Verify that $\bigl(p\to(q\to r)\bigr)\to \bigl((p\to q)\to (p\to r)\bigr)$ is a tautology. Verify that $\bigl(p\to(q\to r)\bigr)\to \bigl((p\to q)\to (p\to r)\bigr)$ is a tautology.


I am confused on this whole tautology even after looking at examples both in my book and on-line. I started a truth table and this is what I have so far. Can someone please explain this to me in a way that I can understand.

\begin{array}{ccc}
p &amp; q &amp; r &amp; ? &amp; ? &amp; ? &amp; ?\\
\hline
T &amp; T &amp; T\\
T &amp; T &amp; F\\
T &amp; F &amp; T\\
T &amp; F &amp; F\\
F &amp; T &amp; T\\
F &amp; T &amp; F\\
F &amp; F &amp; T\\
F &amp; F &amp; F
\end{array}
A completely different approach (called a tableau proof, and a bit on the informal side for one). Feel free to ignore.


  1) Assume $p \implies (q \implies r)$.
  
  
    2) Assume $p\implies q$.
    
    
      3) Assume $p$.
      
      
        4) By (2) and (3) and modus ponens, $q$.
        
        5) By (1) and (3) and modus ponens, $q \implies r$.
        
        6) By (4) and (5) and modus ponens, $r$.
      
      
      7) By (3)&ndash;(6) and implication, $p \implies r$.
    
    
    8) By (2)&ndash;(7) and implication, $(p\implies q)\implies (p\implies r)$.
  
  
  9) By (1)&ndash;(8) and implication, $\bigl(p\implies (q\implies r)\bigr)\implies \bigl((p\implies q)\implies (p\implies r)\bigr)$.
Your well-formed formula is an instance of axioms schema 3 of the Łukasiewicz axiomatization of propositional logic.   By the completeness of propositional logic, your wf is a tautology.