# 8.4 Consistency Results

A consistency result explains that one formal theory does not lead to contradiction, usually relative to the consistency of another theory. In set theory, such results are especially important because many central principles cannot be proved or refuted from the standard axioms alone.

The typical form is a relative consistency statement. Instead of proving absolutely that a theory is consistent, one proves that if a base theory is consistent, then a stronger or different theory is also consistent. This is the natural form of many results in foundations, because Godel's second incompleteness theorem prevents sufficiently strong theories from proving their own consistency.

### Definition 8.67 (Consistency)

A formal theory $T$ is consistent if there is no sentence $A$ such that both:
$$
T \vdash A
$$
and:
$$
T \vdash ¬A
$$

Equivalently, $T$ is consistent if it does not prove a contradiction:
$$
T \nvdash \bot
$$

Here $\bot$ denotes a contradiction or absurdity.

This is a proof theoretic definition. It speaks about what can and cannot be derived from the axioms of the theory.

### Definition 8.68 (Relative Consistency)

Let $T$ and $S$ be formal theories. A relative consistency result from $T$ to $S$ is a statement of the form:
$$
\operatorname{Con}(T) \to \operatorname{Con}(S)
$$

This means that if $T$ is consistent, then $S$ is consistent.

Such a result does not prove $\operatorname{Con}(S)$ absolutely. It transfers consistency from one theory to another.

For example, the statement:
$$
\operatorname{Con}(\mathrm{ZF}) \to \operatorname{Con}(\mathrm{ZFC})
$$
means that if ZF is free of contradiction, then ZFC is also free of contradiction.

### Models and Consistency

Model existence provides a semantic way to prove consistency. If a theory has a model, then every theorem of the theory is true in that model, and therefore the theory cannot prove a contradiction.

### Proposition 8.69 (Models Give Consistency)

If a theory $T$ has a model, then $T$ is consistent.

Proof

Assume that $T$ has a model $\mathcal{M}$. This means that every axiom of $T$ is true in $\mathcal{M}$.

By soundness of first order logic, every theorem provable from the axioms of $T$ is true in every model of $T$, and in particular is true in $\mathcal{M}$.

Suppose for contradiction that $T$ is inconsistent. Then there is a sentence $A$ such that:
$$
T \vdash A
$$
and:
$$
T \vdash ¬A
$$

By soundness:
$$
\mathcal{M} \models A
$$
and:
$$
\mathcal{M} \models ¬A
$$

This is impossible, because a structure cannot satisfy both a sentence and its negation.

Therefore $T$ is consistent.

### Proposition 8.70 (Model Construction Gives Relative Consistency)

Let $T$ and $S$ be theories. Suppose that from any model of $T$ one can construct a model of $S$. Then:
$$
\operatorname{Con}(T) \to \operatorname{Con}(S)
$$

Proof

Assume that $T$ is consistent and suppose that the consistency of $T$ is witnessed by a model $\mathcal{M}$ of $T$.

By hypothesis, from $\mathcal{M}$ one can construct a model $\mathcal{N}$ of $S$.

By Proposition 8.69, since $S$ has a model, $S$ is consistent.

Thus:
$$
\operatorname{Con}(T) \to \operatorname{Con}(S)
$$

This is the common pattern behind many consistency results in set theory.

### Inner Models and Consistency

An inner model is a model of set theory located inside a larger universe. Inner models are useful because they allow one to prove that certain additional axioms are compatible with a base theory.

### Definition 8.71 (Inner Model)

An inner model is a transitive class $M$ such that:

1. $M$ contains all ordinals.

2. $M$ satisfies the axioms of ZF.

3. Membership in $M$ is the ordinary membership relation restricted to $M$.

The transitivity condition ensures that elements of sets in $M$ are also in $M$, so membership inside $M$ behaves like ordinary membership.

### Lemma 8.72

If $M$ is an inner model of ZF and $M$ satisfies a sentence $\varphi$, then:
$$
\operatorname{Con}(\mathrm{ZF}) \to \operatorname{Con}(\mathrm{ZF}+\varphi)
$$
provided the existence of $M$ can be established from any model of ZF.

Proof

Assume there is a model of ZF. By hypothesis, from this model one can construct an inner model $M$ satisfying ZF and also satisfying $\varphi$.

Then:
$$
M \models \mathrm{ZF}+\varphi
$$

By Proposition 8.69, the theory:
$$
\mathrm{ZF}+\varphi
$$
is consistent.

Therefore:
$$
\operatorname{Con}(\mathrm{ZF}) \to \operatorname{Con}(\mathrm{ZF}+\varphi)
$$

### The Constructible Universe and Consistency

The constructible universe $L$ gives the most important early example of an inner model construction. It is built inside any universe of ZF and satisfies ZF. Moreover, it satisfies the axiom of constructibility:
$$
V=L
$$

This gives a relative consistency result for $V=L$.

