Principle of Commutation/Formulation 1
Theorem
- $p \implies \paren {q \implies r} \dashv \vdash q \implies \paren {p \implies r}$
This can of course be expressed as two separate theorems:
Forward Implication
- $p \implies \left({q \implies r}\right) \vdash q \implies \left({p \implies r}\right)$
Reverse Implication
- $q \implies \paren {p \implies r} \vdash p \implies \paren {q \implies r}$
Proof 1
Proof of Forward Implication
By the tableau method of natural deduction:
Line | Pool | Formula | Rule | Depends upon | Notes | |
---|---|---|---|---|---|---|
1 | 1 | $p \implies \paren {q \implies r}$ | Premise | (None) | ||
2 | 2 | $q$ | Assumption | (None) | ||
3 | 3 | $p$ | Assumption | (None) | ||
4 | 1, 3 | $q \implies r$ | Modus Ponendo Ponens: $\implies \mathcal E$ | 1, 3 | ||
5 | 1, 2, 3 | $r$ | Modus Ponendo Ponens: $\implies \mathcal E$ | 2, 4 | ||
6 | 1, 2 | $p \implies r$ | Rule of Implication: $\implies \II$ | 3 – 5 | Assumption 3 has been discharged | |
7 | 1 | $q \implies \paren {p \implies r}$ | Rule of Implication: $\implies \II$ | 2 – 6 | Assumption 2 has been discharged |
$\blacksquare$
Proof of Reverse Implication
By the tableau method of natural deduction:
Line | Pool | Formula | Rule | Depends upon | Notes | |
---|---|---|---|---|---|---|
1 | 1 | $q \implies \left({p \implies r}\right)$ | Premise | (None) | ||
2 | 2 | $p$ | Assumption | (None) | ||
3 | 3 | $q$ | Assumption | (None) | ||
4 | 1, 3 | $p \implies r$ | Modus Ponendo Ponens: $\implies \mathcal E$ | 1, 3 | ||
5 | 1, 2, 3 | $r$ | Modus Ponendo Ponens: $\implies \mathcal E$ | 2, 4 | ||
6 | 1, 2 | $q \implies r$ | Rule of Implication: $\implies \II$ | 3 – 5 | Assumption 3 has been discharged | |
7 | 1 | $p \implies \left({q \implies r}\right)$ | Rule of Implication: $\implies \II$ | 2 – 6 | Assumption 2 has been discharged |
$\blacksquare$
Proof 2
We apply the Method of Truth Tables to the proposition.
As can be seen by inspection, the truth values under the main connectives match for all boolean interpretations.
$\begin{array}{|ccccc||ccccc|} \hline
p & \implies & (q & \implies & r) & q & \implies & (p & \implies & r) \\
\hline
F & T & F & T & F & F & T & F & T & F \\
F & T & F & T & T & F & T & F & T & T \\
F & T & T & F & F & T & T & F & T & F \\
F & T & T & T & T & T & T & F & T & T \\
T & T & F & T & F & F & T & T & F & F \\
T & T & F & T & T & F & T & T & T & T \\
T & F & T & F & F & T & F & T & F & F \\
T & T & T & T & T & T & T & T & T & T \\
\hline
\end{array}$
$\blacksquare$