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Theorem nanbi 1388
Description: Show equivalence between the biconditional and the Nicod version. (Contributed by Jeff Hoffman, 19-Nov-2007.) (Proof shortened by Wolf Lammen, 27-Jun-2020.)
Assertion
Ref Expression
nanbi  |-  ( (
ph 
<->  ps )  <->  ( ( ph  -/\  ps )  -/\  ( ( ph  -/\  ph )  -/\  ( ps  -/\  ps )
) ) )

Proof of Theorem nanbi
StepHypRef Expression
1 dfbi3 901 . . 3  |-  ( (
ph 
<->  ps )  <->  ( ( ph  /\  ps )  \/  ( -.  ph  /\  -.  ps ) ) )
2 df-or 371 . . 3  |-  ( ( ( ph  /\  ps )  \/  ( -.  ph 
/\  -.  ps )
)  <->  ( -.  ( ph  /\  ps )  -> 
( -.  ph  /\  -.  ps ) ) )
3 df-nan 1380 . . . . 5  |-  ( (
ph  -/\  ps )  <->  -.  ( ph  /\  ps ) )
43bicomi 205 . . . 4  |-  ( -.  ( ph  /\  ps ) 
<->  ( ph  -/\  ps )
)
5 nannot 1387 . . . . 5  |-  ( -. 
ph 
<->  ( ph  -/\  ph )
)
6 nannot 1387 . . . . 5  |-  ( -. 
ps 
<->  ( ps  -/\  ps )
)
75, 6anbi12i 701 . . . 4  |-  ( ( -.  ph  /\  -.  ps ) 
<->  ( ( ph  -/\  ph )  /\  ( ps  -/\  ps )
) )
84, 7imbi12i 327 . . 3  |-  ( ( -.  ( ph  /\  ps )  ->  ( -. 
ph  /\  -.  ps )
)  <->  ( ( ph  -/\ 
ps )  ->  (
( ph  -/\  ph )  /\  ( ps  -/\  ps )
) ) )
91, 2, 83bitri 274 . 2  |-  ( (
ph 
<->  ps )  <->  ( ( ph  -/\  ps )  -> 
( ( ph  -/\  ph )  /\  ( ps  -/\  ps )
) ) )
10 nannan 1384 . 2  |-  ( ( ( ph  -/\  ps )  -/\  ( ( ph  -/\  ph )  -/\  ( ps  -/\  ps )
) )  <->  ( ( ph  -/\  ps )  -> 
( ( ph  -/\  ph )  /\  ( ps  -/\  ps )
) ) )
119, 10bitr4i 255 1  |-  ( (
ph 
<->  ps )  <->  ( ( ph  -/\  ps )  -/\  ( ( ph  -/\  ph )  -/\  ( ps  -/\  ps )
) ) )
Colors of variables: wff setvar class
Syntax hints:   -. wn 3    -> wi 4    <-> wb 187    \/ wo 369    /\ wa 370    -/\ wnan 1379
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-or 371  df-an 372  df-nan 1380
This theorem is referenced by:  nic-dfim  1548  nic-dfneg  1549
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