False Statement implies Every Statement/Formulation 2
Theorem
- $\vdash \neg p \implies \paren {p \implies q}$
Proof 1
By the tableau method of natural deduction:
| Line | Pool | Formula | Rule | Depends upon | Notes | |
|---|---|---|---|---|---|---|
| 1 | 1 | $\neg p$ | Assumption | (None) | ||
| 2 | 2 | $p$ | Assumption | (None) | ||
| 3 | 1, 2 | $\bot$ | Principle of Non-Contradiction: $\neg \EE$ | 2, 1 | ||
| 4 | 1, 2 | $q$ | Rule of Explosion: $\bot \EE$ | 3 | ||
| 5 | 1 | $p \implies q$ | Rule of Implication: $\implies \II$ | 2 – 4 | Assumption 2 has been discharged | |
| 6 | $\neg p \implies \left({p \implies q}\right)$ | Rule of Implication: $\implies \II$ | 1 – 5 | Assumption 1 has been discharged |
$\blacksquare$
Proof 2
By the tableau method of natural deduction:
| Line | Pool | Formula | Rule | Depends upon | Notes | |
|---|---|---|---|---|---|---|
| 1 | 1 | $\neg p$ | Assumption | (None) | ||
| 2 | 1 | $p \implies q$ | Sequent Introduction | 1 | False Statement implies Every Statement: Formulation 1 | |
| 3 | $\neg p \implies \left({p \implies q}\right)$ | Rule of Implication: $\implies \II$ | 1 – 2 | Assumption 1 has been discharged |
$\blacksquare$
Sources
- 1946: Alfred Tarski: Introduction to Logic and to the Methodology of Deductive Sciences (2nd ed.) ... (previous) ... (next): $\S \text{II}.13$: Symbolism of sentential calculus
- 1964: Donald Kalish and Richard Montague: Logic: Techniques of Formal Reasoning ... (previous) ... (next): $\text{I}$: 'NOT' and 'IF': $\S 5$: Theorem $\text{T18}$
- 1973: Irving M. Copi: Symbolic Logic (4th ed.) ... (previous) ... (next): $2$ Arguments Containing Compound Statements: $2.4$: Statement Forms