Self-Distributive Law for Conditional/Forward Implication/Formulation 2
Theorem
- $\paren {p \implies \paren {q \implies r} } \implies \paren {\paren {p \implies q} \implies \paren {p \implies r} }$
Proof 1
By the tableau method of natural deduction:
Line | Pool | Formula | Rule | Depends upon | Notes | |
---|---|---|---|---|---|---|
1 | 1 | $p \implies \paren {q \implies r}$ | Assumption | (None) | ||
2 | 1 | $\paren {p \implies q} \implies \paren {p \implies r}$ | Sequent Introduction | 1 | Self-Distributive Law for Conditional: Formulation 1 | |
3 | $\paren {p \implies \paren {q \implies r} } \implies \paren {\paren {p \implies q} \implies \paren {p \implies r} }$ | Rule of Implication: $\implies \II$ | 1 – 2 | Assumption 1 has been discharged |
$\blacksquare$
Proof by Truth Table
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
- 1965: E.J. Lemmon: Beginning Logic ... (previous) ... (next): Chapter $2$: The Propositional Calculus $2$: $2$: Theorems and Derived Rules: Theorem $43$
- 1982: P.M. Cohn: Algebra Volume 1 (2nd ed.) ... (previous) ... (next): Chapter $1$: Sets and mappings: $\S 1.1$: The need for logic: Exercise $(2) \ \text{(ii)}$