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Show whether the following is true or false:

α |= β or β |= α, for any two formulas α and β

I'm assuming here that α and β are formulas, not a set of formulas. My thought is that I can prove that it is false if I show α doesn't entail β and β doesn't entail α. I tried using a truth table to help me with this problem, but it didn't seem to help. I'm not sure how to go about this problem. Any help will be appreciated.

  • 3
    Obviously not; consider p as α and q as β. – Mauro ALLEGRANZA Dec 6 '17 at 7:29
  • 2
    Already asked and answered in MSE. – Mauro ALLEGRANZA Dec 6 '17 at 7:33
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    Sorry, why does setting α and β as new variables make it clearer? – Amous Dec 6 '17 at 7:34
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    You have only to use truth table to check that neither p ⊨ q nor q ⊨ p, as already answered to your question in MSE. – Mauro ALLEGRANZA Dec 6 '17 at 8:39
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A set Γ entails the statement P if and only if there is no truth-value assignment in which every member of Γ is true and P false.

This is a standard def of entailment, check any textbook on formal logic. I would include a reference but that might be a product endorsement?

A simple truth table will show a row where α is true and β is false so ~( α|= β) and another row where β is true and α is false and thus ~( β |= α)

I've seen some debates regarding @FrankHubeny's answer but I believe the problem is that the OP's question was unclear due to a lack of precision.

  1. I assume 'formula' means well-formed formula (wff) because I am not aware of any logic that operates on syntactically incorrect expressions.

  2. Entailment is a relationship between a Set and a statement. Not between two statements/wff. Material implication operates between 2 statements. So talking about entailment between 2 formula is confusing and misleading.

I know that seems petty and pedantic, but Russell's Paradox shows us that it is critical to clearly define the scope of operators and that there can be profound problems if we apply functions incorrectly to sets, statements, classes, etc.

0

The question is how to understand why (A → B) v (B → A) is always true even when A and B are sentences that have nothing to do with each other. The reason is that the truth values for the conditional (→) and the disjunction (v) are defined to be true for three out of the four possible sets of valuations of A and B in such a way that their combination always yields a true result.

Consider the truth table for the conditional:

enter image description here

Note that there is only one way for the conditional to be false, but three ways for it to be true.

Similarly consider the disjunction:

enter image description here

Again, there is only one way for the disjunction to be false.

Furthermore, if we have A → B false then B → A would be true, but the only way for the disjunction to be false is if both of those conditionals are false. The OP noticed this:

My thought is that I can prove that it is false if I show α doesn't entail β and β doesn't entail α.

Since it is not possible for both of those conditionals to be false, that makes the disjunction true for all valuations of A and B. The disjunction (A → B) v (B → A) is a tautology.


Note that the title question asks something different which may be where some of the difficulty comes from:

Do α and β entail each other?

This question perhaps asks whether (A → B) & (B → A) or A ↔ B is always True? However, A and B do not always entail each other as the following truth table shows.

enter image description here

Compare that result with the truth table for the disjunction:

enter image description here

The disjunction is a tautology because all of the valuations of the sentences A and B lead to a True result. The conjunction expressed in the title, however, is not a tautology.


Michael Rieppel. Truth Table Generator. Generated on May 9, 2019 from https://mrieppel.net/prog/truthtable.html

  • The question is how to understand why (A → B) v (B → A) is always true - That is not the question. There's a big difference between implication (→) and entailment (|=). And see Mauro's third comment above, which answers the question correctly. – Eliran May 10 at 2:45
  • @Eliran As the OP notes, "I'm assuming here that α and β are formulas, not a set of formulas." Note the last bullet point in the Wikipedia article that Mauro ALLEGRANZA cited in his third comment.en.wikipedia.org/wiki/… – Frank Hubeny May 10 at 12:26
  • I accept that A→B is logically valid (ie a tautology) iff A⊨B, as Wikipedia says, but that's different from what you did. You showed that (A→B)v(B→A) is a tautology, not that either A→B is a tautology or B→A is a tautology. Those are different things, and the latter does not hold. – Eliran May 10 at 17:09
  • @Eliran Neither A→B nor B→A are tautologies as can be seen from the first truth table in my answer. The OP is asking for the truth value of : α |= β or β |= α, for any two formulas α and β. To check this semantically with a truth table one is verifying that ⊨(A→B)v(B→A) is a tautology, which it is. – Frank Hubeny May 10 at 17:22
  • I think you are misunderstanding the question. The question is whether, for any two formulas, one of the formulas entails the other. The answer is no. Neither of p and ~p entails the other. – Eliran May 10 at 17:34

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