Self-Distributive Law for Conditional

Theorem
The following is known as the Self-Distributive Law:


 * $$p \implies \left({q \implies r}\right) \dashv \vdash \left({p \implies q}\right) \implies \left({p \implies r}\right)$$

We also have, interestingly, this result:


 * $$\left({p \implies q}\right) \implies r \vdash \left({p \implies r}\right) \implies \left({q \implies r}\right)$$

... but:


 * $$\left({p \implies r}\right) \implies \left({q \implies r}\right) \not \vdash \left({p \implies q}\right) \implies r$$

Proof by Natural Deduction
These are proved by the Tableau method.

Proof by Truth Table
We apply the Method of Truth Tables to the proposition: $$p \implies \left({q \implies r}\right) \dashv \vdash \left({p \implies q}\right) \implies \left({p \implies r}\right)$$

As can be seen by inspection, the truth values under the main connectives match for all models.

$$\begin{array}{|ccccc||ccccccc|} \hline p & \implies & (q & \implies & r) & (p & \implies & q) & \implies & (p & \implies & r) \\ \hline F & T & F & T & F & F & T & F & T & F & T & F \\ F & T & F & T & T & F & T & F & T & F & T & T \\ F & T & T & F & F & F & T & T & T & F & T & F \\ F & T & T & T & T & F & T & T & T & F & T & T \\ T & T & F & T & F & T & F & F & T & T & F & F \\ T & T & F & T & T & T & F & F & T & T & T & T \\ T & F & T & F & F & T & T & T & F & T & F & F \\ T & T & T & T & T & T & T & T & T & T & T & T \\ \hline \end{array}$$

Next we apply the Method of Truth Tables to the proposition:
 * $$\left({p \implies q}\right) \implies r \vdash \left({p \implies r}\right) \implies \left({q \implies r}\right)$$

As can be seen for all models by inspection, where the truth value under the main connective on the LHS is $$T$$, that under the one on the RHS is also $$T$$:

$$\begin{array}{|ccccc||ccccccc|} \hline (p & \implies & q) & \implies & r & (p & \implies & r) & \implies & (q & \implies & r) \\ \hline F & T & F & F & F & F & T & F & T & F & T & F \\ F & T & F & T & T & F & T & T & T & F & T & T \\ F & T & T & F & F & F & T & F & F & T & F & F \\ F & T & T & T & T & F & T & T & T & T & T & T \\ T & F & F & T & F & T & F & F & T & F & F & F \\ T & F & F & T & T & T & T & T & T & F & T & T \\ T & T & T & F & F & T & F & F & T & T & F & F \\ T & T & T & T & T & T & T & T & T & T & T & T \\ \hline \end{array}$$

Note that the two formulas are not equivalent, as the relevant columns do not match exactly.

Hence the result:
 * $$\left({p \implies q}\right) \implies r \vdash \left({p \implies r}\right) \implies \left({q \implies r}\right)$$

... but
 * $$\left({p \implies r}\right) \implies \left({q \implies r}\right) \not \vdash \left({p \implies q}\right) \implies r$$