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Theorem pm5.21nd 859
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 113 . 2  |-  ( ph  ->  ( ps  ->  th )
)
3 pm5.21nd.2 . . 3  |-  ( (
ph  /\  ch )  ->  th )
43ex 113 . 2  |-  ( ph  ->  ( ch  ->  th )
)
5 pm5.21nd.3 . . 3  |-  ( th 
->  ( ps  <->  ch )
)
65a1i 9 . 2  |-  ( ph  ->  ( th  ->  ( ps 
<->  ch ) ) )
72, 4, 6pm5.21ndd 654 1  |-  ( ph  ->  ( ps  <->  ch )
)
Colors of variables: wff set class
Syntax hints:    -> wi 4    /\ wa 102    <-> wb 103
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-ia2 105  ax-ia3 106
This theorem depends on definitions:  df-bi 115
This theorem is referenced by:  ideqg  4545  fvelimab  5305  releldm2  5890  relelec  6262  fzrev3  9394  elfzp12  9406
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