research36 min studyLesson 971 in Mathematics

Formal languages, structures, and satisfaction

Separate expressions from their interpretations and calculate quantified truth in a finite structure.

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What you’ll understand

  • Specify a first-order signature, its structures, and variable assignments.
  • Evaluate nested quantifiers by their witness dependencies.
  • Distinguish truth in one structure from validity and derivability.

A formula is a finite expression, while its truth depends on an interpretation. The same expression can hold in one structure and fail in another without any defect in logic. Keeping those two levels separate makes later completeness and incompleteness results intelligible. We will use a small cyclic structure where every function value can be checked, then ask which conclusions concern that structure, every structure, or a chosen formal proof system.

Fix the signature before interpreting an expression

A first-order signature declares constant, function, and relation symbols with fixed finite arities. Terms are built recursively from variables, constants, and function application. Atomic formulas assert relations between terms or their equality; connectives and quantifiers build larger formulas. A structure supplies a nonempty domain, an element for each constant, a total operation for each function symbol, and a relation for each relation symbol. Equality is interpreted as actual equality in the domain.

An assignment maps variables into the domain. Term evaluation follows the term's recursive construction, so a variable's current assigned value is different from its printed name. A sentence has no free variables, and its truth is independent of the assignment. Open formulas usually depend on assignments. Renaming a bound variable preserves meaning, but substituting a term requires avoiding variable capture: inserting a free variable beneath an unrelated quantifier can change its intended dependency.

M,s⊨R(t1,…,tk)  ⟺  (t1M,s,…,tkM,s)∈RM.\mathcal M,s\models R(t_1,\ldots,t_k)\iff\bigl(t_1^{\mathcal M,s},\ldots,t_k^{\mathcal M,s}\bigr)\in R^{\mathcal M}.
Atomic truth uses interpreted terms and the relation selected by the structure.↔ Scroll wide equations horizontally.

Quantifier order records which witness may depend on what

Existential satisfaction permits choosing one domain element, while universal satisfaction requires every domain element. In a universally quantified existential assertion, the existential witness may depend on the universal input. Reversing the order requires one witness that works simultaneously for all inputs. This distinction survives even in tiny finite structures, where checking a table is possible. It should be understood as a dependency statement rather than merely a rule for rearranging printed symbols.

For a domain of six residues, interpret f as addition of two modulo six and P as the three even residues. The operation is a permutation, so every element has a preimage, including the odd elements. Starting at zero and repeatedly applying f visits only the even residues. Reachability from a designated constant is therefore different from quantification over the whole domain. A single displayed orbit cannot by itself establish a claim about all domain elements.

M,s⊨∀x∃y φ  ⟺  for every a∈M there is b∈M:M,s[x↦a,y↦b]⊨φ.\mathcal M,s\models\forall x\exists y\,\varphi\iff\text{for every }a\in M\text{ there is }b\in M:\mathcal M,s[x\mapsto a,y\mapsto b]\models\varphi.
The choice of b may depend on a; reversing the quantifiers forbids that dependence.↔ Scroll wide equations horizontally.

Four different judgments must remain distinct

Satisfaction says that a structure meets specified sentences. Semantic consequence means every model of a theory satisfies a conclusion, and validity is consequence from no assumptions. Syntactic derivability instead asserts the existence of a finite proof in a fixed calculus. Syntactic consistency means no contradiction is derivable; satisfiability means some structure exists. Their equivalence for classical first-order logic needs the soundness and completeness theorems, rather than belonging to the definitions.

Even when a finite structure satisfies every sentence of a theory, those sentences need not describe the intended natural numbers or any other intended object uniquely. Conversely, an expression being false in the residue structure says nothing by itself about its derivability from unrelated arithmetic axioms. Explicitly name the language, premises, calculus, and model class whenever moving between these judgments. Later undecidability statements will additionally require an effective presentation of syntax, which an arbitrary signature need not possess.

T⊨σ  ⟺  ∀M (M⊨T⇒M⊨σ),T⊢σ  ⟺  a finite formal T-proof of σ exists.T\models\sigma\iff\forall\mathcal M\,(\mathcal M\models T\Rightarrow\mathcal M\models\sigma),\qquad T\vdash\sigma\iff\text{a finite formal }T\text{-proof of }\sigma\text{ exists}.
The left statement quantifies over structures; the right quantifies over finite syntactic objects.↔ Scroll wide equations horizontally.

