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Theorem had0 1502
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
StepHypRef Expression
1 had1 1501 . . 3  |-  ( -. 
ph  ->  (hadd ( -. 
ph ,  -.  ps ,  -.  ch )  <->  ( -.  ps 
<->  -.  ch ) ) )
2 hadnot 1500 . . 3  |-  ( -. hadd
( ph ,  ps ,  ch )  <-> hadd ( -.  ph ,  -.  ps ,  -.  ch ) )
3 xnor 1402 . . . 4  |-  ( ( ps  <->  ch )  <->  -.  ( ps  \/_  ch ) )
4 notbi 296 . . . 4  |-  ( ( ps  <->  ch )  <->  ( -.  ps 
<->  -.  ch ) )
53, 4bitr3i 254 . . 3  |-  ( -.  ( ps  \/_  ch ) 
<->  ( -.  ps  <->  -.  ch )
)
61, 2, 53bitr4g 291 . 2  |-  ( -. 
ph  ->  ( -. hadd ( ph ,  ps ,  ch ) 
<->  -.  ( ps  \/_  ch ) ) )
76con4bid 294 1  |-  ( -. 
ph  ->  (hadd ( ph ,  ps ,  ch )  <->  ( ps  \/_  ch )
) )
Colors of variables: wff setvar class
Syntax hints:   -. wn 3    -> wi 4    <-> wb 187    \/_ wxo 1400  haddwhad 1491
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 188  df-xor 1401  df-had 1492
This theorem is referenced by:  hadifp  1503  sadadd2lem2  14398  saddisjlem  14412
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