Rule of Implication/Sequent Form
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Theorem
The Rule of Implication can be symbolised by the sequent:
\(\ds \paren {p \vdash q}\) | \(\) | \(\ds \) | ||||||||||||
\(\ds \vdash \ \ \) | \(\ds p \implies q\) | \(\) | \(\ds \) |
Proof 1
By the tableau method of natural deduction:
Line | Pool | Formula | Rule | Depends upon | Notes | |
---|---|---|---|---|---|---|
1 | 1 | $p$ | Premise | (None) | ||
2 | 1 | $q$ | By hypothesis | 1 | as $p \vdash q$ | |
3 | 1 | $p \implies q$ | Rule of Implication: $\implies \II$ | 1 – 2 | Assumption 1 has been discharged |
$\blacksquare$
Proof by Truth Table
We apply the Method of Truth Tables.
$\begin{array}{|c|c||ccc|} \hline p & q & p & \implies & q\\ \hline \F & \F & \F & \T & \F \\ \F & \T & \F & \T & \T \\ \T & \F & \T & \F & \F \\ \T & \T & \T & \T & \T \\ \hline \end{array}$
As can be seen by inspection, only when $p$ is true and $q$ is false, then so is $p \implies q$ also false.
$\blacksquare$
Proof
- 1973: Irving M. Copi: Symbolic Logic (4th ed.) ... (previous) ... (next): $3$: The Method of Deduction: $3.5$: The Rule of Conditional Proof