Rule of Exportation/Forward Implication/Formulation 2
Theorem
- $\vdash \left({\left ({p \land q}\right) \implies r}\right) \implies \left({p \implies \left ({q \implies r}\right)}\right)$
Proof 1
By the tableau method of natural deduction:
Line | Pool | Formula | Rule | Depends upon | Notes | |
---|---|---|---|---|---|---|
1 | 1 | $\paren {p \land q} \implies r$ | Assumption | (None) | ||
2 | 1 | $p \implies \paren {q \implies r}$ | Sequent Introduction | 1 | Rule of Exportation: Forward Implication: Formulation 1 | |
3 | $\paren {\paren {p \land q} \implies r} \implies \paren {p \implies \paren {q \implies r} }$ | Rule of Implication: $\implies \II$ | 1 – 2 | Assumption 1 has been discharged |
$\blacksquare$
Proof 2
We apply the Method of Truth Tables to the proposition.
As can be seen by inspection, the truth values under the main connective are true for all boolean interpretations.
$\begin{array}{|ccccc|c|ccccc|} \hline ((p & \land & q) & \implies & r) & \implies & (p & \implies & (q & \implies & r)) \\ \hline F & F & F & T & F & T & F & T & F & T & F \\ F & F & F & T & T & T & F & T & F & T & T \\ F & F & T & T & F & T & F & T & T & F & F \\ F & F & T & T & T & T & F & T & T & T & T \\ T & F & F & T & F & T & T & T & F & T & F \\ T & F & F & T & T & T & T & T & F & T & T \\ T & T & T & F & F & T & T & F & T & F & F \\ T & T & T & T & T & T & T & T & T & T & T \\ \hline \end{array}$
$\blacksquare$