0
$\begingroup$

Is the next statement correct: (A∧B)⊨(A⇔B) ?

Formal definition of entailment is this: α⊨β if and only if, in every model in which α is true, β is also true.

I used a truth table to show that there is only one model {A=True, B=True} where (A∧B)=True. (A⇔B) is also True in this model. It means (A∧B)⊨(A⇔B) is correct.

There exist model2 = {A=False, B=False}, (A⇔B) is True in model2. But (A∧B) is not True in model2.

Can you please provide intuition on what does it mean? Does it extend our knowledge base (A∧B) if we add a new statement (A⇔B)?

$\endgroup$

1 Answer 1

1
$\begingroup$

The statement is correct and you have shown it via truth table. Knowing that $A \land B$ implies that both $A$ and $B$ are true, thus $A \iff B$ follows, i.e. it does not extend our knowledge base if $A \iff B$ is added as a new statement, since we already know both $A$ and $B$ are true.

There is a different way to derive $A\iff B$ from $A\land B$, that is, via natural deduction:

\begin{align} 1. &\quad A\land B&\text{Premise}\\ 2. &\quad A &\text{Simplification}\\ 3. &\quad B &\text{Simplification}\\ 4. &\quad \neg B \lor A &\text{Conjunction}\\ 5. &\quad \neg A \lor B &\text{Conjunction}\\ 6. &\quad B \implies A &\text{Material Implication}\\ 7. &\quad A \implies B &\text{Material Implication}\\ 8. &\quad A \iff B &\text{Material Equivalence} \end{align}

or, if your definition of biconditional includes the alternate definition $(A \land B)\lor (\neg A \land \neg B)$ (instead of only the standard $(A \implies B) \land (B \implies A)$), then the conclusion can be drawn from one simple use of the Conjunction rule.

$\endgroup$
5
  • $\begingroup$ From $A$ you can just use weakening to infer $B \to A$, no need for the extra steps. $\endgroup$
    – DanielV
    Jan 6, 2021 at 11:10
  • $\begingroup$ @DanielV I know that the rule you stated is valid, but I can't find any resources regarding this rule in natural deduction/first order logic. Could you be so kind as to provide a link? $\endgroup$
    – player3236
    Jan 6, 2021 at 11:29
  • $\begingroup$ It is fundamental to natural deduction, I think you need to find a resource on the entire logic. $\endgroup$
    – DanielV
    Jan 6, 2021 at 11:50
  • $\begingroup$ @DanielV Is this stub the rule you are referring to? -en.wikipedia.org/wiki/Monotonicity_of_entailment $\endgroup$
    – player3236
    Jan 6, 2021 at 12:21
  • $\begingroup$ It is related. It depends on if you are doing N.D. in Gentzen tree style or Fitch style or Hilbert Style. One way in Gentzen style is $$\dfrac{A \vdash B}{A,X \vdash B}$$ In Fitch style an instance would look like $$\begin{array} {ll} \quad A & \text{Assumption 1} \\ \quad \quad B & \text{Assumption 2} \\ \quad \quad A & \text{Copy} \\ \quad B \to A & \text{discharge assumption 2} \\ A \to (B \to A) & \text{discharge assumption 1} \end{array}$$ In Hilbert style, it is just the first axiom schema, sometimes called "K" or "weakening", it is just $$\vdash A \to (B \to A)$$ $\endgroup$
    – DanielV
    Jan 6, 2021 at 15:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.