# Logic proof help

Can someone give me a proof that,

No true claim can derive a contradiction in a consistent system of axioms,

With out using a proof by contradiction

-
How do you define "a consistent system of axioms"? –  Daniel Fischer Jul 15 '13 at 14:03
A list of axioms, that can't derive two contradictory statements –  user758575 Jul 15 '13 at 14:05
Okay, and what's the definition of a "true claim"? (I'm not trying to be a jerk here, just trying to ascertain what's what.) –  Daniel Fischer Jul 15 '13 at 14:07
True and false are semantic properties of sentences in a given structure (or a class of those). Deriving contradictions and consistency are syntactical properties. –  Asaf Karagila Jul 15 '13 at 14:09
Please let's remain civil, @RGB. –  vadim123 Jul 15 '13 at 14:31

"No true claim can derive a contradiction in a consistent system of axioms."

What does this mean? That given a consistent set of axioms as background assumptions, and presumably some logic to do some derivations with, there is no derivation that leads from a true claim to a contradiction??

But that is just false. Take the single axiom $P$ [if you like, "the moon is made of green cheese"] and let standard propositional logic be your derivation system. Then we have a trivially consistent set of axioms. But, equally trivially, the proposition $\neg P$ (given the background axiom) leads immediately to contradiction. Yet $\neg P$ [on the given interpretation] is true!

The point is a straightforward one: if the axioms are themselves unsound, not true on a given interpretation, then you have a theory which can lead from a true additional premiss to a false conclusion, even a necessarily false one.

-
A trivial example of a consistent system of axioms containing a false axiom is the system consisting of the single axiom $2+2=5$. As long as you don't have other axioms available to deduce the correct facts of arithmetic, $2+2=5$ is consistent.