Rule of Exportation/Formulation 1

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Theorem

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


This can be expressed as two separate theorems:

Forward Implication

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

Reverse Implication

$p \implies \paren {q \implies r} \vdash \paren {p \land q} \implies r$


Proof 1

Proof of Forward Implication

By the tableau method of natural deduction:

$\left ({p \land q}\right) \implies r \vdash p \implies \left ({q \implies r}\right)$
Line Pool Formula Rule Depends upon Notes
1 1 $\left ({p \land q}\right) \implies r$ Premise (None)
2 2 $p$ Assumption (None)
3 3 $q$ Assumption (None)
4 2, 3 $p \land q$ Rule of Conjunction: $\land \mathcal I$ 2, 3
5 1, 2, 3 $r$ Modus Ponendo Ponens: $\implies \mathcal E$ 1, 4
6 1, 2 $q \implies r$ Rule of Implication: $\implies \mathcal I$ 3 – 5 Assumption 3 has been discharged
7 1 $p \implies \left ({q \implies r}\right)$ Rule of Implication: $\implies \mathcal I$ 2 – 6 Assumption 2 has been discharged

$\blacksquare$


Proof of Reverse Implication

By the tableau method of natural deduction:

$p \implies \left ({q \implies r}\right) \vdash \left ({p \land q}\right) \implies r$
Line Pool Formula Rule Depends upon Notes
1 1 $p \implies \left ({q \implies r}\right)$ Premise (None)
2 2 $p \land q$ Assumption (None)
3 2 $p$ Rule of Simplification: $\land \mathcal E_1$ 2
4 2 $q$ Rule of Simplification: $\land \mathcal E_2$ 2
5 1, 2 $q \implies r$ Modus Ponendo Ponens: $\implies \mathcal E$ 1, 3
6 1, 2 $r$ Modus Ponendo Ponens: $\implies \mathcal E$ 5, 4
7 1 $\left ({p \land q}\right) \implies r$ Rule of Implication: $\implies \mathcal I$ 2 – 6 Assumption 2 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 connectives match for all boolean interpretations.


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

$\blacksquare$


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