### Theorem 8.73 (Godel Constructibility Consistency Theorem)

If ZF is consistent, then:
$$
\mathrm{ZF}+V=L
$$
is consistent.

Proof

Assume ZF has a model. Inside that model, construct the constructible universe:
$$
L
$$

By the results of the previous section, $L$ is an inner model of ZF. Also, when the constructible hierarchy is computed inside $L$, every set of $L$ is constructible inside $L$. Hence:
$$
L \models V=L
$$

Therefore:
$$
L \models \mathrm{ZF}+V=L
$$

By soundness, the existence of such a model implies:
$$
\operatorname{Con}(\mathrm{ZF}+V=L)
$$

Thus:
$$
\operatorname{Con}(\mathrm{ZF}) \to \operatorname{Con}(\mathrm{ZF}+V=L)
$$

### Corollary 8.74

If ZF is consistent, then ZFC is consistent.

Proof

By Theorem 8.73, if ZF is consistent, then:
$$
\mathrm{ZF}+V=L
$$
is consistent.

Inside $L$, there is a definable global well order. Therefore $L$ satisfies the axiom of choice.

Thus:
$$
\mathrm{ZF}+V=L
$$
implies:
$$
\mathrm{ZFC}
$$

If the stronger theory $\mathrm{ZF}+V=L$ is consistent, then the weaker theory ZFC is also consistent.

Hence:
$$
\operatorname{Con}(\mathrm{ZF}) \to \operatorname{Con}(\mathrm{ZFC})
$$

### Corollary 8.75

If ZF is consistent, then:
$$
\mathrm{ZFC}+\mathrm{CH}
$$
is consistent.

Proof

Godel's work on the constructible universe shows that:
$$
V=L
$$
implies the continuum hypothesis:
$$
\mathrm{CH}
$$

By Theorem 8.73, if ZF is consistent, then:
$$
\mathrm{ZF}+V=L
$$
is consistent.

Since $V=L$ implies both the axiom of choice and the continuum hypothesis, every model of:
$$
\mathrm{ZF}+V=L
$$
is a model of:
$$
\mathrm{ZFC}+\mathrm{CH}
$$

Therefore:
$$
\operatorname{Con}(\mathrm{ZF}) \to \operatorname{Con}(\mathrm{ZFC}+\mathrm{CH})
$$

### Consistency of the Axiom of Choice

The preceding corollary shows that the axiom of choice cannot be disproved from ZF, assuming ZF itself is consistent.

More precisely, if ZF is consistent, then adding the axiom of choice does not create a contradiction.

This is one half of the independence picture for the axiom of choice. To show full independence, one also needs to show that if ZF is consistent, then:
$$
\mathrm{ZF}+¬\mathrm{AC}
$$
is consistent.

That second direction requires different methods, usually permutation models or forcing, and it is more advanced than the constructible universe argument.

### Definition 8.76 (Independence)

Let $T$ be a theory and let $\varphi$ be a sentence. We say that $\varphi$ is independent of $T$ if:

1. $T \nvdash \varphi$.

2. $T \nvdash ¬\varphi$.

Equivalently, if $T$ is consistent, then both:
$$
T+\varphi
$$
and:
$$
T+¬\varphi
$$
are consistent.

This equivalence is normally understood in the relative consistency sense.

### Example 8.77

Assuming ZF is consistent, the constructible universe shows:
$$
\operatorname{Con}(\mathrm{ZF}+\mathrm{AC})
$$

This means that ZF cannot prove:
$$
¬\mathrm{AC}
$$

If ZF did prove $¬\mathrm{AC}$, then every model of ZF would satisfy $¬\mathrm{AC}$. But the constructible inner model satisfies AC, assuming ZF has a model. Thus the constructible universe blocks any proof of $¬\mathrm{AC}$ from ZF.

### Consistency of the Continuum Hypothesis

The continuum hypothesis states:
$$
2^{\aleph_0}=\aleph_1
$$

Equivalently, every uncountable subset of the real numbers has the same cardinality as the real numbers or has cardinality at most countable, depending on the formulation used.

The constructible universe shows that CH is consistent with ZFC, assuming ZF is consistent.

### Theorem 8.78

If ZF is consistent, then:
$$
\mathrm{ZFC}+\mathrm{CH}
$$
is consistent.

Proof

This is the same result as Corollary 8.75, but we emphasize its significance.

Assume ZF is consistent. Then, by the constructibility theorem:
$$
\mathrm{ZF}+V=L
$$
is consistent.

Inside $L$, the axiom of choice holds, and the continuum hypothesis holds. Therefore:
$$
\mathrm{ZFC}+\mathrm{CH}
$$
has a model whenever ZF has a model.

Thus CH cannot be disproved from ZFC, provided ZFC itself is consistent.

