I seem to find that p↔(q↔r) ≡ (p↔q)↔r ≡ (p↔r)↔q

Question

I seem to find that p↔(q↔r) ≡ (p↔q)↔r ≡ (p↔r)↔q. But I'm not sure it already has had this equivalence or not. Proof :

     p↔(q↔r) ≡ [p→(q↔r)] ∧ [(q↔r)→p]
             ≡ [p→((q→r) ∧ (r→q))] ∧ [( (q→r) ∧ (r→q) ) →p]
             ≡ [¬p∨((¬q∨r) ∧ (¬r∨q))] ∧ [¬((¬q∨r) ∧ (¬r∨q)) ∨p]   <material Implication>
             ≡ [((¬p∨¬q∨r) ∧ (¬p∨q∨¬r))] ∧ [¬((¬q∨r) ∧ (¬r∨q)) ∨p]  <distributive laws>
             ≡ [((¬p∨¬q∨r) ∧ (¬p∨q∨¬r))] ∧ [ ((q∧¬r) ∨ (¬q∧r)) ∨p]  <De Morgan's laws>
             ≡ [((¬p∨¬q∨r) ∧ (¬p∨q∨¬r))] ∧ [ ((q∨¬q) ∧ (q∨r) ∧ (¬r∨¬q) ∧ (¬r∨r)) ∨p ] <distributive laws>
             ≡ [((¬p∨¬q∨r) ∧ (¬p∨q∨¬r))] ∧ [ ((q∨r) ∧ (¬r∨¬q))∨p ]
             ≡ [((¬p∨¬q∨r) ∧ (¬p∨q∨¬r))] ∧ [ ((q∨r) ∧ (¬q∨¬r))∨p ]
             ≡ [((¬p∨¬q∨r) ∧ (¬p∨q∨¬r))] ∧ [ ((p∨q∨r) ∧ (p∨¬q∨¬r))   <distributive laws>
             ≡ (¬p∨¬q∨r) ∧ (¬p∨q∨¬r) ∧ (p∨q∨r) ∧ (p∨¬q∨¬r)
     p↔(q↔r) ≡ (¬p∨¬q∨r) ∧ (¬p∨q∨¬r) ∧ (p∨q∨r) ∧ (p∨¬q∨¬r) ----(1)

     (p↔q)↔r ≡ [(p↔q)→r ] ∧ [r→(p↔q)] 
             ≡ [( (p→q) ∧ (q→p) ) →r] ∧ [ r→( (p→q) ∧ (q→p) )]
             ≡ [¬( (¬p∨q) ∧ (¬q∨p) ) ∨r] ∧ [ ¬r∨( (¬p∨q) ∧ (¬q∨p) )] <material Implication>
             ≡ [( (p∧¬q) ∨ (¬p∧q) ) ∨r] ∧ [ ¬r∨( (¬p∨q) ∧ (p∨¬q) )]  <De Morgan's laws>
             ≡ [( (p∧¬q) ∨ (¬p∧q) ) ∨r] ∧ [( (¬p∨q∨¬r) ∧ (p∨¬q∨¬r) )] <distributive laws>
             ≡ [((p∨¬p) ∧ (p∨q) ∧ (¬p∨¬q) ∧ (¬q∨q) ) ∨r] ∧ [( (¬p∨q∨¬r) ∧ (p∨¬q∨¬r) )] <distributive laws>
             ≡ [((p∨q) ∧ (¬p∨¬q)) ∨r] ∧ [( (¬p∨q∨¬r) ∧ (p∨¬q∨¬r) )]
             ≡ [ (p∨q∨r) ∧ (¬p∨¬q∨r) ]  ∧ [ (¬p∨q∨¬r) ∧ (p∨¬q∨¬r) ]  <distributive laws>
             ≡ (p∨q∨r) ∧ (¬p∨¬q∨r) ∧ (¬p∨q∨¬r) ∧ (p∨¬q∨¬r)
             ≡ (¬p∨¬q∨r) ∧ (¬p∨q∨¬r) ∧ (p∨q∨r) ∧ (p∨¬q∨¬r)
     (p↔q)↔r ≡ (¬p∨¬q∨r) ∧ (¬p∨q∨¬r) ∧ (p∨q∨r) ∧ (p∨¬q∨¬r) ----(2)

