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Completeness and Expressiveness. Consistency: a syntactic definition, related to the proof system. A set of formulas is consistent if there is no formula.

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Presentation on theme: "Completeness and Expressiveness. Consistency: a syntactic definition, related to the proof system. A set of formulas is consistent if there is no formula."— Presentation transcript:

1 Completeness and Expressiveness

2 Consistency: a syntactic definition, related to the proof system. A set of formulas is consistent if there is no formula such that |-- and |--( F). Equivalent definition: |-/-F. Equivalently: is inconsistent if for each, |--.

3 Lemma |-- iff {( F)} is inconsistent. Proof: if {( F)} is inconsistent then {( F)} |--. Also, { } |--.(why?) Therefore, by proof by cases, |--. Conversely, if |-- then from {( F)} we can prove F (using MP), thus it is inconsistent. Consequence: |-/- iff {( F)} consistent.

4 Claim (without proof) Our first order logic proof system is sound and complete. That is: |-- if and only if |== (What is the meaning of |== ?)

5 Therefore: |-- if and only if there is some finite 0 such that 0 |-- (why?) Thus: |== if and only if there is some finite 0 such that 0 |== (why?)

6 More connections |== iff {( F)} is unsatisfiable (why?) Thus: {( F)} is unsatisfiable iff |== iff |-- iff {( F)} is inconsistent. If is consistent then either {( F)} or { } is consistent. Proof: if both are inconsistent, then we can prove from either {( F)} or { }. Hence, we can prove using proof by cases, from. Contradiction.

7 More connections In a similar way: { } is unsatisfiable iff |==( F) iff |--( F) iff { } is inconsistent. Assuming soundness and completeness: is inconsistent iff |--F iff |==F iff is not satisfiable. Thus: given soundness and completeness of the proof system: consistency = satisfiability!

8 Suppose we proved that is consistent is satisfiable We want to prove from this completeness, i.e., |== |--. Proof:By contraposition! |-/- implies {( F)} is consistent, implies {( F)} is satisfiable, implies |=/=

9 Suppose we proved is satisfiable is consistent |-- {( F)} inconsistent {( F)} unsatisfiable |== = Soundness!!!

10 In short: consider the two sentences: 1. A set of first order formulas is satisfiable iff it is consistent. 2. The proof system we gave is sound and complete. … are two manifestations of the same thing!!!

11 We will not show completeness here! One principle is the construction of a maximal consistent structure, as in propositional logic (add a formula or its negation). But there is another problem: what is the interpretation of terms? Solution: initially we interpret a(b(c),d) as the string a(b(c),d). What happens if we have xR(x) (actually (( R(x) F) F)? We invent new constant c x and add R(C x ) to the set of formulas. What happens when we have t 1 =t 2 ? We actually need to interpret a(b(c),d) as the equivalence class of terms according to = that includes a(b(c),d) (denoted [a(b(c),d)]). And we need to show that we maintain a consistent structure after adding all that …

12 We can learn important things from completeness. |-- iff there is a finite subset 0, 0 |--. Thus, |== iff there is a finite subset 0, 0 |==. |--F i.e., is inconsistent iff there is a finite subset of that is inconsistent. is consistent iff every finite subset of it is consistent! --- compactness theorem.

13 We can express some classes of structures (why not sets?) 1. xR(x,x) 2. x y(R(x,y) R(y,x)) 3. x y z ((R(x,y)/\R(y,z)) R(x,z)) We can write this using one formula! This defines the structures with R being an equivalence relation!

14 Provability Suppose we have a class of structures C and we want to prove something about this class. Suppose we can write a set of formulas such that a structure S is in that class iff S satisfies all the formulas in (with having no free variables). Then every property of the structures in C satisfies that |==. Proof: every structure that satisfies all the formulas in is in C and there are supposed to satisfy. Since we have completeness, |==, thus having allows us to prove.

15 A class of structures defines using an infinite set of elements Let i be a sentence that says that we have at least i elements: 2 = x y (x=y), 3 = x y z (( (x=y)/\ (y=z))/\ (x=z)) … etc. Call these collection of sentences.

16 Some inexpressiveness of first order logic We want to say that we have a finite number of elements in our domain. Suppose (for the contradiction) that we say exactly that using. Now, take and any subset of. It is satisfiable (by a structure with at least as many element as the highest index in the subset. Thus, every subset of is also satisfiable. By compactness, is also satisfiable – but it has an infinite model!!! Contradiction.

17 Having an algorithm for doing the proof? If |== then by completeness |--. We can enumerate all the possible proofs from. Say, each time to the i th step. Given enough memeory and time, if there is a proof, it would come out (recursive enumerable). On the other hand, one can write a set of formulas describing the behavior of a model equivalent to Turing machines, and a formula describing termination. Therefore, proving |-- is not decidable.

18 The big picture For some classes of structures C we can find a set of first order formulas that are satisfied exactly by structures in C. For some other structures C, there cannot be such a set of formulas (even an infinite set). Inexpressiveness of FOL. In case we can find the set of formulas and is a property of C, then we can use our (complete) proof system to prove |--. This will imply that is a property of C. But while the proof exists, we do not have a decision procedure that will check whether |-- and there is no limit on the amount of time and space needed for reaching such a proof when it exists.


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