Model Theory Question for Predicate Logic in van Dalen's Logic and Structure. Let L be a language without identity and with at least one constant. Let sigma = exists x1 ··· xn phi (x1,...,xn), where sigma is a formula in predicate logic.

niouzesto

niouzesto

Answered question

2022-09-04

Model Theory Question for Predicate Logic in van Dalen's Logic and Structure
Let L be a language without identity and with at least one constant.
Let σ = x 1 · · · x n ϕ ( x 1 , . . . , x n ), where σ is a formula in predicate logic.
Let = { ϕ ( t 1 , . . . , t n ) | t i closed in L}, where ϕ is quantifier free.
I want to show that:
(i) σ each structure A is a model of at least one sentence in Σ. (Hint: for each A, look at the substructure generated by .)
(ii) Consider as a set of propositions. Show that for each valuation v (in the sense of propositional logic) there is a model A such that v ( ϕ ( t 1 , . . . , t n ) ) = the interpretation of ϕ ( t 1 , . . . , t n ) in A, for all ϕ ( t 1 , . . . t n ) .

Answer & Explanation

ko1la2h1qc

ko1la2h1qc

Beginner2022-09-05Added 18 answers

Step 1
(i) Given a model A, the substructure A 0 generated by is the set of (the A-interpretations of) all terms that only involve constant symbols (these are the 'closed terms' occuring in the definition of Σ).
For example, if the language contains 1 as a constant symbol and + as a binary operation symbol, then ( ( 1 + 1 ) + 1 ) + ( 1 + 1 ) is a closed term. In the usual model on N, this term is interpreted as 5 N .
By hypothesis, A 0 satisfies σ, so there are elements t i A 0 such that φ ( t 1 , , t n ) is true.
But the elements of A 0 all belong to some closed terms, i.e. there are closed terms t i that are interpreted exactly as t i . Note that the interpretation of constants and operations on A 0 is inherited from A, hence
A φ ( t 1 , , t n ) .
Step 2
(ii) Here we introduce a propositional variable P σ for each σ Σ.
Let's assume that a valuation v : { P σ : σ Σ } { , } is given.
We need a first order model A such that A σ v ( P σ ) = .
The trick is that we can naturally define the constants and operations on the syntactic set C of closed terms. We still need to interpret the relation symbols on C to obtain a model.
Since the language contains no = sign, its quantifier-free formulas are built up by the relation symbols of the language and Boolean connectives.One typically proceeds by formula induction on φ.
If φ = R ( t 1 , , t n ) with t i C and R an n-ary relation symbol, then we can simply interpret R in C according to the given valuation v.

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