True Statement is implied by Every Statement/Formulation 1

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Theorem

\(\ds p\) \(\) \(\ds \)
\(\ds \vdash \ \ \) \(\ds q \implies p\) \(\) \(\ds \)


Proof 1

By the tableau method of natural deduction:

$p \vdash q \implies p$
Line Pool Formula Rule Depends upon Notes
1 1 $p$ Premise (None)
2 1 $\neg q \lor p$ Rule of Addition: $\lor \II_2$ 1
3 1 $q \implies p$ Sequent Introduction 1 Rule of Material Implication

$\blacksquare$


Proof by Truth Table

We apply the Method of Truth Tables.

As can be seen by inspection, where the truth value in the relevant column on the left hand side is $\T$, that under the one on the right hand side is also $\T$:

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

$\blacksquare$


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