Quick question on validity of arguments across propositional and predicate logic

Question

Could someone clarify a quick question regarding the following please?

From the text:

Assume Γ ⊭ ϕ. Then there is an L2-structure A such that all members of Γ are true in A and ϕ is false in A. Derive an L1-structure A′ from A by omitting the domain and the interpretations of constants and n-ary predicate letters for n ≥ 1. Since the satisfaction clauses of sentence letters and the propositional connectives of L2 are the same in L1 and L2, all members of Γ are true in A′ and ϕ is false in A′

I understand that arguments that are invalid in L1 are also invalid in L2 but why is the opposite true ?

Starting with a logically invalid L2-Argument with quantified sentences in the premisses say, how would removing interpretations of constants/n-ary predicate letters from the L2-structure account for the fact that L1 cannot deal with quantifiers?

Thanks!

Answer 1

The question is

Starting with a logically invalid L2-Argument with quantified sentences in the premisses say, how would removing interpretations of constants/n-ary predicate letters from the L2-structure account for the fact that L1 cannot deal with quantifiers?

L1 means propositional logic. L2 means predicate logic. A first order language contains logical symbols which when combined into well-formed formulas can be assigned a truth value. It also contains non-logical symbols which require an interpretation prior to assigning a truth value to a formula.

A structure is a way to specify an interpretation. Here is Wikipedia's description of structure and interpretation:

The most common way of specifying an interpretation (especially in mathematics) is to specify a structure (also called a model; see below). The structure consists of a nonempty set D that forms the domain of discourse and an interpretation I of the non-logical terms of the signature.

One can look at the formulas resulting from the interpretation as propositional forumlas that have been assigned a truth value. As an example, this tree proof generator will either generate a tree or provide a countermodel. Consider the following invalid argument:

This is a structure. The domain has only one element 0. P is an empty predicate since no value of the domain makes it true. Therefore, it is false. Q does have a value from the domain, 0, that makes it true. So Q, with the element 0, is true. The disjunction has the value true. The consequent is the predicate P again, but there is no value from the domain that makes P true. So the consequent is false.

If we ignored the domain and interpretation we could construct the following propositional logic result from this result:

Again we have P is false while Q is true giving a similar result. We can ignore the domain in this countermodel. This might be what the text is asking you to do when it says:

Derive an L1-structure A′ from A by omitting the domain and the interpretations of constants and n-ary predicate letters for n ≥ 1.

However, the textbook may be asking for a detailed proof not just an example and a suggestion why it should be true in general. If that is the case, then one should use the precise way the textbook defines language, L1 structure and L2 structure, and any already proven propositions relating these structures to provide a proof of the result. Also imitate the proofs given for those earlier propositions.

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Wikipedia contributors. (2019, June 27). First-order logic. In Wikipedia, The Free Encyclopedia. Retrieved 15:57, July 5, 2019, from https://en.wikipedia.org/w/index.php?title=First-order_logic&oldid=903705274

Tree Proof Generator. Retrieved on July 5, 2019 from https://www.umsu.de/logik/trees/