True Statement is implied by Every Statement/Formulation 2
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Theorem
- $\vdash q \implies \paren {p \implies q}$
Proof 1
By the tableau method of natural deduction:
Line | Pool | Formula | Rule | Depends upon | Notes | |
---|---|---|---|---|---|---|
1 | 1 | $q$ | Assumption | (None) | ||
2 | 1 | $p \implies q$ | Sequent Introduction | 1 | True Statement is implied by Every Statement: Formulation 1 | |
3 | $q \implies \paren {p \implies q}$ | Rule of Implication: $\implies \II$ | 1 – 2 | Assumption 1 has been discharged |
$\blacksquare$
Proof by Truth Table
We apply the Method of Truth Tables.
As can be seen by inspection, the truth value under the main connective, the first instance of $\implies$, is $\T$ for each boolean interpretation.
$\begin{array}{|ccccc|} \hline q & \implies & ( p & \implies & q ) \\ \hline \F & \T & \T & \F & \F \\ \F & \T & \F & \T & \F \\ \T & \T & \T & \T & \T \\ \T & \T & \F & \T & \T \\ \hline \end{array}$
$\blacksquare$
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
- 1973: Irving M. Copi: Symbolic Logic (4th ed.) ... (previous) ... (next): $2$ Arguments Containing Compound Statements: $2.4$: Statement Forms
- 1982: P.M. Cohn: Algebra Volume 1 (2nd ed.) ... (previous) ... (next): Chapter $1$: Sets and mappings: $\S 1.1$: The need for logic: Exercise $(2) \ \text{(i)}$
- 2014: Christopher Clapham and James Nicholson: The Concise Oxford Dictionary of Mathematics (5th ed.) ... (previous) ... (next): tautology