Law of Identity/Formulation 2

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Theorem

Every proposition entails itself:

$\vdash p \implies p$


Proof 1

By the tableau method of natural deduction:

$\vdash p \implies p$
Line Pool Formula Rule Depends upon Notes
1 1 $p$ Premise (None)
2 $p \implies p$ Rule of Implication: $\implies \mathcal I$ 1 – 1 Assumption 1 has been discharged


$\blacksquare$


Proof 2

We apply the Method of Truth Tables to the proposition.

As can be seen by inspection, the truth value under the main connective is $T$ throughout.

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

$\blacksquare$


Proof 3

Using a tableau proof for instance 1 of a Hilbert proof system:


$p \implies p$
Line Pool Formula Rule Depends upon Notes
1 $\left({p \implies \left({\left({p \implies p}\right) \implies p}\right)}\right) \implies \left({\left({p \implies \left({p \implies p}\right)}\right) \implies \left({p \implies p}\right)}\right)$ Axiom 2 $\mathbf A = p, \mathbf B = p \implies p, \mathbf C = p$
2 $p \implies \left({\left({p \implies p}\right) \implies p}\right)$ Axiom 1 $\mathbf A = p, \mathbf B = p \implies p$
3 $\left({p \implies \left({p \implies p}\right)}\right) \implies \left({p \implies p}\right)$ Modus Ponendo Ponens: $\implies \mathcal E$ 1, 2
4 $p \implies \left({p \implies p}\right)$ Axiom 1 $\mathbf A = p, \mathbf B = p$
5 $p \implies p$ Modus Ponendo Ponens: $\implies \mathcal E$ 3, 4


$\blacksquare$


Also see

Some sources, for example 1980: D.J. O'Connor and Betty Powell: Elementary Logic, use the statement:

$\vdash p \implies p$

to be the defining property of a tautology.


Sources