Non-Equivalence as Conjunction of Disjunction with Negation of Conjunction
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Theorem
- $\neg \left ({p \iff q}\right) \dashv \vdash \left({p \lor q} \right) \land \neg \left({p \land q}\right)$
That is, negation of the biconditional means the same thing as either-or but not both, that is, exclusive or.
Proof 1
By the tableau method of natural deduction:
Line | Pool | Formula | Rule | Depends upon | Notes | |
---|---|---|---|---|---|---|
1 | 1 | $\neg \left ({p \iff q}\right)$ | Premise | (None) | ||
2 | 1 | $\left({\neg p \land q}\right) \lor \left({p \land \neg q}\right)$ | Sequent Introduction | 1 | Non-Equivalence as Disjunction of Conjunctions: Formulation 1 | |
3 | 1 | $\left({p \land \neg q}\right) \lor \left({\neg p \land q}\right)$ | Sequent Introduction | 2 | Disjunction is Commutative | |
4 | 1 | $\left({p \land \neg q}\right) \lor \left({q \land \neg p}\right)$ | Sequent Introduction | 3 | Conjunction is Commutative | |
5 | 1 | $\left ({\left({p \lor q}\right) \land \neg q}\right) \lor \left({\left({q \lor p}\right) \land \neg p}\right)$ | Sequent Introduction | 4 | Conjunction of Disjunction with Negation is Conjunction with Negation | |
6 | 1 | $\left ({\left({p \lor q}\right) \land \neg q}\right) \lor \left({\left({p \lor q}\right) \land \neg p}\right)$ | Sequent Introduction | 5 | Disjunction is Commutative | |
7 | 1 | $\left({p \lor q}\right) \land \left({\neg q \lor \neg p}\right)$ | Sequent Introduction | 6 | Conjunction Distributes over Disjunction | |
8 | 1 | $\left({p \lor q}\right) \land \left({\neg p \lor \neg q}\right)$ | Sequent Introduction | 7 | Disjunction is Commutative | |
9 | 1 | $\left({p \lor q}\right) \land \neg \left({\neg \neg p \land \neg \neg q}\right)$ | Sequent Introduction | 8 | De Morgan's Laws: Disjunction | |
10 | 1 | $\left({p \lor q}\right) \land \neg \left({p \land q}\right)$ | Double Negation Elimination: $\neg \neg \mathcal E$ | 9 |
By the tableau method of natural deduction:
Line | Pool | Formula | Rule | Depends upon | Notes | |
---|---|---|---|---|---|---|
1 | 1 | $\left({p \lor q}\right) \land \neg \left({p \land q}\right)$ | Premise | (None) | ||
2 | 1 | $\left({p \lor q}\right) \land \left({\neg p \lor \neg q}\right)$ | Sequent Introduction | 1 | De Morgan's Laws: Disjunction of Negations | |
3 | 1 | $\left({p \lor q}\right) \land \left({\neg p \lor \neg q}\right)$ | Sequent Introduction | 2 | Disjunction is Commutative | |
4 | 1 | $\left ({\left({p \lor q}\right) \land \neg q}\right) \lor \left({\left({p \lor q}\right) \land \neg p}\right)$ | Sequent Introduction | 3 | Conjunction Distributes over Disjunction | |
5 | 1 | $\left({p \land \neg q}\right) \lor \left({q \land \neg p}\right)$ | Sequent Introduction | 4 | Conjunction of Disjunction with Negation is Conjunction with Negation | |
6 | 1 | $\left({q \land \neg p}\right) \lor \left({p \land \neg q}\right)$ | Sequent Introduction | 5 | Disjunction is Commutative | |
7 | 1 | $\left({\neg p \land q}\right) \lor \left({p \land \neg q}\right)$ | Sequent Introduction | 6 | Conjunction is Commutative | |
8 | 1 | $\neg \left ({p \iff q}\right)$ | Sequent Introduction | 6 | Non-Equivalence as Disjunction of Conjunctions: Formulation 1 |
$\blacksquare$
Proof 2
We apply the Method of Truth Tables.
As can be seen by inspection, the truth values under the main connectives match for all boolean interpretations.
$\begin{array}{|cccc||cccccccc|} \hline
\neg & (p & \iff & q) & (p & \lor & q) & \land & \neg & (p & \land & q) \\
\hline
F & F & T & F & F & F & F & F & T & F & F & F \\
T & F & F & T & F & T & T & T & T & F & F & T \\
T & T & F & F & T & T & F & T & T & T & F & F \\
F & T & T & T & T & T & T & F & F & T & T & T \\
\hline
\end{array}$
$\blacksquare$