Does consistency imply completeness?

Question

A deductive theory is consistent if no two asserted statements of this theory contradict each other.

A theory is called complete if of any two contradictory sentences at least one sentence can be proved within this theory.

I could (I guess) rewrite the definition as: "A theory is called complete ↔️ if there are any two contradictory sentences, then at least one sentence can be proved within this theory."

So, by my understanding, if a theory is consistent then there are no statements that contradict each other, and if there are not statements that contradict each other, the antecedent of the conditional "if there are any two contradictory sentences, then at least one sentence can be proved within this theory" is false, which means that the sentence is true, which means that it is consistent. Is this correct? if not how can a theory be consistent and incomplete?

Answer 1

No.

Consider a silly theory which proves $\vdash p$ for every atomic propositional letter $p$ and nothing else.

Clearly, this theory is consistent, because without negative formulas, there is no way a contradiction could be derived.

However, it is also not complete, because we have e.g. neither $\vdash p \to p$ nor $\vdash \neg (p \to p)$, because there is no axiom or rule that allows to introduce complex formulas.

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The definitions could be re-written as follows:

A theory is consistent
$\Leftrightarrow$ For any two contradictory statements of the language, not both are in the theory.
$\Leftrightarrow$ For any statement $\phi$, if $\vdash \phi$ then $\nvdash \neg \phi$ and if $\vdash \neg \phi$ then $\nvdash \phi$.

A theory is complete
$\Leftrightarrow$ For any two contradictory statements of the language, at least one is in the theory.
$\Leftrightarrow$ For any statement $\phi$, if $\nvdash \phi$ then $\vdash \neg \phi$ and if $\nvdash \neg \phi$ then $\vdash \phi$.

The first does not entail the second: "$\nvdash \phi$ and $\nvdash \neg \phi$" for some statement $\phi$ (such as $\phi := p \to p$ above) is compatible with consistency, but not completeness.

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The problem with your reasoning is here:

"if there are any two contradictory sentences"

This means any two contradictory sentences in the language, not in the theory. The antecedent is not restricted to those statements that can already be proven -- then, if the theory is consistent, the implication would indeed be vacuously true -- but to all conceivable statements that are well-formed formulas of the language -- and there are infinitely many contradictory pairs of well-formed statements $\phi, \neg \phi$ that make the antecedent true.

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There is the notion of maximal consistency

A theory is maximally consistent
$\Leftrightarrow$ the theory is consistent and every proper superset of the theory is inconsistent (i.o.w., adding any more statements to the theory would render it inconsistent)

which does entail completeness.