Theorem andnand1 14148

Description: Double and in terms of double nand.

Assertion

Ref	Expression
andnand1	|- ((ph /\ ps /\ ch) <-> ((ph -/\ ps -/\ ch) -/\ (ph -/\ ps -/\ ch)))

Proof of Theorem andnand1

Step	Hyp	Ref	Expression
1		3anass 862	. . 3 |- ((ph /\ ps /\ ch) <-> (ph /\ (ps /\ ch)))
2		pm4.63 245	. . . 4 |- (-. (ps -> -. ch) <-> (ps /\ ch))
3	2	anbi2i 538	. . 3 |- ((ph /\ -. (ps -> -. ch)) <-> (ph /\ (ps /\ ch)))
4		annim 257	. . 3 |- ((ph /\ -. (ps -> -. ch)) <-> -. (ph -> (ps -> -. ch)))
5	1, 3, 4	3bitr2i 196	. 2 |- ((ph /\ ps /\ ch) <-> -. (ph -> (ps -> -. ch)))
6		df-3nand 14145	. . 3 |- ((ph -/\ ps -/\ ch) <-> (ph -> (ps -> -. ch)))
7	6	notbii 204	. 2 |- (-. (ph -/\ ps -/\ ch) <-> -. (ph -> (ps -> -. ch)))
8		nic-justneg 1233	. 2 |- (-. (ph -/\ ps -/\ ch) <-> ((ph -/\ ps -/\ ch) -/\ (ph -/\ ps -/\ ch)))
9	5, 7, 8	3bitr2i 196	1 |- ((ph /\ ps /\ ch) <-> ((ph -/\ ps -/\ ch) -/\ (ph -/\ ps -/\ ch)))

Syntax hints:  -. wn 2   -> wi 3   <-> wb 163   /\ wa 240   /\ w3a 858   -/\ wnand 1229   -/\ w3nand 14144

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7

This theorem depends on definitions:  df-bi 164  df-an 242  df-3an 860  df-nand 1230  df-3nand 14145

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