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Theorem List for Metamath Proof Explorer - 38401-38500   *Has distinct variable group(s)
TypeLabelDescription
Statement
 
Theoremconfun4 38401 An attempt at derivative. Resisted simplest path to a proof. (Contributed by Jarvin Udandy, 6-Sep-2020.)
 |-  ph   &    |-  ( ( ph  ->  ps )  ->  ps )   &    |-  ( ps  ->  ( ph  ->  ch ) )   &    |-  ( ( ch 
 ->  th )  ->  (
 ( ph  ->  th )  <->  ps ) )   &    |-  ( ta  <->  ( ch  ->  th ) )   &    |-  ( et  <->  -.  ( ch  ->  ( ch  /\  -.  ch ) ) )   &    |-  ps   &    |-  ( ch  ->  th )   =>    |-  ( ch  ->  ( ps  ->  ta ) )
 
Theoremconfun5 38402 An attempt at derivative. Resisted simplest path to a proof. Interesting that ch, th, ta, et were all provable. (Contributed by Jarvin Udandy, 7-Sep-2020.)
 |-  ph   &    |-  ( ( ph  ->  ps )  ->  ps )   &    |-  ( ps  ->  ( ph  ->  ch ) )   &    |-  ( ( ch 
 ->  th )  ->  (
 ( ph  ->  th )  <->  ps ) )   &    |-  ( ta  <->  ( ch  ->  th ) )   &    |-  ( et  <->  -.  ( ch  ->  ( ch  /\  -.  ch ) ) )   &    |-  ps   &    |-  ( ch  ->  th )   =>    |-  ( ch  ->  ( et 
 <->  ta ) )
 
Theoremplcofph 38403 Given, a,b and a "definition" for c, c is demonstrated. (Contributed by Jarvin Udandy, 8-Sep-2020.)
 |-  ( ch 
 <->  ( ( ( (
 ph  /\  ps )  <->  ph )  ->  ( ph  /\ 
 -.  ( ph  /\  -.  ph ) ) )  /\  ( ph  /\  -.  ( ph  /\  -.  ph )
 ) ) )   &    |-  ph   &    |-  ps   =>    |- 
 ch
 
Theorempldofph 38404 Given, a,b c, d, "definition" for e, e is demonstrated. (Contributed by Jarvin Udandy, 8-Sep-2020.)
 |-  ( ta 
 <->  ( ( ch  ->  th )  /\  ( ph  <->  ch )  /\  ( ( ph  ->  ps )  ->  ( ps 
 <-> 
 th ) ) ) )   &    |-  ph   &    |-  ps   &    |-  ch   &    |-  th   =>    |- 
 ta
 
Theoremplvcofph 38405 Given, a,b,d, and "definitions" for c, e, f: f is demonstrated. (Contributed by Jarvin Udandy, 8-Sep-2020.)
 |-  ( ch 
 <->  ( ( ( (
 ph  /\  ps )  <->  ph )  ->  ( ph  /\ 
 -.  ( ph  /\  -.  ph ) ) )  /\  ( ph  /\  -.  ( ph  /\  -.  ph )
 ) ) )   &    |-  ( ta 
 <->  ( ( ch  ->  th )  /\  ( ph  <->  ch )  /\  ( ( ph  ->  ps )  ->  ( ps 
 <-> 
 th ) ) ) )   &    |-  ( et  <->  ( ch  /\  ta ) )   &    |-  ph   &    |-  ps   &    |-  th   =>    |- 
 et
 
Theoremplvcofphax 38406 Given, a,b,d, and "definitions" for c, e, f, g: g is demonstrated. (Contributed by Jarvin Udandy, 8-Sep-2020.)
 |-  ( ch 
 <->  ( ( ( (
 ph  /\  ps )  <->  ph )  ->  ( ph  /\ 
 -.  ( ph  /\  -.  ph ) ) )  /\  ( ph  /\  -.  ( ph  /\  -.  ph )
 ) ) )   &    |-  ( ta 
 <->  ( ( ch  ->  th )  /\  ( ph  <->  ch )  /\  ( ( ph  ->  ps )  ->  ( ps 
 <-> 
 th ) ) ) )   &    |-  ( et  <->  ( ch  /\  ta ) )   &    |-  ph   &    |-  ps   &    |-  th   &    |-  ( ze  <->  -.  ( ps  /\  -. 
 ta ) )   =>    |-  ze
 
