Can I interpret $\models$ as $\implies$ while proving logical expressions using different techniques?

Question

In the book by Artificial Intelligence by Norvig and Russel, I came across following problem:

Prove if correct: $(A ∧ B) \models (A ⇔ B).$

I quickly interpreted $\models$ as $\implies$ and tried to prove it using the truth table:

It seems that at least while interpreting $\models$ as $\implies$, the statement is true. Then I gave second thought and did some more reading to come across this thread. Now I know that the two are not same. But, it turns out, when we don't interpret $\models$ as $\implies$, the statement is still true (my course TA uploaded answers without proof).

So I am wondering:

Q1. How exactly the given statement is correct (given that $\models$ and $\implies$ are not same)?
Q2. Is my method to interpret both same and then forming truth table, a correct method for such problems? If not then how I should solve it?
Q3. If answer to Q2 is no, then will above method of interpreting $\models$ as $\implies$ and forming truth table always given correct answer? If not, when it will fail to give correct answer?
Q4. I also tried to solve the same using resolution:

$$\neg (A\Longleftrightarrow B)\equiv \neg((A\wedge B)\vee(\neg A\wedge\neg B))\equiv\neg(A\wedge B)\wedge \neg(\neg A\wedge \neg B)\equiv (\neg A\vee\neg B)\wedge (A \vee B)$$ So my clauses will be:

• $A$ (from $A\wedge B$)
• $B$ (from $A\wedge B$)
• $(\neg A\vee\neg B)$ from $(\neg A\vee\neg B)\wedge (A \vee B)$
• $(A \vee B)$ from $(\neg A\vee\neg B)\wedge (A \vee B)$

So I was able to derive empty clause, so my assumption $\neg (A\Longleftrightarrow B)$ was incorrect. So $(A ∧ B) \implies (A ⇔ B)$. Will application of resolution technique for $\models$ be exactly same?

Update

[This is my updated understanding based on Graham's answer]

(a) After reading Graham's answer, I felt that the truth table above is proving the "tautology" $(A ∧ B) \implies (A ⇔ B)$, but not $(A ∧ B) \models (A ⇔ B)$.

(b) To prove $(A ∧ B) \models (A ⇔ B)$, I need another truth table, something like this:

(c) Also, I guess the resolution technique is used to prove tautology $(A ∧ B) \implies (A ⇔ B)$ and not $(A ∧ B) \models (A ⇔ B)$. However, I feel we can use the (first) truth table and resolution proving $\implies$ to also prove $\models$, because of the following fact:

$\varphi\vDash \psi$ iff $M(\varphi)\subseteq M(\psi)$: that is, iff every truth assignment that makes $\varphi$ true also make $\psi$ true. This is the case iff $\vDash \varphi\Rightarrow\psi$, i.e., if the formula $\varphi\Rightarrow\psi$ is true in all truth assignments (is a tautology). - source

Can someone please confirm if my understanding in the above points is correct?

Answer 1

$\mathcal A \vDash \varphi$ says: "statement $\varphi$ is a semantic consequence of the list of statements $\mathcal A$ (often called the premises, or assumptions)" and a semantic consequence must be true when all the premises are interpreted as true. [Note: this list may be of zero, one, or several statements.]

In propositional logic, you can use a truth table to establish such a semantic consequence: and $(A\land B)\vDash(A\leftrightarrow B)$ since indeed "every assignment that values $A\land B$ as true will also value $A\leftrightarrow B$ as true," will be shown by a table of every assignment of $A,B$ and the corresponding evaluations of the statements.

$$\boxed{\begin{array}{c4|c}A&B & A\land B& A\leftrightarrow B\\\hline\top&\top &\top &\top&\star\\\top&\bot&\bot &\bot\\\bot&\top&\bot&\bot\\\bot&\bot&\bot&\top\\\end{array}}\\\text{the consequent is true in the only row where the premise is true}$$

[Note: the trend this century is to use single bar arrows, $\to,\leftrightarrow$, for logical connectives, and double edge for metalogic inferences, though not all authors do so.]

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$(A\land B)\to(A\leftrightarrow B)$ is a statement, and moreover one which is a tautology.   That means that it is a logical consequence of no premises at all. Thus:$$\vDash (A\land B)\to(A\leftrightarrow B)$$

This can also be demonstrated by a truth table, where we show that "every assignment of $A,B$ will value the whole statement as true."

$$\boxed{\begin{array}{c4|c}A&B & (A\land B)\to(A\leftrightarrow B)\\\hline\top&\top &\top\\\top&\bot&\top\\\bot&\top&\top\\\bot&\bot&\top\\\end{array}}\\\text{the consequent is true in every row}$$

That is not exactly the same test, but it is clearly the case that where one works the other will too.

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It should be noted, truth tables work as semantic proofs only in classic propositional logic. They are less applicable in higher order logics, modal logics, among others.

Answer 2

(a) After reading Graham's answer, I felt that the truth table above is proving the "tautology" $(A ∧ B) \implies (A ⇔ B)$, but not $(A ∧ B) \models (A ⇔ B)$.
(b) To prove $(A ∧ B) \models (A ⇔ B)$, I need another truth table, something like this:

To be absolutely clear: the difference is this (here, $\psi$ and $\theta$ are compound propositional formulae):

1. using a truth table to verify that $(\psi\to \theta)$ is a tautology: construct the truth table of $(\psi\to \theta),$ then verify that the column with the main connective is filled only with 1's;
2. using a truth table to verify that $(\psi\models \theta):$ construct a truth table containing both $\psi$ and $\theta,$ then verify that in every row where $\psi$ has value 1, $\theta$ also has value 1.

(1) and (2) are equivalent (the Deduction Theorem and its converse formally justifies this).