NOR with Equal Arguments/Proof 1
Jump to navigation
Jump to search
Theorem
- $p \downarrow p \dashv \vdash \neg p$
That is, the NOR of a proposition with itself corresponds to the negation operator.
Proof
By the tableau method of natural deduction:
Line | Pool | Formula | Rule | Depends upon | Notes | |
---|---|---|---|---|---|---|
1 | 1 | $p \downarrow p$ | Premise | (None) | ||
2 | 1 | $\neg \paren {p \lor p}$ | Sequent Introduction | 1 | Definition of Logical NOR | |
3 | 1 | $\neg p$ | Sequent Introduction | 2 | Rule of Idempotence: Disjunction |
$\Box$
By the tableau method of natural deduction:
Line | Pool | Formula | Rule | Depends upon | Notes | |
---|---|---|---|---|---|---|
1 | 1 | $\neg p$ | Premise | (None) | ||
2 | 1 | $\neg \paren {p \lor p}$ | Sequent Introduction | 1 | Rule of Idempotence: Disjunction | |
3 | 1 | $p \downarrow p$ | Sequent Introduction | 2 | Definition of Logical NOR |
$\blacksquare$