Conjunction with Negative is Equivalent to Negation of Conditional/Formulation 2/Proof 1
Jump to navigation
Jump to search
Theorem
- $\vdash \paren {p \land \neg q} \iff \paren {\neg \paren {p \implies q} }$
Proof
By the tableau method of natural deduction:
| Line | Pool | Formula | Rule | Depends upon | Notes | |
|---|---|---|---|---|---|---|
| 1 | 1 | $p \land \neg q$ | Assumption | (None) | ||
| 2 | 1 | $\neg \paren {p \implies q}$ | Sequent Introduction | 1 | Conjunction with Negative is Equivalent to Negation of Conditional: Formulation 1: Forward Implication | |
| 3 | $\paren {p \land \neg q} \implies \paren {\neg \paren {p \implies q} }$ | Rule of Implication: $\implies \II$ | 1 – 2 | Assumption 1 has been discharged | ||
| 4 | 4 | $\neg \paren {p \implies q}$ | Assumption | (None) | ||
| 5 | 4 | $p \land \neg q$ | Sequent Introduction | 4 | Conjunction with Negative is Equivalent to Negation of Conditional: Formulation 1:Reverse Implication | |
| 6 | $\paren {\neg \paren {p \implies q} } \implies \paren {p \land \neg q}$ | Rule of Implication: $\implies \II$ | 4 – 5 | Assumption 4 has been discharged | ||
| 7 | $\paren {p \land \neg q} \iff \paren {\neg \paren {p \implies q} }$ | Biconditional Introduction: $\iff \II$ | 3, 6 |
$\blacksquare$
Sources
- 1964: Donald Kalish and Richard Montague: Logic: Techniques of Formal Reasoning ... (previous) ... (next): $\text{II}$: 'AND', 'OR', 'IF AND ONLY IF': $\S 3$: Theorem $\text{T40}$