2

Provide an example of two sentences , the first formally implying the second, the second analytically(but not formally) implying the first. Are two such sentences formally equivalent? Are they analytically equivalent?

This is what I have 1) The White house is White. 2) The Obama family lives in the White house only if it is White.

1

I'll prove the general claims and leave you with the task of finding examples. What I said earlier about analyticity being relative to a set of meaning postulates is to be inherited in this discussion as well. To keep the models very simple, we're going to assume that logical truths are the same as tautologies.

Definition 1. (Logical Truth) Sentence φ in language L is logically true iff every truth-assignment to the sentential symbols in φ makes φ true (i.e. φ is assigned ⊤ by all rows of the truth-table).

Definition 2. (Analyticity) Sentence φ in language L is analytically true with respect to meaning postulates Π iff (Π → φ) is logically true in L.

Let's abbreviate "φ is logically true" as L(φ) and "φ is analytically true" as A(φ). Making use of those abbreviations, we want to decide whether the following two claims are true or false:

Claim 1.φ, ψ, Π : [ L(φ → ψ) ∧ A(ψ → φ, Π) ∧ ¬L(ψ → φ) ] → L(φ ↔ ψ).

Claim 2.φ, ψ, Π : [ L(φ → ψ) ∧ A(ψ → φ, Π) ∧ ¬L(ψ → φ) ] → A(φ ↔ ψ, Π).

Both claims begin with the same (pretty ugly) antecedent:

(Δ)                                           L(φ → ψ)   ∧   A(ψ → φ, Π)   ∧   ¬L(ψ → φ)

which says of a given arbitrary sentences φ, ψ and a meaning postulate Π, that: (ψ is a logical consequence of φ), (φ is a Π-analytic consequence of ψ), and it's not the case that (φ is a logical consequence of ψ). Given the (Definitions 1–2), we can get rid of analyticity by defining it in terms of L:

(Δ)                                         L(φ → ψ)   ∧   L(Π → (ψ → φ))   ∧   ¬L(ψ → φ)

The general form of what we want to prove/disprove is now easier to see:

Claim 1.φ, ψ, Π : Δ → L(φ ↔ ψ).

Claim 2.φ, ψ, Π : Δ → L(Π → (φ ↔ ψ)).

Fact 1. Claim (1) is false.

Proof. We want to show that given arbitrary sentences φ, ψ and postulates Π, it is possible to satisfy Δ but falsify L(φ ↔ ψ). To satisfy L(φ ↔ ψ) it would be necessary to satisfy L(φ → ψ) and L(ψ → φ), but Δ contains the conjunct ¬L(ψ → φ), so if Δ is satisfied then L(ψ → φ) will be falsified. Since by hypothesis Δ is satisfied, we know that ¬L(ψ → φ) is satisfied, so L(ψ → φ) is falsified, and therefore L(φ ↔ ψ) is falsified. Since φ, ψ, and Π were arbitrary, Claim (1) must be false.                                  ■

Fact 2. Claim (2) is true.

Proof. Suppose arbitrary sentences φ, ψ and meaning postulates Π are such that Δ is true. We want to show that L(Π → (φ ↔ ψ)) is then also true. L(Π → (φ ↔ ψ)) is true just in case L(Π → (φ → ψ)) and L(Π → (ψ → φ)) are true. Luckily both are implicit in Δ! The first conjunct of Δ is L(φ → ψ), which is logically equivalent to L(Π → (φ → ψ)). The second conjunct of Δ is L(Π → (ψ → φ)), which we take as it is and combining with the previous result, obtain the required conclusion: L(Π → (φ ↔ ψ)). And since φ, ψ, and Π were arbitrary, we know that Claim (2) is generally true.                                         ■

As usual, feel free to leave a comment if something, especially the notation, isn't clear.

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.