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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?

Robert
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    Obviously not; see G's Incompl Th: IF consistent, PA is not complete. – Mauro ALLEGRANZA Feb 14 '20 at 11:03
  • @MauroALLEGRANZA: Well, the Incompleteness Theorems are not really "obvious" truths. –  Feb 14 '20 at 14:05
  • @user170039 - we are discussing Ch.41 of Tarski's book called "Consistency and completeness of a deductive theory; the decision problem" that ends (three pages) with "Certain profound studies have shown that these [arithmetical] theories are incomplete in a very deep sense: it will never be possible to construct a consistent and complete deductive theory which contains as its theorems all true sentences of arithmetic [ref. to Kurt Goedel in footnote]" and the OP is asking "Does consistency imply completeness?" – Mauro ALLEGRANZA Feb 14 '20 at 14:22
  • @MauroALLEGRANZA: Well, my point simply was that your answer to the question is not really "obvious" to a beginner in Logic (which I assumed OP to be). –  Feb 17 '20 at 16:46

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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.

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.

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.

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.

Natalie Clarius
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  • thank you for your answer, but I'm still a bit confused, does: "of any two contradictory sentences at least one sentence can be proved" mean "for any sentence x and y, if x contradicts y then at least one of them can be proved"? because if it does and there are not contradictions (so the theory is consistent) the antecedent "x contradicts y" is false and the conditional is true which means that every time the theory is consistent it would even be complete.. im so confused. – Robert Feb 14 '20 at 14:39
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    @Robert This is the confusion I'm trying to resolve in my third paragraph. – Natalie Clarius Feb 14 '20 at 14:41
  • There always are contradictions -- in fact, infinitely many of them. In the language. Not necessarily in the theory, but that doesn't matter. That's the whole point. – Natalie Clarius Feb 14 '20 at 14:43
  • I understand the meaning of complete, which means that every sentence that can be built from the terms of an axiomatic system can be either proved or disproved. this definition (that I've taken from another book) is easy to understand, the other one to me is the one that is confusing – Robert Feb 14 '20 at 14:46
  • No, your last comment shows that the definition of completeness is apparently what you did not understand. "there are no contradictions" can't happen -- because the definition is talking about the contradictions of the language, of which there are infinitely many, not the ones that are provable, of which there are none provided the theory is consistent. – Natalie Clarius Feb 14 '20 at 14:48
  • I'm sorry for my confusion, but this is my first logic book and i'm not so experienced, what do you mean with "language, not in the theory"? with theory you mean "primitive terms, axioms, and theorems (formulated from the axioms and theory preceding this theory)"? – Robert Feb 14 '20 at 14:52
  • Let us continue this discussion in chat. – Robert Feb 14 '20 at 14:54
  • How can the theory from your example exist? Won't we have logical axioms as theorems and also $\vdash p\to p$ by the tautology theorem? – Ansar Feb 21 '20 at 07:15
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    @Ansar No, not if we decide deliberately not to include these logical axioms. I can just make up a proof system with exactly one axiom, namely $\vdash p$, and no other axioms or rules (nothing says a proof system must have modus ponens). This proof system will do exactly what I outlined above. A proof system for classical logic will behave like you said, and classical propositional logic is not complete; but this is just one out of many possible systems, and the point of my example was to demonstrate that in general it is possible to have a proof system which has the behavior outlined above. – Natalie Clarius Feb 21 '20 at 13:38
  • Oh, I see. So we are not necessarily talking about first-order theories, as I was assumed. – Ansar Feb 21 '20 at 13:52
  • Not necessarily, but there are also first-order theories which are consistent but incomplete. Peano arithmetic is the straightforward example here (at least, it is widely believed to be consistent, an proven to be incomplete). – Natalie Clarius Feb 21 '20 at 14:11
  • Yes, I understand. Also, isn't PA proven to be consistent since it has a model? – Ansar Feb 21 '20 at 17:48
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    @Ansar See https://math.stackexchange.com/q/2229914/344246. – Natalie Clarius Feb 21 '20 at 21:49
  • (If you have more questions, consider opening a new post, rather than comments on an existing answer.) – Natalie Clarius Feb 21 '20 at 21:50