Non-Equivalence as Conjunction of Disjunction with Negation of Conjunction
Theorem
- $\neg \paren {p \iff q} \dashv \vdash \paren {p \lor q} \land \neg \paren {p \land q}$
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 | $\neg \paren {p \iff q}$ | Premise | (None) | ||
2 | 1 | $\paren {\neg p \land q} \lor \paren {p \land \neg q}$ | Sequent Introduction | 1 | Non-Equivalence as Disjunction of Conjunctions: Formulation 1 | |
3 | 1 | $\paren {p \land \neg q} \lor \paren {\neg p \land q}$ | Sequent Introduction | 2 | Disjunction is Commutative | |
4 | 1 | $\paren {p \land \neg q} \lor \paren {q \land \neg p}$ | Sequent Introduction | 3 | Conjunction is Commutative | |
5 | 1 | $\paren {\paren {p \lor q} \land \neg q} \lor \paren {\paren {q \lor p} \land \neg p}$ | Sequent Introduction | 4 | Conjunction of Disjunction with Negation is Conjunction with Negation | |
6 | 1 | $\paren {\paren {p \lor q} \land \neg q} \lor \paren {\paren {p \lor q} \land \neg p}$ | Sequent Introduction | 5 | Disjunction is Commutative | |
7 | 1 | $\paren {p \lor q} \land \paren {\neg q \lor \neg p}$ | Sequent Introduction | 6 | Conjunction Distributes over Disjunction | |
8 | 1 | $\paren {p \lor q} \land \paren {\neg p \lor \neg q}$ | Sequent Introduction | 7 | Disjunction is Commutative | |
9 | 1 | $\paren {p \lor q} \land \neg \paren {\neg \neg p \land \neg \neg q}$ | Sequent Introduction | 8 | De Morgan's Laws: Disjunction | |
10 | 1 | $\paren {p \lor q} \land \neg \paren {p \land q}$ | Double Negation Elimination: $\neg \neg \EE$ | 9 |
By the tableau method of natural deduction:
Line | Pool | Formula | Rule | Depends upon | Notes | |
---|---|---|---|---|---|---|
1 | 1 | $\paren {p \lor q} \land \neg \paren {p \land q}$ | Premise | (None) | ||
2 | 1 | $\paren {p \lor q} \land \paren {\neg p \lor \neg q}$ | Sequent Introduction | 1 | De Morgan's Laws: Disjunction of Negations | |
3 | 1 | $\paren {p \lor q} \land \paren {\neg p \lor \neg q}$ | Sequent Introduction | 2 | Disjunction is Commutative | |
4 | 1 | $\paren {\paren {p \lor q} \land \neg q} \lor \paren {\paren {p \lor q} \land \neg p}$ | Sequent Introduction | 3 | Conjunction Distributes over Disjunction | |
5 | 1 | $\paren {p \land \neg q} \lor \paren {q \land \neg p}$ | Sequent Introduction | 4 | Conjunction of Disjunction with Negation is Conjunction with Negation | |
6 | 1 | $\paren {q \land \neg p} \lor \paren {p \land \neg q}$ | Sequent Introduction | 5 | Disjunction is Commutative | |
7 | 1 | $\paren {\neg p \land q} \lor \paren {p \land \neg q}$ | Sequent Introduction | 6 | Conjunction is Commutative | |
8 | 1 | $\neg \paren {p \iff q}$ | Sequent Introduction | 6 | Non-Equivalence as Disjunction of Conjunctions: Formulation 1 |
$\blacksquare$
Proof by Truth Table
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.
$\begin{array}{|cccc||cccccccc|} \hline
\neg & (p & \iff & q) & (p & \lor & q) & \land & \neg & (p & \land & q) \\
\hline
\F & \F & \T & \F & \F & \F & \F & \F & \T & \F & \F & \F \\
\T & \F & \F & \T & \F & \T & \T & \T & \T & \F & \F & \T \\
\T & \T & \F & \F & \T & \T & \F & \T & \T & \T & \F & \F \\
\F & \T & \T & \T & \T & \T & \T & \F & \F & \T & \T & \T \\
\hline
\end{array}$
$\blacksquare$