# Can a valid argument with formally consistent premises have an analytically impossible conclusion? What about the converse?

Can a valid argument with formally consistent premises have an analytically impossible conclusion? What about the converse; analytically compossible premises, but a conclusion that is a formal contradiction? ( Compossible = possible to be jointly true)

Okay I think the answer is yes, something can be formally consistent but analytically impossible , reasoning because formal consistency is not analyzing, just truth values.

The converse im not sure about , the lack of information I have to solve this stuff is frustrating thanks for the help.

• The word valid means something in logic. Valid necessarily means consistent with the conclusion so it is impossible for the conclusion to be false once the premises are true. By definition you can't have validity with inconsistencies. – Logikal Dec 5 '17 at 1:25
• What do you mean by analytic. This word may differ between philosophy and mathematics. What you sound like you are expressing is a conclusion that is self contradictory which is analytically impossible to be true in some circles. The complement to self contradictory premises is logically necessary which are impossible to be false. – Logikal Dec 5 '17 at 1:27

In an earlier answer I suggested a Carnapian definition of analyticity, which was the following:

Definition 1. (Analyticity) Sentence φ (in language L) is analytically true with respect to meaning postulates Π iff (Π → φ) is logically true.

1. This definition of analyticity is a relative one: a sentence is analytic "with respect to a set of meaning postulates", not analytic simpliciter. For example, the sentence (1 + 1 = 2) is not analytic simpliciter, but is analytic with respect to, say, Peano's Axioms. So we always need to qualify which set of postulates (e.g. Peano's Axioms) makes a sentence in a language analytically true.

2. The definition relies on the notion of logical truth, which is also a relative notion but in a different sense. If L is a propositional language, then logical truth will coincide with truth under all truth-assignments to the sentential symbols; if L is a first-order language, then logical truth will coincide with the full-fledged Tarskian definition in terms of truth in a model under an assignment; and so on. So we need to know what the underlying logic is if we want to prove things involving analyticity.

Those points mean that since your problem is underspecified, we can only proceed informally and by making assumptions. The assumption will be that we're working with a propositional language, so logical truth will mean tautology for the rest of our discussion. Let's continue by recalling the basic definition of what makes an argument valid, making use of the notion of logical truth as tautology.

Definition 2. (Argument Validity) An argument of the form Γ ⊢ φ is valid iff every truth-assignment to the propositional symbols occurring in (Γ ∪ φ) that makes all sentences in Γ true, make φ true.

If propositional logic is not what you're working with in class, the definition needs to be altered to say, for example, that the argument is valid just in case all models of Γ are models of φ, and so on. Now we're ready to address the questions.

Q1. Can a valid argument with consistent premises have an analytically impossible conclusion?

For concreteness, let's consider a particular analytically (with respect to Peano Axioms) impossible sentence, say: (0 = 1). It's impossible because it means that 0 is equal to the successor of some number, and Peano's Axioms exclude that possibility. The question, more concretely is, can a valid argument Γ ⊢ φ with consistent Γ be such that φ ≡ (0 = 1)? The consistency condition is meant to rule-out a trivial 'yes', because if Γ is inconsistent, then Γ ⊢ φ is valid for all sentences φ.

Answer. Let Γ be the singleton set {(0 = 1)}. Since (0 = 1) is satisfiable (e.g. assign '0' and '1' both to some element a in the domain of interpretation), Γ is consistent. We already agreed that (0 = 1) is analytically impossible (with respect to PA), so what remains is to show that {(0 = 1)} ⊢ (0 = 1) is a valid argument. It's obvious that any interpretation that makes all sentences in {(0 = 1)} true, makes (0 = 1). So yes, a valid argument with consistent premises can have an analytically impossible conclusion.

Q2. Can a valid argument with analytically compossible premises have a contradictory conclusion?

Consider an arbitrary Γ such that φ1, φ2 are both in Γ (possibly with other sentences compossible with them) and φ1 and φ2 are analytically compossible (with respect to some set of meaning postulates). To define analytic compossibility we need to define analytic possibility first, which we can do as follows:

Definition 3. (Analytic Possibility) Sentence φ (in language L) is analytically possible with respect to meaning postulates Π iff ¬(φ) is not analytic with respect to Π in language L.

We can then say that sentences φ1 and φ2 are analytically compossible (in language L) (with respect to meaning postulates Π) just in case (φ1 ∧ φ2) is analytically possible (in L) (with respect to Π). From this it follows that at least one truth-assignment jointly satisfies φ1 and φ1, so the argument {φ1, φ2} ⊢ ⊥ will be valid just in case those assignments that jointly satisfy φ1 and φ1 also satisfy ⊥. But since ⊥ is a not satisfiable (since it's a contradiction), the answer to (Q2) must be a 'no'.

Hope this helps. Just be aware that a lot hangs on the definition of analyticity and the richness of the underlying language/logic. You may need to tweak the definitions of logical truth and/or analyticity to suit your particular needs. For corrections/suggestions leave a comment or simply edit the post.

• thanks very helpful as usual but I do get quite lost starting at def. 3. – Achilles Apr 13 '14 at 23:42
• Let's see if this helps. A is necessarily true iff A is true in all possible worlds (don't worry about the actual meaning of that). Given that, we have two options for defining possibility. We can "waste" another definition and be repetitive and say: A is possibly true iff A is true in some possible world. Or alternatively we can use the definition of necessity to define possibility as follows: A is possibly true iff ~A is not necessarily true. In our case, I've re-used the definition of analytic necessity to define analytic possibility. Can you see a way of re-writing Def 3 to avoid that? – Hunan Rostomyan Apr 13 '14 at 23:47
• Alright, I'm impatient, here's how: Def 3 alternative. Sentence p (in language L) is analytically possible with respect to meaning postulates G iff (G -> p) is not a "formal contradiction" (as you would call it) . – Hunan Rostomyan Apr 13 '14 at 23:52
• It's not possible to rewrite Def 3 to avoid it ? is it? – Achilles Apr 13 '14 at 23:53
• Sorry I'm here I was just trying to figure out a way to rewrite it. – Achilles Apr 13 '14 at 23:54