Self-Distributive Law for Conditional/Formulation 1

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Theorem

The following is known as the Self-Distributive Law:

$p \implies \paren {q \implies r} \dashv \vdash \paren {p \implies q} \implies \paren {p \implies r}$


This can of course be expressed as two separate theorems:

Forward Implication

$p \implies \paren {q \implies r} \vdash \paren {p \implies q} \implies \paren {p \implies r}$

Reverse Implication

$\paren {p \implies q} \implies \paren {p \implies r} \vdash p \implies \paren {q \implies r}$


Proof 1

Proof of Forward Implication

By the tableau method of natural deduction:

$p \implies \paren {q \implies r} \vdash \paren {p \implies q} \implies \paren {p \implies r} $
Line Pool Formula Rule Depends upon Notes
1 1 $p \implies \paren {q \implies r}$ Premise (None)
2 2 $p \implies q$ Assumption (None)
3 3 $p$ Assumption (None)
4 1, 3 $q \implies r$ Modus Ponendo Ponens: $\implies \mathcal E$ 1, 3
5 2, 3 $q$ Modus Ponendo Ponens: $\implies \mathcal E$ 2, 3
6 1, 2, 3 $r$ Modus Ponendo Ponens: $\implies \mathcal E$ 4, 5
7 1, 2 $p \implies r$ Rule of Implication: $\implies \II$ 3 – 6 Assumption 3 has been discharged
8 1 $\paren {p \implies q} \implies \paren {p \implies r}$ Rule of Implication: $\implies \II$ 2 – 7 Assumption 2 has been discharged

$\blacksquare$


Proof of Reverse Implication

By the tableau method of natural deduction:

$\paren {p \implies q} \implies \paren {p \implies r} \vdash p \implies \paren {q \implies r} $
Line Pool Formula Rule Depends upon Notes
1 1 $\paren {p \implies q} \implies \paren {p \implies r}$ Premise (None)
2 2 $p$ Assumption (None)
3 3 $q$ Assumption (None)
4 3 $p \implies q$ Sequent Introduction 3 True Statement is implied by Every Statement
5 1, 3 $p \implies r$ Modus Ponendo Ponens: $\implies \mathcal E$ 1, 4
6 1, 2, 3 $r$ Modus Ponendo Ponens: $\implies \mathcal E$ 5, 2
7 1, 2 $q \implies r$ Rule of Implication: $\implies \II$ 3 – 6 Assumption 3 has been discharged
8 1 $p \implies \paren {q \implies r}$ Rule of Implication: $\implies \II$ 2 – 7 Assumption 2 has been discharged

$\blacksquare$


Proof by Truth Table

We apply the Method of Truth Tables to the proposition: $p \implies \paren {q \implies r} \dashv \vdash \paren {p \implies q} \implies \paren {p \implies r}$

As can be seen by inspection, the truth values under the main connectives match for all boolean interpretations.

$\begin{array}{|ccccc||ccccccc|} \hline p & \implies & (q & \implies & r) & (p & \implies & q) & \implies & (p & \implies & r) \\ \hline \F & \T & \F & \T & \F & \F & \T & \F & \T & \F & \T & \F \\ \F & \T & \F & \T & \T & \F & \T & \F & \T & \F & \T & \T \\ \F & \T & \T & \F & \F & \F & \T & \T & \T & \F & \T & \F \\ \F & \T & \T & \T & \T & \F & \T & \T & \T & \F & \T & \T \\ \T & \T & \F & \T & \F & \T & \F & \F & \T & \T & \F & \F \\ \T & \T & \F & \T & \T & \T & \F & \F & \T & \T & \T & \T \\ \T & \F & \T & \F & \F & \T & \T & \T & \F & \T & \F & \F \\ \T & \T & \T & \T & \T & \T & \T & \T & \T & \T & \T & \T \\ \hline \end{array}$

$\blacksquare$


Also see

$\paren {p \implies q} \implies r \vdash \paren {p \implies r} \implies \paren {q \implies r}$

it is not the case that:

$\paren {p \implies r} \implies \paren {q \implies r} \not \vdash \paren {p \implies q} \implies r$