     (p↔r)↔q ≡ [(p↔r)→q] ∧ [q→(p↔r)]
             ≡ [( (p→r) ∧ (r→p) ) →q] ∧ [q→( (p→r) ∧ (r→p) )]
             ≡ [¬( (¬p∨r) ∧ (¬r∨p) ) ∨q] ∧ [¬q ∨ ((¬p∨r) ∧ (¬r∨p))] <material implication>
             ≡ [ ( (p∧¬r) ∨ (¬p∧r) ) ∨q] ∧ [¬q ∨ ((¬p∨r) ∧ (¬r∨p))] <De Morgan's laws>
             ≡ [((p∨¬p) ∧ (p∨r) ∧ (¬p∨¬r) ∧ (¬r∨r))∨q] ∧ [((¬q∨¬p∨r) ∧ (¬q∨p∨¬r))] <distributive laws>
             ≡ [ ( (p∨r) ∧ (¬p∨¬r) ) ∨q] ∧ [((¬q∨¬p∨r) ∧ (¬q∨p∨¬r))]
             ≡ [ ((p∨q∨r) ∧ (¬p∨q∨¬r)) ] ∧ [((¬q∨¬p∨r) ∧ (¬q∨p∨¬r))] <distributive laws>
             ≡ (p∨q∨r) ∧ (¬p∨q∨¬r) ∧ (¬q∨¬p∨r) ∧ (¬q∨p∨¬r)
             ≡ (p∨q∨r) ∧ (¬p∨q∨¬r) ∧ (¬p∨¬q∨r) ∧ (p∨¬q∨¬r)
             ≡ (¬p∨¬q∨r) ∧ (¬p∨q∨¬r) ∧ (p∨q∨r) ∧ (p∨¬q∨¬r) 
     (p↔r)↔q ≡ (¬p∨¬q∨r) ∧ (¬p∨q∨¬r) ∧ (p∨q∨r) ∧ (p∨¬q∨¬r) ----(3)                  

                          (1) ≡ (2) ≡ (3)
   Therefore        p↔(q↔r) ≡ (p↔q)↔r ≡ (p↔r)↔q
                                                       by Punnawit Kasien
                                                           16/4/2018
                                                             11:23

Tag : new study, logic, logical equivalence , biconditional

Answer 1

Indeed the observation that biconditional is associative and commutative is not anything new.   But it is interesting just the same.

$p\leftrightarrow(q\leftrightarrow r)$ is true iff:

• $p$ is true and $q$ and $r$ have the same value, or
• $p$ is false and $q$ and $r$ have complementary values

Which is to say

• all three from $p,q,r$ are true, or
• any two from the three are false and the third true.

And so by symmetry, the positions of the literals are interchangable.

$$\begin{array}{c} p\leftrightarrow(q\leftrightarrow r) &\iff& q\leftrightarrow(p\leftrightarrow r)&\iff& r\leftrightarrow(p\leftrightarrow q)\\ \Updownarrow && \Updownarrow&& \Updownarrow\\ p\leftrightarrow(r\leftrightarrow q) &\iff& q\leftrightarrow(r\leftrightarrow p)&\iff& r\leftrightarrow(q\leftrightarrow p)\\ \Updownarrow && \Updownarrow&& \Updownarrow\\ (p\leftrightarrow q)\leftrightarrow r &\iff& (q\leftrightarrow p)\leftrightarrow r&\iff& (r\leftrightarrow p)\leftrightarrow q\\ \Updownarrow && \Updownarrow&& \Updownarrow\\ (p\leftrightarrow r)\leftrightarrow q &\iff& (q\leftrightarrow r)\leftrightarrow p&\iff& (r\leftrightarrow q)\leftrightarrow p \end{array}$$

---

Now consider $ (p\leftrightarrow q)\leftrightarrow (r\leftrightarrow s) $ is true when:

• $p$ and $q$ have the same value, and $r$ and $s$ have the same value (possibly not the same-same value), or
• $p$ and $q$ have complementary values, and $r$ and $s$ have complementary values.

Which is to say:

• All four literals are true, or
• All four literals are false, or
• Any two literals are true and the other two literals are false.

And $ p\leftrightarrow (q \leftrightarrow (r\leftrightarrow s)) $ is true for the exact same evaluations.

---

A pattern is emergent.