Exclusive Or is Commutative/Proof 1
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Theorem
- $p \oplus q \dashv \vdash q \oplus p$
Proof
By the tableau method of natural deduction:
| Line | Pool | Formula | Rule | Depends upon | Notes | |
|---|---|---|---|---|---|---|
| 1 | 1 | $p \oplus q$ | Premise | (None) | ||
| 2 | 1 | $\left({p \lor q} \right) \land \neg \left({p \land q}\right)$ | Sequent Introduction | 1 | Definition of Exclusive Or | |
| 3 | 1 | $\left({q \lor p} \right) \land \neg \left({p \land q}\right)$ | Sequent Introduction | 2 | Disjunction is Commutative | |
| 4 | 1 | $\left({q \lor p} \right) \land \neg \left({q \land p}\right)$ | Sequent Introduction | 3 | Conjunction is Commutative | |
| 5 | 1 | $q \oplus p$ | Sequent Introduction | 4 | Definition of Exclusive Or |
By the tableau method of natural deduction:
| Line | Pool | Formula | Rule | Depends upon | Notes | |
|---|---|---|---|---|---|---|
| 1 | 1 | $q \oplus p$ | Premise | (None) | ||
| 2 | 1 | $\left({q \lor p} \right) \land \neg \left({q \land p}\right)$ | Sequent Introduction | 1 | Definition of Exclusive Or | |
| 3 | 1 | $\left({q \lor p} \right) \land \neg \left({p \land q}\right)$ | Sequent Introduction | 2 | Conjunction is Commutative | |
| 4 | 1 | $\left({p \lor q} \right) \land \neg \left({p \land q}\right)$ | Sequent Introduction | 3 | Disjunction is Commutative | |
| 5 | 1 | $p \oplus q$ | Sequent Introduction | 4 | Definition of Exclusive Or |
$\blacksquare$