Self-Distributive Law for Conditional/Forward Implication/Formulation 2/Proof by Truth Table
Theorem
- $\paren {p \implies \paren {q \implies r} } \implies \paren {\paren {p \implies q} \implies \paren {p \implies r} }$
Proof
We apply the Method of Truth Tables.
As can be seen by inspection, the truth value under the main connective is true for all boolean interpretations.
$\begin{array}{|ccccc|c|ccccccc|} \hline (p & \implies & (q & \implies & r)) & \implies & ((p & \implies & q) & \implies & (p & \implies & r)) \\ \hline \F & \T & \F & \T & \F & \T & \F & \T & \F & \T & \F & \T & \F \\ \F & \T & \F & \T & \T & \T & \F & \T & \F & \T & \F & \T & \T \\ \F & \T & \T & \F & \F & \T & \F & \T & \T & \T & \F & \T & \F \\ \F & \T & \T & \T & \T & \T & \F & \T & \T & \T & \F & \T & \T \\ \T & \T & \F & \T & \F & \T & \T & \F & \F & \T & \T & \F & \F \\ \T & \T & \F & \T & \T & \T & \T & \F & \F & \T & \T & \T & \T \\ \T & \F & \T & \F & \F & \T & \T & \T & \T & \F & \T & \F & \F \\ \T & \T & \T & \T & \T & \T & \T & \T & \T & \T & \T & \T & \T \\ \hline \end{array}$
Sources
- 1988: Alan G. Hamilton: Logic for Mathematicians (2nd ed.) ... (previous) ... (next): $\S 1$: Informal statement calculus: $\S 1.2$: Truth functions and truth tables: Exercises $3 \ \text{(h)}$