Is it possible to generate sentences that are made up of "atomic contradictions", but which remain logically valid as a whole? By "atomic contradictions", I mean atomic propositions that are not logically valid.

My question could be restated as follows: consider any formal system pertaining to classical logic or perhaps, think of classical logic as being a formal system in itself. Then, an axiom (any axiom) such as

A → (B → A)
or, in English: "if A, then B only if A"

could be constructed from a "contradictory version" of that very same axiom.

If substituting A for

(A → (B → ¬A))

and B for

(B → (B → A))

we would get an axiom made up of two contradictions—the instances of A and B stated above.

So is the statement

(A → (B → ¬A)) → ((B → (B → A)) → (A → (B → ¬A)))

a valid statement? Is it still an axiom?

  • 1
    It is not clear to me what an "atomic contradiction" is. Is it simply a falsehood?
    – Mitch
    Aug 26, 2011 at 13:01

5 Answers 5


Yes, your last statement is still a valid statement, and still an axiom, or at least can get considered an axiom for some formal system, since you can always join any theorem say in a natural deduction context as an axiom if you wish. There exists no question, that your last statement comes as a "valid" statement or "tautology" in classical logic, and consequently for the complete system of classical propositional logic it will also come as a theorem. Why? Basically because of truth-functionality. You've basically applied "the rule of (uniform) substitution" which can get proven as a metatheorem of classical and, I think, any truth-functional logic.

Since (p->(q->p)) comes as valid, it holds for all truth values in {T, F} or equivalently {1, 0}. The material conditional "->" comes as a truth function. This means that for any ordered pair of inputs (x, y), x, y belonging to {1, 0}, the material conditional "->" assigns a unique member of {1, 0}. "¬" comes as a unary truth-function also taking any member of {1, 0} uniquely to a member of {1, 0}. So, given any values of A, and B in {1, 0}, (B → ¬A) evaluates to single value in {1, 0}. If you continue on like this, you can show that as long as "A" and "B" get assigned truth values consistently, that the entire formula, and every subformula of it (once properly parenthesized with parentheses around the entire formula in this case) will have some unique truth value. Now, since ((p->(q->p)) always comes as valid for all truth values in {1, 0}, and the subformulas of (A → (B → ¬A)) → ((B → (B → A)) → (A → (B → ¬A))) only take on truth values in {1, 0}, we have the larger formula as always taking on truth value of 1, so it also comes as valid.

You might also want to see S. C. Kleene's Mathematical Logic and The Schaum's Outline of Boolean Algebra for proofs of the rule of (uniform) substitution in general.


Classical logic is explosive; once you accept a contradiction, any other proposition follows as a consequence: ex falso quodlibet. So, if you are axiomatically accepting a contradiction, all statements would become valid. Not terribly useful.

If you want to be able accept contradictions without explosion, you're outside of most classical logics, and into the domain of Paraconsistent Logic.

  • Okay. The logical proposition I wrote above is not explosive because it is not a contradiction. I specified the formal system of classical logic because that is where my interest resides, not in many valued logics, modal logics, or paraconsisten logics amongst other types of logics. So, let me ask another question now: if a logically valid statement can consist of a given number of contradictions substituted into the structure of an
    – Gabriella
    Aug 26, 2011 at 11:27
  • axiom or theorem, would it make sense to say that then, classical logic is to some extent paraconsistent as it seems to be able to deal with contradicton-made valid propositional formulas?
    – Gabriella
    Aug 26, 2011 at 11:29
  • Let me see if I am following you correctly. What if we take a proposition like "If A, then B" and substitute in (A and not A) for A, and (B and not B) for B? It would seem to me that the general proposition (If A then B) still holds, as long as we don't evaluate A or B-- but I am not a logician. Aug 26, 2011 at 11:39
  • Yes, you are following me correctly; that is what I meant to say. But, there is a weirdness about constructing or accepting logically valid sentences made up of contradictions, right?
    – Gabriella
    Aug 27, 2011 at 15:52

Yes, but with qualification of your terms.

F -> F

is a theorem of propositional logic (where F is false), so that anything that evaluates always to false will also be a theorem.

A contradiction is presence of two hypotheses that have opposite truth value. A proposition is not itself a possible contradiction, it's just the collection of all its valuations.


P ^ -P

a contradiction or just a truth function that is always false? Either way, you can substitute it into any axiom or theorem of propositional logic to get another theorem. It seems strange but even if you substitute a falsehood (or maybe contradiction, in your terms), you wil get another theorem, a true-in-all-valuations statement.

A -> (B -> A)

is a theorem, and substituting P^-P in for A and Q^-Q for B, you still get another theorem (if you can read through all the parentheses):

(P ^ -P) -> ((Q ^ -Q) -> (P ^ -P))

That is a theorem (do the truth table) and it is built up out of falsehoods, in a manner of speaking.

So yes you can construct a theorem out of falsehoods or contradictions if you like.

  • What you say provides me with a new or different way of seeing the problem, which is why I joined this site, group, or whatever you might call this in the first place. So thanks! I have confirmed for myself that this is indeed a logical problem. I don't think that you understood my question, nonetheless your answer is useful.
    – Gabriella
    Aug 27, 2011 at 16:01
  • @Gabriella: I answered too shortly. I extended my answer with an example inspired by Michael's comment. See if that explains things better.
    – Mitch
    Aug 27, 2011 at 17:20
  • That was very useful, thanks Mitch. Yet, the concept is rather strange, I must say.
    – Gabriella
    Aug 27, 2011 at 18:23
  • Sorry, I am still getting a hang of this; just found the upper right box called chat. I checked out the Schrodinger chat room: looks like a virtual Vienna Circle.
    – Gabriella
    Aug 27, 2011 at 18:40
  • @Gabriela: re concepts...propositional logic is a different beast than ...um...working with words. It's mathematical, that is, rule based (even though those rules were created out of the problems working with words. With mathematical logic, when in doubt, do it mechanically (by truth-table, or syntactic rule manipulation like DeMorgan's law).
    – Mitch
    Aug 27, 2011 at 21:26

Is it possible to generate logically valid sentences made up of “atomic contradictions”?

There are no atomic contradictions at all.

If p is atomic then the contradiction (p ^ ~p) is composite not atomic.

Only contradiction implies contradiction.

(p ^ ~p) -> (p ^ ~p) is logically valid for all p.


You define an 'atomic contradiction' as an atomic proposition that is not logically valid (or to be pedantic not a tautology ('valid' is the word used for the similar concept in predicate calculus/FOL). by the stipulations of classical propositional logic, the only atomic propositions are the propositional variables, in practice usually given labels like P or Q (that is, 'P ^ Q' is a proposition that is not atomic).

A proposition that is true for all valuations (assignments to its variables) is called 'valid'. Since a variable can be assigned either true or false, it is not true for all valuations, and so is not valid/not a tautology. So I think


is an atomic contradiction in your terms (and likewise any propositional behavior.

P -> P

is, by truth table analysis or accepting it as is common and natural as an axiom of propositional logic, a tautology (true under all valuations of its variables).

So from an 'atomic contradiction' one can construct a proposition that is a tautology.

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