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Theorem pm5.21nd 901
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 114 . 2  |-  ( ph  ->  ( ps  ->  th )
)
3 pm5.21nd.2 . . 3  |-  ( (
ph  /\  ch )  ->  th )
43ex 114 . 2  |-  ( ph  ->  ( ch  ->  th )
)
5 pm5.21nd.3 . . 3  |-  ( th 
->  ( ps  <->  ch )
)
65a1i 9 . 2  |-  ( ph  ->  ( th  ->  ( ps 
<->  ch ) ) )
72, 4, 6pm5.21ndd 694 1  |-  ( ph  ->  ( ps  <->  ch )
)
Colors of variables: wff set class
Syntax hints:    -> wi 4    /\ wa 103    <-> wb 104
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107
This theorem depends on definitions:  df-bi 116
This theorem is referenced by:  ideqg  4690  fvelimab  5477  releldm2  6083  relelec  6469  fzrev3  9879  elfzp12  9891  eltg  12235  eltg2  12236  cncnp2m  12414
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