Rule of Exportation/Formulation 1

From ProofWiki
Jump to navigation Jump to search

Theorem

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


This can be expressed as two separate theorems:

Forward Implication

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

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:

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

$\blacksquare$


Proof of Reverse Implication

By the tableau method of natural deduction:

$p \implies \paren {q \implies r} \vdash \paren {p \land q} \implies r$
Line Pool Formula Rule Depends upon Notes
1 1 $p \implies \paren {q \implies r}$ Premise (None)
2 2 $p \land q$ Assumption (None)
3 2 $p$ Rule of Simplification: $\land \EE_1$ 2
4 2 $q$ Rule of Simplification: $\land \EE_2$ 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 $\paren {p \land q} \implies r$ Rule of Implication: $\implies \II$ 2 – 6 Assumption 2 has been discharged

$\blacksquare$


Proof by Truth Table

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.


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

$\blacksquare$


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Rule_of_Exportation/Formulation_1&oldid=587849"