Can a statement in FOL be equivalent to two separate independent statements?

Question

This may seem like a dumb question, and it certainly seems dumb to me asking it, but I'm running into a contradiction. I'm looking at the problem of finding a statement $\phi$ such that $\psi$ and $\chi$ together are equivalent to $\phi$ but neither one on their own implies $\phi$:

$\phi$ $\longleftrightarrow$ ($\psi$ $\land$ $\chi$)

$\lnot$ ($\psi$ $\rightarrow$ $\phi$)

$\lnot$ ($\chi$ $\rightarrow$ $\phi$)

The problem is, when I evaluate the conjunction of all these statements in propositional calculus, it adds up to a contradiction, which seems a bit strange to me. I didn't expect that you couldn't compose a statement of two smaller statements that on their own don't imply $\phi$ but together are equal to it. I expect I'm doing something wrong in my formulation of my problem, or somehow I evaluated the statements wrong.

Say for example $\phi$ is the Axiom of Choice and $\psi$ is the Axiom of Dependent Choice, this implies there's no statement $\chi$ that doesn't fully imply the Axiom of Choice on its own to satisfy the first statement. I feel I must be doing something wrong, because it seems weird that I can't take two weak statements to make a strong statement. It seems so wrong I'm more than a little embarrassed asking this.

I don't think there's any magical translation that lets you turn invalid propositional statements into valid first order statements. Is there something I'm missing here? The conclusion seems surprising, but I feel like I'm forgetting something obvious.

Answer 1

I think you are looking for an example of statements here. That is $\phi, \psi, \chi$ s.t. these properties hold. Let $L=\{\cdot\}$. Let $T=\text{group axioms}\cup \text{there are only twelve distinct elements}$. Now Let $\phi$ say that "I'm is abelian and I'm not is not cyclic" (i.e. $\forall a,b, a\cdot b = b\cdot a$ and a fairly long sentence that says $\neg{\exists{x}\forall{b}, x=b \lor x^2=b \lor\ldots }$). Then let $\psi$ say that "I'm abelian " and let $\chi$ say that "I'm not cyclic". Now these have all the properties you want.

EDIT: Based on what you've said, I think the issue arises because you are thinking in terms of provability as opposed to using a specific models to show counter examples. I'm guessing your contradiction arises by way of saying that $\chi, \psi$ have to be true and $\phi$ has to be false for the conditionals to hold. But then the bi-conditional fails. However your formulation of the problem is not correct.

What you want to say is that $(\chi\not\vdash\phi$ or equivalentely $\not\vdash\chi\rightarrow{\phi}$. But this you have taken to say $\vdash\neg(\chi\rightarrow\phi)$. This is just not true as can be seen from the example.

Edit: The formulation as I see it should be:

$\vdash\phi\leftrightarrow{(\psi\wedge\chi)}$, and $\not\vdash\chi\rightarrow{\phi}$ and $\not\vdash\psi\rightarrow{\phi}$

Answer 2

We have that : $\lnot (ψ \rightarrow ϕ)$ is equivalent to : $\lnot (\lnot ψ \lor ϕ)$ i.e. to :

$ψ \land \lnot ϕ$.

And $\lnot (χ \rightarrow ϕ)$ is equivalent to :

$χ \land \lnot ϕ$.

Thus, there is no way to satisfy both formuale together with :

$ϕ \leftrightarrow (ψ \land χ)$.

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Comment

The "problem" in your argument is the bi-implication.

If we state only :

$ϕ \rightarrow (ψ \land χ)$

now the three formulae are simultaneously satisfiable : let $v(ψ)=v(χ)=T$ and $v(ϕ)=F$.

In this case, $ϕ \rightarrow (ψ \land χ)$ is $F \rightarrow (T \land T)$, which evaluates to $T$.

In classical logic : $(A \rightarrow B) \lor (B \rightarrow A)$ [Dummett's law] is a tautology.

Thus, it seeme to me, your argument is :

1) assume $\lnot (A \rightarrow B)$

2) by disjunctive syllogism : $(B \rightarrow A)$ follows.

Thus, by the second and third premises in your argument, we have :

$ϕ \rightarrow ψ$ and $ϕ \rightarrow χ$

and from them, obviously : $ϕ \rightarrow (ψ \land χ)$ follows.

What is wrong with the intuition that :

we couldn't compose a statement of two smaller statements that on their own don't imply $ϕ$ but together are equal to it ?

It seems to me that the we can "reverse" the above expectation :

if $ψ$ does not "licence" to assert $ϕ$ nor $χ$ does, why we expect that "composing" them we are licensed to assert $ϕ$ ?

Answer 3

If ¬ (ψ → ϕ), then ψ, and ¬ϕ.

If ¬ (χ → ϕ), then χ, and ¬ϕ.

If ψ, and χ, then (ψ ∧ χ).

If (ψ ∧ χ), and ¬ϕ, then ¬[(ψ ∧ χ)→ϕ].

If [ϕ ⟷ (ψ ∧ χ)], then [(ψ ∧ χ)→ϕ]. Thus,

ϕ ⟷ (ψ ∧ χ), ¬ (ψ → ϕ), and ¬ (χ → ϕ) do not simultaneously hold true.