TITLE: How do we know that our definitions/axioms are not contradictory?
QUESTION [7 upvotes]: 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?

REPLY [7 votes]: 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:

$T \neq T'$
$T \cup \{\psi\} \nvdash \bot$

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

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 $T' \vdash \psi$.
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.