True Statement is implied by Every Statement/Formulation 2

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Theorem

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


Proof 1

By the tableau method of natural deduction:

$\vdash q \implies \paren {p \implies q} $
Line Pool Formula Rule Depends upon Notes
1 1 $q$ Assumption (None)
2 1 $p \implies q$ Sequent Introduction 1 True Statement is implied by Every Statement: Formulation 1
3 $q \implies \paren {p \implies q}$ Rule of Implication: $\implies \II$ 1 – 2 Assumption 1 has been discharged

$\blacksquare$


Proof by Truth Table

We apply the Method of Truth Tables.

As can be seen by inspection, the truth value under the main connective, the first instance of $\implies$, is $\T$ for each boolean interpretation.

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

$\blacksquare$


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