Is there an axiom that prevents other axioms from contradicting each other?

Question

i.e. Does an axiom already exist, which prevents the addition of those new axioms which can contradict already existing axioms?

Also, who decides that something is an axiom?

Answer 1

No, adding more axioms can't remove a contradiction, it can only produce more contradictions.

Answer 2

The Law of Non-Contradiction is a fundamental axiom to logical reasoning, pioneered and formalized by Plato (I think), and logic is integral to mathematical and geometrical reasoning.

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The result is that we build a system of knowledge and calculi that is self-consistent... but not necessarily true. The truth of all of mathematics (and of science that rests on mathematics) is ultimately dependent on the truth of the axioms it rests on.

You can build up two mutually exclusive logics which are self-consistent within themselves, but not when you compare or try to integrate the two.

The point is, when you have two assumptions (axioms) that produce a contradiction... within a certain "world" of logic... we can conclude that one of the two, if not both, are incorrect.

Answer 3

All of mathematics operates in a "world" (e.g. Euclidean Geometry) defined by a set of axioms (e.g. the celebrated 5 axioms, including the Parallel Postulate), which are statements that are assumed to be true. All the of the investigations in that world boil down to deciding if certain statements can be derived logically from the axioms.

If an axiom is contradicted by others (i.e. the other axioms can derive the negation of the axiom), the entire theory would make no sense at all, and nobody discusses these theories.

Anyone can define their own world by setting their own axioms, and the commonplace worlds we see (e.g. Linear Algebra, Real Analysis, Number Theory etc) are axioms that have been chosen well to give rise to extremely interesting worlds.

Answer 4

In monotonic logic we have $ A \vDash B \Rightarrow A \cup A' \vDash B$. If we have contradiction then $A \vDash \bot$. Hence if we add axioms ($A' \neq \emptyset$) by monotonicity $A \cup A' \vDash \bot$. Therefore adding more would not allow to remove the contradiction.

While non-monotonic logic exists usually in formal systems a monotonic one is used.

Answer 5

Since nobody would be forced to actually use that purported No-Contradiction-Axiom in a proof, any proof of a contradiction $P\land\neg P$ would still be valid.

Or: Assume you have a model for axioms $\mathbf A_1, \mathbf A_2, \ldots $ and $\mathbf{AxiomOfNoContradiction}$. Then it is also a model of $\mathbf A_1, \mathbf A_2, \ldots $ alone.