NAND is not Associative/Proof 1

Theorem

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

Proof

By the tableau method of natural deduction:

$\neg \paren {\paren {p \uparrow q} \uparrow r \implies p \uparrow \paren {q \uparrow r} } $

Line		Pool	Formula	Rule	Depends upon	Notes
1		1	$\paren {p \uparrow q} \uparrow r \implies p \uparrow \paren {q \uparrow r}$	Assumption	(None)
2		2	$p \land \neg r$	Assumption	(None)
3		2	$p$	Rule of Simplification: $\land \EE_1$	2
4		2	$\neg r$	Rule of Simplification: $\land \EE_2$	2
5		2	$\paren {\neg q} \lor \paren {\neg r}$	Rule of Addition: $\lor \II_2$	4
6		2	$\neg \paren {q \land r}$	Sequent Introduction	5	De Morgan's Laws: Disjunction of Negations
7		2	$q \uparrow r$	Sequent Introduction	6	Definition of Logical NAND
8		2	$p \land \paren {q \uparrow r}$	Rule of Conjunction: $\land \II$	3, 7
9		2	$\neg \neg \paren {p \land \paren {q \uparrow r} }$	Double Negation Introduction: $\neg \neg \II$	8
10		2	$\neg \paren {p \uparrow \paren {q \uparrow r} }$	Sequent Introduction	9	Definition of Logical NAND
11		2	$\paren {\neg \paren {p \uparrow q} } \lor \paren {\neg r}$	Rule of Addition: $\lor \II_2$	4
12		2	$\neg \paren {\paren {p \uparrow q} \land r}$	Sequent Introduction	11	De Morgan's Laws: Disjunction of Negations
13		2	$\paren {p \uparrow q} \uparrow r$	Sequent Introduction	12	Definition of Logical NAND
14		2	$\neg \neg \paren {\paren {p \uparrow q} \uparrow r}$	Double Negation Introduction: $\neg \neg \II$	13
15		2	$\neg \neg \paren {\paren {p \uparrow q} \uparrow r} \land \neg \paren {p \uparrow \paren {q \uparrow r} }$	Rule of Conjunction: $\land \II$	13, 10
16		2	$\neg \paren {\neg \paren {\paren {p \uparrow q} \uparrow r} \lor \paren {p \uparrow \paren {q \uparrow r} } }$	Sequent Introduction	15	De Morgan's Laws: Conjunction of Negations
17		2	$\neg \paren {\paren {p \uparrow q} \uparrow r \implies p \uparrow \paren {q \uparrow r} }$	Sequent Introduction	16	Disjunction and Conditional
18		1, 2	$\bot$	Principle of Non-Contradiction: $\neg \EE$	1, 17

Taking $p = \top$, $r = \bot$, we find $\vdash p \land \neg r$, and conclude our initial assumption was false.

$\blacksquare$