Theorem had0 1543

Description: If the first input is false, then the adder sum is equivalent to the exclusive disjunction of the other two inputs. (Contributed by Mario Carneiro, 4-Sep-2016.) (Proof shortened by Wolf Lammen, 12-Jul-2020.)

Assertion

Ref	Expression
had0	|- ( -. ph -> (hadd ( ph , ps , ch ) <-> ( ps \/_ ch ) ) )

Proof of Theorem had0

Step	Hyp	Ref	Expression
1		had1 1542	. . 3 |- ( -. ph -> (hadd ( -. ph , -. ps , -. ch ) <-> ( -. ps <-> -. ch ) ) )
2		hadnot 1541	. . 3 |- ( -. hadd ( ph , ps , ch ) <-> hadd ( -. ph , -. ps , -. ch ) )
3		xnor 1466	. . . 4 |- ( ( ps <-> ch ) <-> -. ( ps \/_ ch ) )
4		notbi 309	. . . 4 |- ( ( ps <-> ch ) <-> ( -. ps <-> -. ch ) )
5	3, 4	bitr3i 266	. . 3 |- ( -. ( ps \/_ ch ) <-> ( -. ps <-> -. ch ) )
6	1, 2, 5	3bitr4g 303	. 2 |- ( -. ph -> ( -. hadd ( ph , ps , ch ) <-> -. ( ps \/_ ch ) ) )
7	6	con4bid 307	1 |- ( -. ph -> (hadd ( ph , ps , ch ) <-> ( ps \/_ ch ) ) )

Syntax hints:   -. wn 3   -> wi 4   <-> wb 196   \/_ wxo 1464  haddwhad 1532

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 197  df-xor 1465  df-had 1533

This theorem is referenced by:  hadifp  1544  sadadd2lem2  15172  saddisjlem  15186