### Why These Results Are Relative

One might ask why the results are stated as:
$$
\operatorname{Con}(\mathrm{ZF}) \to \operatorname{Con}(\mathrm{ZFC}+\mathrm{CH})
$$

rather than simply:
$$
\operatorname{Con}(\mathrm{ZFC}+\mathrm{CH})
$$

The reason is that consistency is itself a mathematical statement. Strong enough theories, such as ZF and ZFC, cannot prove their own consistency if they are consistent.

Thus set theory usually proves relative consistency statements. These results show that one theory is no more contradictory than another.

### Theorem 8.79 (Informal Form of Godel's Second Incompleteness Theorem)

A sufficiently strong consistent formal theory cannot prove its own consistency.

In particular, theories such as PA, ZF, and ZFC cannot prove their own consistency, assuming they are consistent.

This explains why consistency results in set theory are normally relative. One proves that if a base theory is consistent, then an extension is consistent.

### Conservative and Nonconservative Extensions

A theory $S$ extending a theory $T$ is conservative over $T$ for a class of sentences if every sentence of that class provable in $S$ is already provable in $T$.

Relative consistency is weaker than conservativity. If:
$$
\operatorname{Con}(T) \to \operatorname{Con}(S)
$$

then $S$ does not introduce contradiction, assuming $T$ had none. However, $S$ may prove new statements that $T$ cannot prove.

For example:
$$
\mathrm{ZFC}+\mathrm{CH}
$$
proves CH, while ZFC does not prove CH, assuming the usual independence results. Thus it is not conservative over ZFC for all set theoretic sentences.

### Definition 8.80 (Equiconsistency)

Two theories $T$ and $S$ are equiconsistent if:
$$
\operatorname{Con}(T) \leftrightarrow \operatorname{Con}(S)
$$

Equiconsistency means that each theory is consistent exactly when the other is consistent.

It does not mean that the theories prove the same sentences.

### Example 8.81

ZF and ZFC are equiconsistent.

One direction is immediate because ZFC contains all axioms of ZF. If ZFC is consistent, then ZF is consistent, since any contradiction provable from ZF would also be provable from ZFC.

The other direction follows from the constructible universe:
$$
\operatorname{Con}(\mathrm{ZF}) \to \operatorname{Con}(\mathrm{ZFC})
$$

Therefore:
$$
\operatorname{Con}(\mathrm{ZF}) \leftrightarrow \operatorname{Con}(\mathrm{ZFC})
$$

### Example 8.82

ZF and:
$$
\mathrm{ZF}+V=L
$$
are equiconsistent.

One direction is immediate because:
$$
\mathrm{ZF}+V=L
$$
extends ZF.

The other direction is Godel's constructibility theorem:
$$
\operatorname{Con}(\mathrm{ZF}) \to \operatorname{Con}(\mathrm{ZF}+V=L)
$$

Thus:
$$
\operatorname{Con}(\mathrm{ZF}) \leftrightarrow \operatorname{Con}(\mathrm{ZF}+V=L)
$$

### How to Read Consistency Results

A consistency result should be read carefully. The statement:
$$
\operatorname{Con}(\mathrm{ZFC}+\mathrm{CH})
$$
relative to ZF does not say that CH is true. It says that CH can be added without contradiction if the base theory is consistent.

Likewise:
$$
\operatorname{Con}(\mathrm{ZFC}+¬\mathrm{CH})
$$
does not say that CH is false. It says that the negation of CH can also be added without contradiction, assuming suitable background consistency.

When both directions are established, the conclusion is independence, not truth or falsity.

### Example 8.83 (A Typical Consistency Argument)

Suppose we want to show:
$$
\operatorname{Con}(T) \to \operatorname{Con}(T+\varphi)
$$

A common strategy is:

1. Start with a model $\mathcal{M}$ of $T$.

2. Define inside $\mathcal{M}$ a substructure or class $\mathcal{N}$.

3. Prove that $\mathcal{N}$ satisfies $T$.

4. Prove that $\mathcal{N}$ also satisfies $\varphi$.

5. Conclude that $T+\varphi$ has a model.

The constructible universe follows this pattern with:
$$
\mathcal{N}=L^\mathcal{M}
$$

where $L^\mathcal{M}$ is the constructible universe computed inside the model $\mathcal{M}$.

### Limits of Constructibility

Constructibility gives consistency results for principles that hold in $L$, such as AC and CH. It cannot show consistency for principles that fail in $L$.

For example, to show:
$$
\operatorname{Con}(\mathrm{ZFC}+¬\mathrm{CH})
$$

one needs a method that can build models where CH fails. The standard method is forcing.

Similarly, to show that the axiom of choice can fail over ZF, one needs techniques that produce models of ZF in which no global choice function exists.

These methods appear later in the study of forcing and permutation models.