Conjunction with Negative is Equivalent to Negation of Conditional/Formulation 1/Forward Implication
Jump to navigation
Jump to search
Theorem
- $p \land \neg q \vdash \neg \paren {p \implies q}$
Proof
By the tableau method of natural deduction:
| Line | Pool | Formula | Rule | Depends upon | Notes | |
|---|---|---|---|---|---|---|
| 1 | 1 | $p \land \neg q$ | Premise | (None) | ||
| 2 | 2 | $p \implies q$ | Assumption | (None) | Assume the opposite of what is to be proved ... | |
| 3 | 1 | $p$ | Rule of Simplification: $\land \EE_1$ | 1 | ||
| 4 | 1 | $\neg q$ | Rule of Simplification: $\land \EE_2$ | 1 | ||
| 5 | 1, 2 | $q$ | Modus Ponendo Ponens: $\implies \mathcal E$ | 2, 3 | ||
| 6 | 1, 2 | $\bot$ | Principle of Non-Contradiction: $\neg \EE$ | 5, 4 | ... and demonstrate a contradiction | |
| 7 | 1 | $\neg \paren {p \implies q}$ | Proof by Contradiction: $\neg \II$ | 2 – 6 | Assumption 2 has been discharged |
$\blacksquare$