Conditional is Left Distributive over Conjunction/Formulation 2
Theorem
- $\vdash \paren {p \implies \paren {q \land r} } \iff \paren {\paren {p \implies q} \land \paren {p \implies r} }$
This can be expressed as two separate theorems:
Forward Implication
- $\vdash \paren {p \implies \paren {q \land r} } \implies \paren {\paren {p \implies q} \land \paren {p \implies r} }$
Reverse Implication
- $\vdash \paren {\paren {p \implies q} \land \paren {p \implies r} } \implies \paren {p \implies \paren {q \land r} }$
Proof 1
Proof of Forward Implication
Let us use the following abbreviations
| \(\ds \phi\) | \(\text {for}\) | \(\ds p \implies \paren {q \land r}\) | ||||||||||||
| \(\ds \psi\) | \(\text {for}\) | \(\ds \paren {p \implies q} \land \paren {p \implies r}\) |
By the tableau method of natural deduction:
| Line | Pool | Formula | Rule | Depends upon | Notes | |
|---|---|---|---|---|---|---|
| 1 | 1 | $\phi$ | Assumption | (None) | ||
| 2 | 1 | $\psi$ | Sequent Introduction | 1 | Conditional is Left Distributive over Conjunction: Formulation 1 | |
| 3 | $\phi \implies \psi$ | Rule of Implication: $\implies \II$ | 1 – 2 | Assumption 1 has been discharged |
Expanding the abbreviations leads us back to:
- $\paren {p \implies \paren {q \land r} } \implies \paren {\paren {p \implies q} \land \paren {p \implies r} }$
$\blacksquare$
Proof of Reverse Implication
Let us use the following abbreviations
| \(\ds \phi\) | \(\text {for}\) | \(\ds \paren {p \implies q} \land \paren {p \implies r}\) | ||||||||||||
| \(\ds \psi\) | \(\text {for}\) | \(\ds p \implies \paren {q \land r}\) |
By the tableau method of natural deduction:
| Line | Pool | Formula | Rule | Depends upon | Notes | |
|---|---|---|---|---|---|---|
| 1 | 1 | $\phi$ | Assumption | (None) | ||
| 2 | 1 | $\psi$ | Sequent Introduction | 1 | Conditional is Left Distributive over Conjunction: Formulation 1 | |
| 3 | $\phi \implies \psi$ | Rule of Implication: $\implies \II$ | 1 – 2 | Assumption 1 has been discharged |
Expanding the abbreviations leads us back to:
- $\paren {\paren {p \implies q} \land \paren {p \implies r} } \implies \paren {p \implies \paren {q \land r} }$
$\blacksquare$
Proof by Truth Table
We apply the Method of Truth Tables.
As can be seen by inspection, the truth values under the main connective is true for all boolean interpretations.
$\begin{array}{|ccccc|c|ccccccc|} \hline (p & \implies & (q & \land & r)) & \iff & ((p & \implies & q) & \land & (p & \implies & r)) \\ \hline \F & \T & \F & \F & \F & \T & \F & \T & \F & \T & \F & \T & \F \\ \F & \T & \F & \F & \T & \T & \F & \T & \F & \T & \F & \T & \T \\ \F & \T & \T & \F & \F & \T & \F & \T & \T & \T & \F & \T & \F \\ \F & \T & \T & \T & \T & \T & \F & \T & \T & \T & \F & \T & \T \\ \T & \F & \F & \F & \F & \T & \T & \F & \F & \F & \T & \F & \F \\ \T & \F & \F & \F & \T & \T & \T & \F & \F & \F & \T & \T & \T \\ \T & \F & \T & \F & \F & \T & \T & \T & \T & \F & \T & \F & \F \\ \T & \T & \T & \T & \T & \T & \T & \T & \T & \T & \T & \T & \T \\ \hline \end{array}$
$\blacksquare$
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 $\text{T29}$