Skip to content

2.5 Validity and Entailment

Validity, semantic entailment, satisfiability, countermodels, and logical consequence in first order logic.

Validity and entailment are semantic notions that compare formulas across all structures, rather than inside one fixed structure. Satisfaction tells us whether a formula is true in a particular structure under a particular assignment, while validity and entailment ask whether truth is forced in every possible interpretation of the language.

Validity

A sentence is valid if it is true in every structure for its language.

Definition 2.41 (Validity)

Let AA be a sentence. We say that AA is valid if:

MA \mathcal{M} \models A

for every structure M\mathcal{M} for the language of AA.

In this case we write:

A \models A

Validity means that the truth of AA depends only on its logical form, not on any special property of a particular structure.

Example 2.42

The sentence:

x(x=x) \forall x \, (x = x)

is valid, because equality is reflexive in every structure.

The sentence:

xP(x) \forall x \, P(x)

is not valid, because we may choose a structure in which the predicate PP does not hold for every element of the domain.

For example, let the domain be:

M={0,1} M = \{0,1\}

and let:

PM={0} P^{\mathcal{M}} = \{0\}

Then:

M⊭xP(x) \mathcal{M} \not\models \forall x \, P(x)

because 11 is not in PMP^{\mathcal{M}}.

Countermodels

To show that a sentence is not valid, it is enough to give one structure in which the sentence is false. Such a structure is called a countermodel.

Definition 2.43 (Countermodel)

Let AA be a sentence. A countermodel to AA is a structure M\mathcal{M} such that:

M⊭A \mathcal{M} \not\models A

Countermodels are important because they provide concrete evidence that a sentence does not follow from logic alone.

Semantic Entailment

Entailment generalizes validity by allowing assumptions.

Definition 2.44 (Semantic Entailment)

Let Γ\Gamma be a set of sentences, and let AA be a sentence in the same language.

We say that Γ\Gamma semantically entails AA if every model of Γ\Gamma is also a model of AA.

In this case we write:

ΓA \Gamma \models A

Equivalently:

ΓA \Gamma \models A

means that for every structure M\mathcal{M}, if:

MB \mathcal{M} \models B

for every BΓB \in \Gamma, then:

MA \mathcal{M} \models A

Thus AA is a semantic consequence of Γ\Gamma exactly when AA is true in every structure where all assumptions in Γ\Gamma are true.

Example 2.45

Let Γ\Gamma contain the two sentences:

xy(x<y¬(y<x)) \forall x \, \forall y \, (x < y \to ¬(y < x))

and:

xyz((x<yy<z)x<z) \forall x \, \forall y \, \forall z \, ((x < y \land y < z) \to x < z)

The first sentence says that << is asymmetric, and the second sentence says that << is transitive.

Then:

Γx¬(x<x) \Gamma \models \forall x \, ¬(x < x)

To see this, let M\mathcal{M} be any model of Γ\Gamma, and suppose for contradiction that there is an element aa such that:

a<a a < a

By asymmetry, from:

a<a a < a

we obtain:

¬(a<a) ¬(a < a)

This is impossible, so no such aa exists, and therefore:

Mx¬(x<x) \mathcal{M} \models \forall x \, ¬(x < x)

Since M\mathcal{M} was arbitrary, the entailment holds.

Validity as Entailment from No Assumptions

Validity is a special case of entailment.

For a sentence AA:

A \models A

means the same thing as:

A \varnothing \models A

This says that every structure satisfying all sentences in the empty set also satisfies AA. Since every structure satisfies the empty set of assumptions, this is exactly the statement that AA is true in every structure.

Logical Equivalence

Two sentences are logically equivalent if they entail each other.

Definition 2.46 (Logical Equivalence)

Let AA and BB be sentences. We say that AA and BB are logically equivalent if:

AB A \models B

and:

BA B \models A

Equivalently:

AB \models A \leftrightarrow B

This means that AA and BB have the same truth value in every structure.

