$((p\to q) \land (q \to p)) \to (p \leftrightarrow q)$ is tautology

Question

$((p\to q) \land (q \to p)) \to (p \leftrightarrow q)$ is tautology

I prove this by using proof by contradiction:

So I defined A = [(p $\implies$ q) $\land$ (q $\implies$ p)] and B = (p $\iff$ q) , and that will be A $\implies$ B. For this to happen A needs to be T and B needs to be F.

Therefore, I went with my cases:

Case I: If p is T and q is F; A = F and B is F $\therefore$ A $\implies$ B is T

Case II: If p is F and q is T; A = F and B is F $\therefore$ A $\implies$ B is T

Case III: If p and q are the same; A = F (If p and q are both F) and B = T $\therefore$ A $\implies$ B is T; or A = T (If p and q are both T) and B = T $\therefore$ A $\implies$ B is T.

By the proof of contradiction this is invalid therefore is tautology.

Then I got curious and tried to prove. this by using this rules only and failed completely, how can this be done by using this rules?:

Answer 1

You can employ the equivalence from table 8, rule 1 to transform it into a tautology with the addition of meta-logical laws:

$\begin{align} & \quad \ p \leftrightarrow q \equiv (p \to q) \land (q \to p) \quad & \text{(table 8 rule 1)} \\ & \rightsquigarrow p \leftrightarrow q \vDash (p \to q) \land (q \to p) \ \text{ and }\ (p \to q) \land (q \to p) \vDash (p \leftrightarrow q) \quad & \text{(def. } \equiv \text{)}\\ & \rightsquigarrow (p \to q) \land (q \to p) \vDash (p \leftrightarrow q) \quad & \text{(simplification)}\\ & \rightsquigarrow\ \! \vDash ((p \to q) \land (q \to p)) \to (p \leftrightarrow q) \quad & \text{(import-export theorem)} \end{align}$

where "$\vDash$" means logical consequence/tautologicity.

Answer 2

,

What you did, by cases (there are four cases, two of which you covered in your Case III), is essentially describe the results of the truth table for the the truth value of each side of the implication in the title.

By your rules, we have $$(p\leftrightarrow q) \equiv ((p \to q) \land (q\to p)$$ which means, $$((p\to q) \land (q\to p))\equiv (p \leftrightarrow q).$$

And this means $$[((p \to q) \land (q \to p)) \rightarrow (p\leftrightarrow q)] \land [(p\leftrightarrow q) \to ((p \to q) \land (q \to p))].$$

Answer 3

$$((p \to q) \land (q \to p)) \to (p \leftrightarrow q) \overset{Equivalence \ \text{(this is one of the rules from Table 8)}}{\equiv}$$

$$(p \leftrightarrow q)\to (p \leftrightarrow q) \overset{Implication \ \text{( this is one of the rules from Table 7)}}{\equiv}$$

$$\neg ( p \leftrightarrow q) \lor (p \leftrightarrow q) \overset{Commutation}{\equiv}$$

$$(p \leftrightarrow q)\lor \neg (p \leftrightarrow q)\overset{Complement \ \text{(one of the Negation Laws)}}{\equiv}$$

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