This is not a question about "if a false statement with ZF(C) is considered true, what happens." (The answer for that is trivial, as falsity in classical logic proves everything.) What I am rather asking is what happens if we negate axioms of ZF(C) - in other words, axioms of the new set theory, call it $ZF(C)'''$, are exact negation of each axiom of ZF(C).

When I think of such $ZF'''$ (I am going to drop axiom of choice being negated for now), it has to mirror $ZF$ in some way, so I would expect that all false statements with ZF would now be true statements with $ZF'''$ and true statement with ZF would now seem to be false statement in $ZF'''$. Yet this does not seem to be true also, because I really have not negated laws of natural deduction system in propositional logic, contained in first-order logic.

What really then would be consequences of $ZF'''$?

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    $\begingroup$ Boy, you must really hate ZFC. $\endgroup$ – Rene Schipperus May 6 '18 at 14:36
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    $\begingroup$ So there exist some sets that have the same elements, but are still different? There exist sets $x,y$ such that there is no set having only $x,y$ as elements? There exists a set that we cannot take the union of? and so on ...? I guess that a universe of two distinct empty sets and a Quine atom does most of it $\endgroup$ – Hagen von Eitzen May 6 '18 at 14:46
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    $\begingroup$ I think you're going to have a (serious) issue with negating each axiom in the axiom schema of specification. There's an axiom in that schema that just says $\forall z \exists y \forall x[x\in y \Leftrightarrow x\in z]$. That's definitely not a statement you can safely negate. (Negating the axiom schema of replacement is also an issue) $\endgroup$ – Milo Brandt May 6 '18 at 15:03
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    $\begingroup$ Your intuition that ZF' should "mirror" ZF is flawed: in general (indeed, usually) we can't transition from "$\{ p\}$ proves $q$" to "$\{\neg p\}$ proves $\neg q$." $\endgroup$ – Noah Schweber May 6 '18 at 15:26
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    $\begingroup$ I'd argue that if you negate all the axioms of ZFC, then what you have is not a theory of sets in any meaningful sense. (At least not in any sense more meaningful than taking any other theory with a single binary relation symbol and pretending it's a theory of sets by interpreting the relation symbol to mean 'is an element of'.) $\endgroup$ – Clive Newstead May 6 '18 at 15:26

As remarked below under Replacement and Comprehension, it is impossible to negate all axioms of ZF. Nevertheless, here's a small model that defies most of ZF (many different models with totally unrelated idiosyncrasies are possible):

There are four sets $A,B,C,D$ and the only element-relations that hold are $B\in C$, $C\in C$, $C\in D$, $D\in D$.

Axiom of Extensionality falsified by $A$ and $B$, which have the same elements (namely none) without being equal.

Axiom of Regularity falsified by $D$ because it is a non-empty set that has an element in common with each of its elements

Axiom Schema of Comprehension is a separate axiom for every predicate. It is impossible to falsify the instance of the schema that is obtained from a tautology. However, some instances are readily falsified. For example, with $\phi(x)\equiv x=D$, thee is no such thing as $\{\,x\in D\mid \phi(x)\,\}$.

Axiom of Pairing falsified by $A$ because no set has $A$ as element

Axiom Schema of Replacement is a separate axiom for every function predicate. It is impossible to falsify the instance of the schema that is obtained from the identity function. However, some instances are readily falsified: If $f(C)=A$, then these is no such thing as $\{\,f(x)\mid x\in C\,\}$

Axiom of Infinite is obviously false.

Axiom of Power Set is falsified because clearyly $A$ and $B$ are subsets of $A$, but there is no set having either of these as element.

Axiom of Union is falsified by $C$ because only $C$ is an element of an element of $C$ and there is no set having only $C$ as element.

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    $\begingroup$ Be glad you don't have an axiom of the empty set. $\endgroup$ – Henning Makholm May 6 '18 at 15:44

Well, the big change (as Clive Newstead points out in the comments above) is that it stops being what one could intuitively call a "set theory." One could certainly call it a "model theory," but it stops acting anything like what we'd expect of sets. Let's just deal with the immediate consequences.

  • Since Extensionality is negated, then there exist "sets" $x$ and $y$ with exactly the same "elements," but such that $x\neq y.$ What are "sets" and "elements" that this should be so? Or, perhaps, what does "=" mean that this should be so?
  • Since Replacement is negated, then there is a "set" $x$ and a "function" $f$ on $x$ such that the "image" of $f$ is not a "set," but a "proper class." But typically, a proper class is a class that is (in a sense) too big to be a set, but images of functions are intuitively no larger than the domains, so images of functions on sets won't be too big to be sets. So what in the world are "sets"/"functions"/"images"/"proper classes"?
  • Since Comprehension is negated, then there is a "set" $x$ and a formula $\phi$ such that the "class" of "elements" of $x$ satisfying $\phi$ is not a "set," but a "proper class." Why?
  • Since Pairing is negated, then there exist "sets" $x$ and $y$ such that any "class" having both $x$ and $y$ as "elements" is a "proper class." Why?
  • Since Power Set is negated, then there is a "set" $x$ such that any "class" containing all "subsets" of $x$ are "proper classes." Why?
  • Since Union is negated, then there is a "set" $x$ whose "elements" are all "sets," but such that any "class" containing all "elements" of the "elements" of $x$ is a "proper class." Why?
  • Since Infinity is negated, then either $\emptyset$ isn't a "set," or $\emptyset$ is a "set" but the von Neumann ordinal $\omega$ is not a "set." The latter isn't a huge deal, but why should the former be true?
  • Since Foundation is negated, then there is a "non-empty" "set" $x$ such that every "element" of $x$ has an "element" in common with $x.$ Again, this isn't too objectionable.
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    $\begingroup$ To be fair, the negation of powerset isn't so unreasonable: models of ZF$^-$ are important objects, and powerset is rejected in pocket set theories. $\endgroup$ – Noah Schweber May 6 '18 at 16:57
  • $\begingroup$ @NoahSchweber You could say the same about Replacement and Comprehension. Studying how they fail in a fine structural setting is an important part of all of inner model theory. $\endgroup$ – Stefan Mesken May 7 '18 at 8:54
  • $\begingroup$ @Noah: That's certainly true. I'm getting more at the intuition that led to standard set theories. The OP is suggesting we throw all that intuition out, and I'm wondering how we can think of these things that would make that sensible. $\endgroup$ – Cameron Buie May 7 '18 at 12:19

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