A six-element structure separates two existential claims

Take domain {0,1,2,3,4,5}\{0,1,2,3,4,5\}, constant c interpreted as zero, f(x)=x+2(mod6)f(x)=x+2\pmod6, and P={0,2,4}P=\{0,2,4\}. Evaluate preservation of P, surjectivity of f, and the assertion that every element has a P-preimage.

  1. Compute the complete function table: the images of 0,1,2,3,4,5 are 2,3,4,5,0,1. The P-inputs 0,2,4 map to 2,4,0, all in P. Thus ∀x(P(x)→P(f(x)))\forall x(P(x)\to P(f(x))) holds.

  2. Every residue occurs exactly once in the image table. Given x, choose y=x−2(mod6)y=x-2\pmod6. This proves ∀x∃y(f(y)=x)\forall x\exists y(f(y)=x) in this particular structure, using a witness that depends on x.

  3. The P-inputs have only even images. For x equal to one, its unique preimage is five, which is outside P. Therefore ∀x∃y(P(y)∧f(y)=x)\forall x\exists y(P(y)\land f(y)=x) fails. One counterexample input suffices despite the successful orbit from c.

f(P)=P={0,2,4},f(M)=M,1∉f(P).f(P)=P=\{0,2,4\},\qquad f(M)=M,\qquad 1\notin f(P).
The restricted image and full image answer different quantified questions.↔ Scroll wide equations horizontally.
The result

Preservation and surjectivity are true; surjectivity restricted to P-preimages is false. These are model evaluations, not claims of logical validity.

Explore the model

Interactive lab

Interactive lab · modular

Compare a reachable orbit with the declared domain

Pause and Reset, then set modulus to 6 and additive step to 2. Predict orbit 0,2,4 with period three; take three Steps to return to zero. For a fresh run Reset and set both modulus 6 and step 1: the period becomes six. Each Step performs one addition. The orbit represents iterations from zero, not every variable assignment or an arbitrary first-order structure. Parameter edits retain the clock, so Reset before comparing runs.

Compare a reachable orbit with the declared domainAddition modulo m starts at zero: rₙ = na mod m. Playback advances two additions per second. The orbit has period m/gcd(m,a), including period one for a = 0. This is not a cryptographic security test or a primality algorithm. Current readings: Residue r0: 0; Orbit period · m/gcd(m,a): 12; Distinct residues visited: 1 of 12. All parameters are adjustable with the labeled controls below.Repeated addition modulo 12: the current residue is 0, and the orbit period is 12.01234567891011+ 5mod 12An additive orbitn = 0rₙ = 0Period = 12Cycle starts at 0:0 → 5 → 10 → 3 →8 → 1 → 6 → 11 →4 → 9 → 2 → 7 → 0Exact residues · two additions per second
Additive orbitCurrent residue / next edge
Residue r0
0
Orbit period · m/gcd(m,a)
12
Distinct residues visited
1 of 12
t = 0 s

Addition modulo m starts at zero: rₙ = na mod m. Playback advances two additions per second. The orbit has period m/gcd(m,a), including period one for a = 0. This is not a cryptographic security test or a primality algorithm. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forA finite reachable component can be a proper subset of the domain quantified over.

Check your understanding

In the six-element worked structure, why does every x have an f-preimage although the orbit from zero has only three elements?

Try it yourself

On residues modulo five, let f(x)=x+2(mod5)f(x)=x+2\pmod5 and P={0,2,4}P=\{0,2,4\}. Decide whether P is preserved and whether every x has an f-preimage. Then decide whether one y can satisfy f(y)=x for every x.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Check the image of four first. For the last sentence, keep the same y while comparing two different x values.

Compare with the solution

Preservation fails because f(4)=1 and one is outside P. The images are 2,3,4,0,1, so f is surjective and every x has the preimage x minus two modulo five. The sentence ∃y∀x(f(y)=x)\exists y\forall x(f(y)=x) is false: its chosen f(y) cannot equal both zero and one. Changing the modulus altered preservation but did not erase the difference between dependent and fixed witnesses.

Take it with you

  • A structure interprets symbols; an assignment supplies values for free variables.
  • Quantifier order specifies the permitted dependence of witnesses.
  • Satisfaction, validity, derivability, consistency, and satisfiability are distinct definitions.

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