Theorem hadbi 1387

Description: The half adder is the same as the triple biconditional. (Contributed by Mario Carneiro, 4-Sep-2016.)

Assertion

Ref	Expression
hadbi	|- (hadd(ph, ps, ch) <-> ((ph <-> ps) <-> ch))

Proof of Theorem hadbi

Step	Hyp	Ref	Expression
1		df-xor 1305	. 2 |- (((ph \/_ ps) \/_ ch) <-> -. ((ph \/_ ps) <-> ch))
2		df-had 1380	. 2 |- (hadd(ph, ps, ch) <-> ((ph \/_ ps) \/_ ch))
3		xnor 1306	. . . 4 |- ((ph <-> ps) <-> -. (ph \/_ ps))
4	3	bibi1i 305	. . 3 |- (((ph <-> ps) <-> ch) <-> (-. (ph \/_ ps) <-> ch))
5		nbbn 347	. . 3 |- ((-. (ph \/_ ps) <-> ch) <-> -. ((ph \/_ ps) <-> ch))
6	4, 5	bitri 240	. 2 |- (((ph <-> ps) <-> ch) <-> -. ((ph \/_ ps) <-> ch))
7	1, 2, 6	3bitr4i 268	1 |- (hadd(ph, ps, ch) <-> ((ph <-> ps) <-> ch))

Syntax hints:  -. wn 3   <-> wb 176   \/_ wxo 1304  haddwhad 1378

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8

This theorem depends on definitions:  df-bi 177  df-xor 1305  df-had 1380

This theorem is referenced by:  had1  1402