Are there multiple definitions of validity?


I have recently started learning the basics of propositional logic. According to, a sentence is valid if and only if it is satisfied by every truth assignment.

As discussed later in the same document, the Equivalence Theorem states that a sentence Φ and a sentence Ψ are logically equivalent if and only if the sentence (Φ ↔︎ Ψ) is valid. Based on the definition of validity in this document, the biconditional cannot be valid. If one of the propositions is true and the other false, then the biconditional yields false. Because the biconditional yields false, then the sentence cannot be valid.

Do I misunderstand the definition of validity provided here, or is there a second definition of validity? Thanks in advance.


Posted 2020-07-31T11:25:22.457

Reputation: 45



No, it's the same definition for validity, and you seem to mistake the notation "Φ ↔︎ Ψ" for a sentence.

In Φ ↔︎ Ψ, Φ and Ψ are not variables for truth assignments, but variables for sentences (else the book would have used "p" and "q"). So it is not (only) possible to truth-assign "true" to Φ and "false" to Ψ (unless as sentences). Instead, they first need to each be replaced by actual sentences, and then truth assignment can happen to those sentences. As an example, when Φ is "p" and Ψ is also "p", this yields the sentence "p ↔︎ p", which is satisfied by all truth assignments (true ↔︎ true, false ↔︎ false). Other examples for equivalent statements: "p ∧ q ↔︎ q ∧ p". Not equivalent: "p ↔︎ q", "p ∧ q ↔︎ p ∧ r"


Posted 2020-07-31T11:25:22.457

Reputation: 727


As you say, a sentence of the propositional logic is valid if and only if it is satisfied by every truth assignment. So the sentence "Φ ↔︎ Ψ" might be valid if the sentences Φ and Ψ are such that there is no valuation under which one of them is true and the other false. For example, if Φ is ¬(A ⋀ B), and Ψ is (¬A ⋁ ¬B) then Φ ↔︎ Ψ is valid, since there is no way to assign truth values to A and B that makes Φ ↔︎ Ψ come out false. Other examples of Φ ↔︎ Ψ might be invalid.

A more general answer to your question would be to point out that each logic comes with its own specification of what constitutes validity. Validity in propositional logic is simply a matter of assigning truth values to the propositions. In predicate logic things become more complex and we speak of propositions having interpretations under which they may be true. Then there are different logics entirely, such as intuitionistic logic, which have different rules for validity. There is also an important distinction between syntactical and semantic validity. In simple terms, syntactical validity is concerned with the rules that determine whether something is provable just by manipulating formulas, while semantic validity is concerned with whether a semantic property (usually truth) holds under all interpretations, or all possible worlds, or some other generality.

Edit: it is worth clarifying that it is obvious from your question that you are asking about the concept of validity as it occurs in the context of logic and reasoning. In ordinary English 'valid' has other uses, as one can speak of a valid will, or a valid ticket for a journey, or a valid contract. Statisticians also describe a data set as valid, meaning that it is unbiased and correctly represents what they are trying to measure. Those are quite different uses of 'valid' from the way logicians use the term.


Posted 2020-07-31T11:25:22.457

Reputation: 11 885


It is the same definition.

Φ ↔︎ Ψ is not in general a valid formula because, as you say, it is false for some Φ and Ψ.

However, if Φ ↔︎ Ψ is valid for a specific Φ and a specific Ψ, then Φ and Ψ are equivalent. Consider, for instance

Φ = (p OR q)
Ψ = (q OR p)

Φ ↔︎ Ψ is certainly valid in this case. Hence, (p OR q) and (q OR p) are equivalent.

Mr. White

Posted 2020-07-31T11:25:22.457

Reputation: 835

I think this approach is muddying the waters - that statement isn't just "certainly valid", until the reader understands that this is because that however the variables p and q are assigned truth values, given a truth value assignment that follows the standard interpretation of the OR operator, that same assignment will give phi and psi the same truth values, and this is sufficient to ground assignment of the bi-implication. – Paul Ross – 2020-08-01T08:43:33.327

It might just be enough for the reader to understand. Or not, and then she will ask. – Mr. White – 2020-08-01T08:51:44.510