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)?


1 Answer 1


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.

  • $\begingroup$ 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

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