proof for propositional logic I am unable to prove the following proposition logic.
$(p \lor \neg r) \land (r \lor \neg p) \leftrightarrow (p \leftrightarrow q) \land (q \leftrightarrow r)$
My solution is given in the image. Please correct it and let me know where i am wrong.

 A: The formula you exhibited is not a tautology.  It is false when $p$ and $r$ are true while $q$ is false.  (It's also false when $p,r$ are false and $q$ is true.)
Perhaps the last $\land$ was supposed to be $\leftrightarrow$; then the formula would be a tautology.
A: It's easier to just list out the component implications to show that one direction is not true and therefore they are not logically equivalent.
$(p\leftrightarrow q)\wedge (q\leftrightarrow r)\Leftrightarrow
(p\rightarrow q)\wedge (q\rightarrow p)\wedge (q\rightarrow r)\wedge (r\rightarrow q)$
Trying to show:
$(p\rightarrow r)\wedge (r\rightarrow p) \Rightarrow (p\rightarrow q)\wedge (q\rightarrow p)\wedge (q\rightarrow r)\wedge (r\rightarrow q)$
Not true.  I think you're kind of arguing by the converse.
$(p\rightarrow r)\wedge (r\rightarrow p)\Rightarrow (p\leftrightarrow r)$
$(p\rightarrow r)\wedge (r\rightarrow p)\Leftrightarrow (p\leftrightarrow r)$
However $(p\leftrightarrow r)\nRightarrow (p\leftrightarrow r)\wedge (q\leftrightarrow r)$(the opposite is true by conjunctive 'simplification' as I will now show.)
However, $(p\rightarrow q)\wedge (q\rightarrow p)\wedge (q\rightarrow r)\wedge (r\rightarrow q)\Rightarrow (p\rightarrow r)\wedge (r\rightarrow p)$
$(p\rightarrow q)\wedge (q\rightarrow p)\wedge (q\rightarrow r)\wedge (r\rightarrow q)$
$(p\rightarrow q)\wedge (q\rightarrow r)$ conjunctive 'simplification.'(just stating something as true)
$(p\rightarrow r)$ transitivity.
$(r\rightarrow q)\wedge (q\rightarrow p)$ conjunctive 'simplification.'(just stating something as true)
$(r\rightarrow p)$ transitivity.
$\therefore (p\rightarrow r)\wedge (r\rightarrow p)$ conjunction.
Or with only 3 variables could just use truth table.
