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
|show 3 more comments|
There is no proof of the sort asked for in the question because the assertion to be proved isn't correct. A consistent system of axioms may well contain a false axiom. The negation of that false axiom would be a true statement which, in the system under consideration, leads to a contradiction.
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.
A more reasonable but more complicated example would be to take the usual ZFC axiom system for set theory (whose consistency I take for granted) and adjoin to it the false axiom that expresses "there is a formal proof of a contradiction in ZFC" (i.e., it expresses "ZFC is inconsistent"). Despite the false axiom, this system is consistent, by Gödel's second incompleteness theorem.
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.