Conditional is not Right Self-Distributive/Formulation 1
Theorem
While this holds:
- $\paren {p \implies q} \implies r \vdash \paren {p \implies r} \implies \paren {q \implies r}$
its converse does not:
- $\paren {p \implies r} \implies \paren {q \implies r} \not \vdash \paren {p \implies q} \implies r$
Proof
We apply the Method of Truth Tables to the proposition:
- $\paren {p \implies q} \implies r \vdash \paren {p \implies r} \implies \paren {q \implies r}$
As can be seen for all boolean interpretations by inspection, where the truth value under the main connective on the left hand side is $\T$, that under the one on the right hand side is also $\T$:
$\begin{array}{|ccccc||ccccccc|} \hline (p & \implies & q) & \implies & r & (p & \implies & r) & \implies & (q & \implies & r) \\ \hline \F & \T & \F & \F & \F & \F & \T & \F & \T & \F & \T & \F \\ \F & \T & \F & \T & \T & \F & \T & \T & \T & \F & \T & \T \\ \F & \T & \T & \F & \F & \F & \T & \F & \F & \T & \F & \F \\ \F & \T & \T & \T & \T & \F & \T & \T & \T & \T & \T & \T \\ \T & \F & \F & \T & \F & \T & \F & \F & \T & \F & \F & \F \\ \T & \F & \F & \T & \T & \T & \T & \T & \T & \F & \T & \T \\ \T & \T & \T & \F & \F & \T & \F & \F & \T & \T & \F & \F \\ \T & \T & \T & \T & \T & \T & \T & \T & \T & \T & \T & \T \\ \hline \end{array}$
$\Box$
The two formulas are not equivalent, as the relevant columns do not match exactly.
For example, when $p = q = r = \F$ we have that:
- $\paren {p \implies q} \implies r = \F$
but:
- $\paren {p \implies r} \implies \paren {q \implies r} = \T$
Hence the result:
- $\paren {p \implies q} \implies r \vdash \paren {p \implies r} \implies \paren {q \implies r}$
but:
- $\paren {p \implies r} \implies \paren {q \implies r} \not \vdash \paren {p \implies q} \implies r$
$\blacksquare$