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Theorem pm5.21nd 898
Description: Eliminate an antecedent implied by each side of a biconditional. (Contributed by NM, 20-Nov-2005.) (Proof shortened by Wolf Lammen, 4-Nov-2013.)
Hypotheses
Ref Expression
pm5.21nd.1  |-  ( (
ph  /\  ps )  ->  th )
pm5.21nd.2  |-  ( (
ph  /\  ch )  ->  th )
pm5.21nd.3  |-  ( th 
->  ( ps  <->  ch )
)
Assertion
Ref Expression
pm5.21nd  |-  ( ph  ->  ( ps  <->  ch )
)

Proof of Theorem pm5.21nd
StepHypRef Expression
1 pm5.21nd.1 . . 3  |-  ( (
ph  /\  ps )  ->  th )
21ex 434 . 2  |-  ( ph  ->  ( ps  ->  th )
)
3 pm5.21nd.2 . . 3  |-  ( (
ph  /\  ch )  ->  th )
43ex 434 . 2  |-  ( ph  ->  ( ch  ->  th )
)
5 pm5.21nd.3 . . 3  |-  ( th 
->  ( ps  <->  ch )
)
65a1i 11 . 2  |-  ( ph  ->  ( th  ->  ( ps 
<->  ch ) ) )
72, 4, 6pm5.21ndd 354 1  |-  ( ph  ->  ( ps  <->  ch )
)
Colors of variables: wff setvar class
Syntax hints:    -> wi 4    <-> wb 184    /\ wa 369
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 185  df-an 371
This theorem is referenced by:  ideqg  5160  fvelimab  5930  brrpssg  6577  ordsucelsuc  6652  releldm2  6845  relbrtpos  6978  relelec  7364  elfiun  7902  fpwwe2lem2  9022  fpwwelem  9035  fzrev3  11757  elfzp12  11769  eqgval  16122  eltg  19327  eltg2  19328  cncnp2  19650  isref  19878  islocfin  19886  isdivrngo  25256  isfne  30084  opelopab3  30134  islshpkrN  34318  dihatexv2  36537
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