where and are distinct. The following derivation uses the instance of the substitution axiom, whose general conditions remain to be specified:

$Misplaced & \begin{array}{rll} 1.&\forall y\,(A\to P(y))&\text{universally closed substitution instance}\\ 2.&\forall y\,(A\to P(y))\to(B\to C)&\text{distribution}\\ 3.&B\to C&\text{MP on 1, 2}\\ 4.&A\to B&\text{generalisation axiom; }y\text{ not free in }A\\ 5.&(B\to C)\to((A\to B)\to(A\to C))&\text{tautology}\\ 6.&(A\to B)\to(A\to C)&\text{MP on 3, 5}\\ 7.&A\to C&\text{MP on 4, 6} \end{array}         $

- A useful proof strategy is to handle the quantifiers first, then use a propositional tautology and modus ponens to combine the resulting implications. The abbreviations $A,B,C$ stand for entire formulas.

Substitution in terms
- For a variable and terms , write for the term obtained by replacing with . Recursively:

$Misplaced & \begin{aligned} c[x:=t]&=c,&x[x:=t]&=t,&y[x:=t]&=y\quad(y\neq x),\\ f(s_1,\ldots,s_k)[x:=t] &=f\bigl(s_1[x:=t],\ldots,s_k[x:=t]\bigr). \end{aligned}         $

- The term substitution lemma relates syntactic replacement to changing the assignment:

$Misplaced & \bar b\bigl(s[x:=t]\bigr) =\overline{b_{x\mapsto\bar b(t)}}(s).         $

	Evaluate $t$ under the original $b$, assign that value to $x$, and evaluate the original term $s$. The identity can be proved by induction on $s$.

Substitution in formulas
- Only free occurrences of the substituted variable are replaced. For and ,

$Misplaced & \begin{aligned} \varphi[x:=t]&=\exists y\,(y+y=z\cdot z+1),\\ \varphi[y:=t]&=\exists y\,(y+y=x). \end{aligned}         $

	The occurrences of $y$ are bound by $\exists y$ and are therefore not replaced.
- The intended substitution axiom is $\forall x\,\varphi\to\varphi[x:=t]$, but it is not valid for unrestricted textual replacement: a variable in $t$ may become bound after substitution. The admissibility condition and its repair are left for the next lecture.
- **Boundary case:** With $\varphi:=\exists y\,(x\neq y)$ and $t:=y$, naive substitution gives $\exists y\,(y\neq y)$. In a structure with at least two elements, $\forall x\,\exists y\,(x\neq y)$ is true while this result is false. The free $y$ in the substituted term has been captured by the quantifier.