Example 2.47

The sentences:

¬xP(x) ¬\forall x \, P(x)

and:

x¬P(x) \exists x \, ¬P(x)

are logically equivalent.

Indeed, a structure satisfies:

¬xP(x) ¬\forall x \, P(x)

exactly when not every element satisfies PP, and this holds exactly when some element fails to satisfy PP.

Thus:

¬xP(x)x¬P(x) \models ¬\forall x \, P(x) \leftrightarrow \exists x \, ¬P(x)

Satisfiability

A set of sentences is satisfiable if it has at least one model.

Definition 2.48 (Satisfiability)

Let Γ\Gamma be a set of sentences. We say that Γ\Gamma is satisfiable if there exists a structure M\mathcal{M} such that:

MΓ \mathcal{M} \models \Gamma

If no such structure exists, then Γ\Gamma is unsatisfiable.

Satisfiability asks whether the assumptions can all be made true together in some structure.

Example 2.49

The set:

Γ={xP(x),x¬P(x)} \Gamma = \{\exists x \, P(x), \forall x \, ¬P(x)\}

is unsatisfiable.

The first sentence says that at least one object satisfies PP, while the second sentence says that no object satisfies PP.

No structure can make both sentences true at the same time.

Lemma 2.50 (Entailment and Unsatisfiability)

Let Γ\Gamma be a set of sentences, and let AA be a sentence. Then:

ΓA \Gamma \models A

if and only if:

Γ{¬A} \Gamma \cup \{¬A\}

is unsatisfiable.

Proof

Suppose first that:

ΓA \Gamma \models A

Then every structure satisfying all sentences in Γ\Gamma also satisfies AA.

Therefore no structure can satisfy all sentences in Γ\Gamma and also satisfy ¬A¬A, because satisfying ¬A¬A means that AA is false.

Hence:

Γ{¬A} \Gamma \cup \{¬A\}

is unsatisfiable.

Conversely, suppose that:

Γ{¬A} \Gamma \cup \{¬A\}

is unsatisfiable.

Let M\mathcal{M} be any structure such that:

MΓ \mathcal{M} \models \Gamma

If M⊭A\mathcal{M} \not\models A, then:

M¬A \mathcal{M} \models ¬A

and therefore:

MΓ{¬A} \mathcal{M} \models \Gamma \cup \{¬A\}

This contradicts the assumption that Γ{¬A}\Gamma \cup \{¬A\} is unsatisfiable.

Thus every model of Γ\Gamma satisfies AA, and so:

ΓA \Gamma \models A

Example 2.51

We show:

{pq,p}q \{p \to q, p\} \models q

using unsatisfiability.

It is enough to show that:

{pq,p,¬q} \{p \to q, p, ¬q\}

is unsatisfiable.

Suppose a valuation satisfies all three formulas. Then:

p p

is true, and:

q q

is false.

But if pp is true and qq is false, then:

pq p \to q

is false.

This contradicts the assumption that:

pq p \to q

is true.

Therefore no valuation satisfies all three formulas, and hence:

{pq,p}q \{p \to q, p\} \models q

Entailment with Free Variables

For formulas with free variables, entailment is defined using both structures and assignments.

Let Γ\Gamma be a set of formulas and let AA be a formula. We write:

ΓA \Gamma \models A

if for every structure M\mathcal{M} and every assignment ss, whenever:

M,sB \mathcal{M}, s \models B

for every BΓB \in \Gamma, then:

M,sA \mathcal{M}, s \models A

This definition extends the earlier definition for sentences, since sentences do not depend on assignments.

Example 2.52

The entailment:

x<y,y<zx<z x < y, \quad y < z \models x < z

does not hold in all structures with an arbitrary binary relation <<.

It becomes valid if transitivity is included as an assumption:

xyz((x<yy<z)x<z),x<y,y<zx<z \forall x \, \forall y \, \forall z \, ((x < y \land y < z) \to x < z), \quad x < y, \quad y < z \models x < z

This example shows that entailment depends on the assumptions that have been included, and that relation symbols have no fixed meaning unless axioms constrain their behavior.