Disjunction of Conditional and Converse

From ProofWiki
Jump to navigation Jump to search

Theorem

Given any two statements, one of them implies the other.

$\vdash \left({p \implies q}\right) \lor \left({q \implies p}\right)$


That is, given any conditional, either it is true or its converse is.


Proof 1

By the tableau method of natural deduction:

$\vdash \left({p \implies q}\right) \lor \left({q \implies p}\right)$
Line Pool Formula Rule Depends upon Notes
1 $p \lor \neg p$ Law of Excluded Middle (None)
2 2 $p$ Assumption (None)
3 2 $q \implies p$ Sequent Introduction 2 True Statement is implied by Every Statement
4 2 $\left({p \implies q}\right) \lor \left({q \implies p}\right)$ Rule of Addition: $\lor \mathcal I_2$ 2
5 5 $\neg p$ Assumption (None)
6 5 $p \implies q$ Sequent Introduction 5 False Statement implies Every Statement
7 5 $\left({p \implies q}\right) \lor \left({q \implies p}\right)$ Rule of Addition: $\lor \mathcal I_1$ 7
8 $\left({p \implies q}\right) \lor \left({q \implies p}\right)$ Proof by Cases: $\text{PBC}$ 1, 2 – 4, 5 – 7 Assumptions 2 and 5 have been discharged

$\blacksquare$


Proof 2

We apply the Method of Truth Tables to the proposition.

As can be seen by inspection, the truth values under the main connective is true for all boolean interpretations, proving a tautology.


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

$\blacksquare$

Law of the Excluded Middle

This theorem depends on the Law of the Excluded Middle.

This is one of the axioms of logic that was determined by Aristotle, and forms part of the backbone of classical (Aristotelian) logic.

However, the intuitionist school rejects the Law of the Excluded Middle as a valid logical axiom. This in turn invalidates this theorem from an intuitionistic perspective.


Also see


Sources