Peirce's Law/Formulation 2

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Theorem

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


Proof 1

By the tableau method of natural deduction:

$\vdash \paren {\paren {p \implies q} \implies p} \implies p$
Line Pool Formula Rule Depends upon Notes
1 1 $\paren {p \implies q} \implies p$ Assumption (None)
2 1 $p$ Sequent Introduction 1 Peirce's Law: Formulation 1: $\paren {p \implies q} \implies p \vdash p$
3 $\paren {\paren {p \implies q} \implies p} \implies p$ Rule of Implication: $\implies \II$ 1 – 2 Assumption 1 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 connective are $\T$ for all boolean interpretations.

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

$\blacksquare$


Source of Name

This entry was named for Charles Sanders Peirce.


Sources