Biconditional with Contradiction
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Theorem
A biconditional with a contradiction:
- $p \iff \bot \dashv \vdash \neg p$
Proof by Natural Deduction
By the tableau method of natural deduction:
Line | Pool | Formula | Rule | Depends upon | Notes | |
---|---|---|---|---|---|---|
1 | 1 | $p \iff \bot$ | Premise | (None) | ||
2 | 1 | $p \implies \bot$ | Biconditional Elimination: $\iff \EE_1$ | 1 | ||
3 | 1 | $\neg p$ | Sequent Introduction | 2 | Contradictory Consequent |
$\Box$
By the tableau method of natural deduction:
Line | Pool | Formula | Rule | Depends upon | Notes | |
---|---|---|---|---|---|---|
1 | 1 | $\neg p$ | Assumption | (None) | ||
2 | 1 | $p \implies \bot$ | Sequent Introduction | 1 | Contradictory Consequent | |
3 | $\top$ | Rule of Top-Introduction: $\top \II$ | (None) | |||
4 | $\bot \implies p$ | Sequent Introduction | 3 | Contradictory Antecedent | ||
5 | 1 | $p \iff \bot$ | Biconditional Introduction: $\iff \II$ | 2, 4 |
$\blacksquare$
Proof by Truth Table
We apply the Method of Truth Tables to the proposition.
As can be seen by inspection, in each case, the truth values in the appropriate columns match for all boolean interpretations.
$\begin{array}{|c|ccc||cc|} \hline p & p & \iff & \bot & \neg & p \\ \hline F & F & T & F & T & F \\ T & T & F & F & F & T \\ \hline \end{array}$
$\blacksquare$
Sources
- 2012: M. Ben-Ari: Mathematical Logic for Computer Science (3rd ed.) ... (previous) ... (next): $\S 2.3.3$