The following is my attempt to formalize Berkeley's argument that it's not possible for a sensible object to exist without conceiving it. Line $3$ involves a meta-argument which I believe might be formalized in a better way than how I did it.

Although this proof seems to have the form of a valid proof, I believe there is something wrong with my application of Universal Introduction on line $8$. It seems intuitively invalid to assert that all things are conceived of simply in virtue of being able to assume their existence as on line $2$. The potential to conceive is not the same as actually conceiving, so there seems to be no valid basis to universalize a particular instance of assuming in this way. I believe the problem has to do with my unorthodox meta-argument on line $3$.

My question: Is there some formal reason to explain why this application of Universal Introduction seems invalid? Or is there an alternative way to formalize this argument to make its validity/invalidity more clear?

  • Ax = x is an assumption
  • Cx = x is conceived of

\begin{array}{l} & \{1\} & 1. & \forall x[Ax \to Cx] & \text{ Prem. }\\ & \{2\} & 2. & \neg Ca & \text{ Assum. }\\ & \{2\} & 3. & Aa & \text{ (a) is assumed on 2 }\\ & \{1\} & 4. & Aa \to Ca & \text{ 1 UE }\\ & \{1,2\} & 5. & Ca & \text{ 3,4 MP }\\ & \{1,2\} & 6. & Ca \land \neg Ca & \text{ 2,5 $\land$I }\\ & \{1\} & 7. & Ca & \text{ 2,6 RAA }\\ & \{1\} & 8. & \forall x[Cx] & \text{ 7 UI - invalid }\\ & \{1\} & 9. & \neg \neg \forall x[Cx] & \text{ 8 DNI }\\ & \{1\} & 10. & \neg \exists x[\neg Cx] & \text{ 9 QI }\\ \end{array}


2 Answers 2


The issue is with $a$.

The rule of $\forall$-intro is:

$$\dfrac {\varphi(x)}{\forall x \varphi(x)}$$

where in $x$ may not occur free in any hypothesis on which $\varphi(x)$ depends,

while $\forall$-elim is:

$$\dfrac {\forall x \varphi(x)}{\varphi(t)}$$

where $t$ is free for $x$.

Thus, if $a$ is a free variable in 2, you cannot "generalize" it in 8, due to 3.

Intuitively, it is quite obvious: if you stop at 8, we can say: "all fishes live in the sea" (by 1) and thus "everything (in the universe) lives in the sea" (by 8), that "sounds wrong"...

  • $\begingroup$ It seems my mistake was thinking that line 3 only depended on 2 whose dependency would be eliminated by RAA. $\endgroup$
    – User4407
    Jul 1, 2016 at 12:23
  • $\begingroup$ @PédeLeão - exactly; 3 is an assumption and all assumptions stand "on the same level". With it, you are assuming that there exists something that is $A$, but this is not licensed by 1; 1 can still hold with no $A$s at all. $\endgroup$ Jul 1, 2016 at 12:33

If what you are trying to derive from the premise $\forall x(Ax\to Cx)$ is $\neg\exists x \neg Cx$, this tree proof generator provides the following countermodel:

enter image description here

A domain containing one element, $0$, which does not make either $A$ or $C$ true would make the conclusion false since there does exist $x=0$ for which $\neg C$ is true. However, the antecedent, a conditional, is true for $x=0$ since false implies false is true.

For such problems it may be helpful to use a proof checker. Here is how a proof might progress using the OP's steps. Since "A" is used to code the universal quantifier symbol in this proof checker, I replaced the predicate $A$ with $B$.

enter image description here

Note that there are two problems with the proof before we reach a line where we could use universal introduction:

  1. The assumption on line 3 has not been discharged.
  2. The conversion of quantifiers rule taking one from line 10 to line 11 is not valid at least as this proof checker expects it to be used.

Regardless, the proof should not succeed because a countermodel has been found. That doesn't mean that Berkeley's argument is wrong. It only means that what was formulated using classical logic is invalid.

Kevin Klement's JavaScript/PHP Fitch-style natural deduction proof editor and checker http://proofs.openlogicproject.org/

Tree Proof Generator. https://www.umsu.de/trees/


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