# How do we know that our definitions/axioms are not contradictory?

Let us assume that we have declared some axioms. Now, we wish to declare a new axiom too. How do we establish that the new axiom is not a consequence of, nor a contradiction to the original set of axioms?

Ditto for definitions. How do we know that the criteria are not contradictory?

For instance, when creating the set of real numbers axiomatically, our definition says that 1 is not equal to 0, where 1 is the multiplicative identity and 0 is the additive identity.

But, how do we know/prove that 1 not equal to 0 is not a consequence of the field axioms, nor is contradictory to the field axioms?

• Not an answer to the question but aside: $1 \ne 0$ is among the axioms of definition of any field. Commented Jun 8, 2015 at 18:02
• I don't think so. I think 1=0 would also yield a field. Just that such a field would be trivial. i.e: it would have exactly one element. Commented Jun 8, 2015 at 18:03
• @OnkarSinghGujral: That is not usually considered a "field", however. Commented Jun 8, 2015 at 18:06
• Search Google for "field with one element" and you'll see that it is not considered a field. The $1\neq 0$ axiom is fundamental. Commented Jun 8, 2015 at 18:22
• Is it possible to have a similar structure with $1=0$? Commented Jun 8, 2015 at 19:14

Let's take an example: the axioms of a group. Now we wish to declare a new axiom: all elements commute.

How do we establish that this new axiom is not a consequence of the old ones? By constructing a noncommutative group.

How do we establish that it is not in contradiction to the original ones? By constructing a commutative group.

Here are a couple of other examples, using exactly the same proof scheme.

The parallel postulate is neither a consequence of nor a contradiction to the other axioms of Euclidean geometry: construct the Cartesian coordinate plane with distances and angles etc., to show that it is not in contradiction; and construct the hyperbolic plane to show that it is not a consequence.

The axiom of choice AC is neither a consequence of nor a contradiction to the other axioms ZF of set theory: construct the set theoretic model of constructible numbers to show that AC is not in contradiction to ZF; use forcing to construct models which show that AC is not a consequence of ZF.

• Of course you are handwaving away here the step from exhibiting an obviously valid model of a noncommutative group with a friendly low number of six elements towads exhibiting a model of ZF with/without AC, where the foundational question of what really makes up a model comes into play ... Commented Jun 16, 2015 at 6:30
• @HagenvonEitzen: My intention is not to handwave, but to emphasize the unity of the proof scheme. Yes, the mathematics of this group theory problem is very, very simple; and the mathematics of this geometry problem is simple to us today although it stumped mathematicians for almost two millennia; and the mathematics of model theory is only a little more than a century old and is quite sophisticated next to the previous two examples. But even Cohen's little book on the independence of AC and of the continuum hypothesis emphasizes this proof scheme. Commented Jun 16, 2015 at 13:13

In a logical approach:

Assume we have a theory $T$ and its set of axioms $\Gamma$, such that $T=\{\varphi|\Gamma \vdash \varphi\}$ (a theory is closed under logical consequence). Now, for some reason, we want to postulate a new axiom to $T$, call it $\psi$. Clearly, we have a new axiom set $\Gamma'=\Gamma \cup \{\psi\}$ and so a new theory $T'=\{\varphi|\Gamma' \vdash \varphi\}$.

Your question is how to proof that:

1. $T \neq T'$
2. $T \cup \{\psi\} \nvdash \bot$

is the case, assuming that $T$ is consistent.

Then:

1. We need to establish an independence proof of $\psi$ in $T$, that is, we need to show that $T \nvdash \psi$. Mathematicians usually do this using models (see the completeness theorem), and the matter turns out to find an interpretation $\mathcal{M}$ of $T$ where $\psi$ is false. This shows that $T \nvdash \psi$ and that $T \subset T'$, since $T' \vdash \psi$.

2. The matter is how to give a consistency proof of $T \cup \{\psi\}$. In other words, this means that adding $\psi$ is consistent with the axiom set of $T$. Since $T$ is consistent, it suffices to show that $T \nvdash \neg\psi$. In this case, we need to show that there is a model $\mathcal{M}$ of $T$ where $\psi$ is true.

• Thanks @Bruno. But, this seems too complicated for me to understand. I've just done very introductory logic, sets. Is there any way by which you can simplify it a bit (Even if that means that some rigour will be lost)? Commented Jun 9, 2015 at 18:41
• @OnkarSinghGujral Roughly, a theory is the set of all sentences that can be proven from the theory axioms. Formally, we represent the provability relation as '$\vdash$' such that '$\Gamma \vdash \varphi$' means that "there is a proof of $\varphi$ from the set of sentences $\Gamma$" given a theory. Then, if $\Gamma$ is the set of axioms of a theory $T$, then $T=\{ \varphi | \Gamma \vdash \varphi \}$. And if we want to add a new axiom to this set $\Gamma$, we will have a new theory. Commented Jun 16, 2015 at 5:53
• @OnkarSinghGujral Now, to establish that the new axiom, say, $\psi$ is not a consequence of the others, we need to show that $\psi$ is independent in $T$. This is shown by demonstrating that $T$ does not prove $\psi$, i.e. $T\nvdash \psi$. (2) To establish that $\psi$'s introduction is not contradictory we must show that $T \cup \{ \psi \}$ is consistent, that is, that it does not contain a contradiction. Commented Jun 16, 2015 at 6:00