Theoremplvofpos 38407 rh is derivable because ONLY one of ch, th, ta, et is implied by mu. (Contributed by Jarvin Udandy, 11-Sep-2020.)
 |-  ( ch 
 <->  ( -.  ph  /\  -.  ps ) )   &    |-  ( th  <->  ( -.  ph  /\ 
 ps ) )   &    |-  ( ta 
 <->  ( ph  /\  -.  ps ) )   &    |-  ( et  <->  ( ph  /\  ps ) )   &    |-  ( ze  <->  ( ( ( ( ( -.  (
 ( mu  ->  ch )  /\  ( mu  ->  th )
 )  /\  -.  (
 ( mu  ->  ch )  /\  ( mu  ->  ta )
 ) )  /\  -.  ( ( mu  ->  ch )  /\  ( ch 
 ->  et ) ) ) 
 /\  -.  ( ( mu  ->  th )  /\  ( mu  ->  ta ) ) ) 
 /\  -.  ( ( mu  ->  th )  /\  ( mu  ->  et ) ) )  /\  -.  (
 ( mu  ->  ta )  /\  ( mu  ->  et )
 ) ) )   &    |-  ( si 
 <->  ( ( ( mu 
 ->  ch )  \/  ( mu  ->  th ) )  \/  ( ( mu  ->  ta )  \/  ( mu 
 ->  et ) ) ) )   &    |-  ( rh  <->  ( ze  /\  si ) )   &    |-  ze   &    |-  si   =>    |- 
 rh
 
Theoremmdandyv0 38408 Given the equivalences set in the hypotheses, there exist a proof where ch, th, ta, et match ph, ps accordingly (Contributed by Jarvin Udandy, 6-Sep-2016.)
 |-  ( ph 
 <-> F.  )   &    |-  ( ps  <-> T.  )   &    |-  ( ch 
 <-> F.  )   &    |-  ( th  <-> F.  )   &    |-  ( ta 
 <-> F.  )   &    |-  ( et  <-> F.  )   =>    |-  ( ( ( ( ch  <->  ph )  /\  ( th 
 <-> 
 ph ) )  /\  ( ta  <->  ph ) )  /\  ( et  <->  ph ) )
 
Theoremmdandyv1 38409 Given the equivalences set in the hypotheses, there exist a proof where ch, th, ta, et match ph, ps accordingly (Contributed by Jarvin Udandy, 6-Sep-2016.)
 |-  ( ph 
 <-> F.  )   &    |-  ( ps  <-> T.  )   &    |-  ( ch 
 <-> T.  )   &    |-  ( th  <-> F.  )   &    |-  ( ta 
 <-> F.  )   &    |-  ( et  <-> F.  )   =>    |-  ( ( ( ( ch  <->  ps )  /\  ( th 
 <-> 
 ph ) )  /\  ( ta  <->  ph ) )  /\  ( et  <->  ph ) )
 
Theoremmdandyv2 38410 Given the equivalences set in the hypotheses, there exist a proof where ch, th, ta, et match ph, ps accordingly (Contributed by Jarvin Udandy, 6-Sep-2016.)
 |-  ( ph 
 <-> F.  )   &    |-  ( ps  <-> T.  )   &    |-  ( ch 
 <-> F.  )   &    |-  ( th  <-> T.  )   &    |-  ( ta 
 <-> F.  )   &    |-  ( et  <-> F.  )   =>    |-  ( ( ( ( ch  <->  ph )  /\  ( th 
 <->  ps ) )  /\  ( ta  <->  ph ) )  /\  ( et  <->  ph ) )
 
Theoremmdandyv3 38411 Given the equivalences set in the hypotheses, there exist a proof where ch, th, ta, et match ph, ps accordingly (Contributed by Jarvin Udandy, 6-Sep-2016.)
 |-  ( ph 
 <-> F.  )   &    |-  ( ps  <-> T.  )   &    |-  ( ch 
 <-> T.  )   &    |-  ( th  <-> T.  )   &    |-  ( ta 
 <-> F.  )   &    |-  ( et  <-> F.  )   =>    |-  ( ( ( ( ch  <->  ps )  /\  ( th 
 <->  ps ) )  /\  ( ta  <->  ph ) )  /\  ( et  <->  ph ) )
 
