Truth conditions, roughly, are the way things should be in order for a sentence to be true.

For instance, the condition for the sentence "Paul is a cat" is that the individual denoted by "Paul" is a memeber of the cat set.

Following Tarski's theory of truth, it should be possibile to say that:

1)"p V (p ∧ q)" is true iff it obtains that p or it obtains that p and it obtains that q.

2)"p ∧ (q V not-q)" is true iff it obtains that p and it obtains that q or q does not obtain.

Following Tarski, it seems to me that the two stataments have different truth conditions, altough their truth table is identical and they are both equivalent to just the statament "p".

Where is my error in the construction of the truth conditions of the stataments in 1 e 2 following Tarski's recursive definition?

  • But if truth conditions are "the way the world is", due to the fact that q V not-q is always true (a tautology) there is no possible "configuration" of reality where it does not hold. Oct 26, 2020 at 11:08
  • This means that, the formula q V not-q (being "everywhere true) does not cut-out, from the collection of all possible states of the world, any situation. Thus, the formula containing it is equivalent to p because it is only p that can do it. Oct 26, 2020 at 11:09
  • 1
    Both 1 and 2 are logically equivalent to p, as you say, but I don't understand why you think they have different truth conditions. Both require that p obtains but are indifferent as to whether q obtains.
    – Bumble
    Oct 26, 2020 at 11:20
  • Shouldn't 1 be equivalent to p? If you say that "p V (p ∧ q) iff p" you end up with a tautology, therefore the two should have the same truth conditions, i.e. being true in the exact same worlds. But it does not seem that the T-biconditional of "p V (p ∧ q)" is the same T-schema of "p". Shouldn't I arrive to the same biconditional, since the two sentences have the same truth conditions, being logically equivalent?
    – PwNzDust
    Oct 26, 2020 at 11:22
  • 2
    You should I arrive at equivalent statements, not necessarily statements that look identical. All they have to do is deliver the same verdict on any case presented. And they will. Whether one can see that by reading the two statements is beside the point. In complex enough cases no human may be able to detect that, Boolean satisfiability problem is NP complete, i.e. intractable for long formulas.
    – Conifold
    Oct 26, 2020 at 12:10

1 Answer 1


For a propositional logic formula, truth conditions are simply valuations that satisfy the formula.

Thus, formula "p ∧ (q V not-q)" is true iff it obtains that p and it obtains that (q or not-q).

But (q or not-q) is always satisfied: there is no possible "state of affair" where it is false., i.e. there is no valuation that falsifies it.

This amounts to concluding that: formula "p ∧ (q V not-q)" is true iff it obtains that p.

A propositional formula that is satisfied by every valuation is a tautology.

Quoting Ludwig Wittgenstein's Tractatus Logico-Philosophicus (1921):

4.46 Among the possible groups of truth-conditions there are two extreme cases. In one of these cases the proposition is true for all the truth-possibilities of the elementary propositions. We say that the truth-conditions are tautological. In the second case the proposition is false for all the truth-possibilities: the truth-conditions are contradictory. In the first case we call the proposition a tautology; in the second, a contradiction.

4.462 Tautologies and contradictions are not pictures of reality. They do not represent any possible situations. For the former admit all possible situations, and the latter none.

4.463 The truth-conditions of a proposition determine the range that it leaves open to the facts. [...] A tautology leaves open to reality the whole —the infinite whole— of logical space: a contradiction fills the whole of logical space leaving no point of it for reality. Thus neither of them can determine reality in any way.

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