Self-Distributive Law for Conditional/Formulation 1/Proof 1

Theorem

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

Proof

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

$\blacksquare$