Theoremmdandyv4 38412 Given the equivalences set in the hypotheses, there exist a proof where ch, th, ta, et match ph, ps accordingly (Contributed by Jarvin Udandy, 6-Sep-2016.)
 |-  ( ph 
 <-> F.  )   &    |-  ( ps  <-> T.  )   &    |-  ( ch 
 <-> F.  )   &    |-  ( th  <-> F.  )   &    |-  ( ta 
 <-> T.  )   &    |-  ( et  <-> F.  )   =>    |-  ( ( ( ( ch  <->  ph )  /\  ( th 
 <-> 
 ph ) )  /\  ( ta  <->  ps ) )  /\  ( et  <->  ph ) )
 
Theoremmdandyv5 38413 Given the equivalences set in the hypotheses, there exist a proof where ch, th, ta, et match ph, ps accordingly (Contributed by Jarvin Udandy, 6-Sep-2016.)
 |-  ( ph 
 <-> F.  )   &    |-  ( ps  <-> T.  )   &    |-  ( ch 
 <-> T.  )   &    |-  ( th  <-> F.  )   &    |-  ( ta 
 <-> T.  )   &    |-  ( et  <-> F.  )   =>    |-  ( ( ( ( ch  <->  ps )  /\  ( th 
 <-> 
 ph ) )  /\  ( ta  <->  ps ) )  /\  ( et  <->  ph ) )
 
Theoremmdandyv6 38414 Given the equivalences set in the hypotheses, there exist a proof where ch, th, ta, et match ph, ps accordingly (Contributed by Jarvin Udandy, 6-Sep-2016.)
 |-  ( ph 
 <-> F.  )   &    |-  ( ps  <-> T.  )   &    |-  ( ch 
 <-> F.  )   &    |-  ( th  <-> T.  )   &    |-  ( ta 
 <-> T.  )   &    |-  ( et  <-> F.  )   =>    |-  ( ( ( ( ch  <->  ph )  /\  ( th 
 <->  ps ) )  /\  ( ta  <->  ps ) )  /\  ( et  <->  ph ) )
 
Theoremmdandyv7 38415 Given the equivalences set in the hypotheses, there exist a proof where ch, th, ta, et match ph, ps accordingly (Contributed by Jarvin Udandy, 6-Sep-2016.)
 |-  ( ph 
 <-> F.  )   &    |-  ( ps  <-> T.  )   &    |-  ( ch 
 <-> T.  )   &    |-  ( th  <-> T.  )   &    |-  ( ta 
 <-> T.  )   &    |-  ( et  <-> F.  )   =>    |-  ( ( ( ( ch  <->  ps )  /\  ( th 
 <->  ps ) )  /\  ( ta  <->  ps ) )  /\  ( et  <->  ph ) )
 
Theoremmdandyv8 38416 Given the equivalences set in the hypotheses, there exist a proof where ch, th, ta, et match ph, ps accordingly (Contributed by Jarvin Udandy, 6-Sep-2016.)
 |-  ( ph 
 <-> F.  )   &    |-  ( ps  <-> T.  )   &    |-  ( ch 
 <-> F.  )   &    |-  ( th  <-> F.  )   &    |-  ( ta 
 <-> F.  )   &    |-  ( et  <-> T.  )   =>    |-  ( ( ( ( ch  <->  ph )  /\  ( th 
 <-> 
 ph ) )  /\  ( ta  <->  ph ) )  /\  ( et  <->  ps ) )
 
Theoremmdandyv9 38417 Given the equivalences set in the hypotheses, there exist a proof where ch, th, ta, et match ph, ps accordingly (Contributed by Jarvin Udandy, 6-Sep-2016.)
 |-  ( ph 
 <-> F.  )   &    |-  ( ps  <-> T.  )   &    |-  ( ch 
 <-> T.  )   &    |-  ( th  <-> F.  )   &    |-  ( ta 
 <-> F.  )   &    |-  ( et  <-> T.  )   =>    |-  ( ( ( ( ch  <->  ps )  /\  ( th 
 <-> 
 ph ) )  /\  ( ta  <->  ph ) )  /\  ( et  <->  ps ) )
 
Theoremmdandyv10 38418 Given the equivalences set in the hypotheses, there exist a proof where ch, th, ta, et match ph, ps accordingly (Contributed by Jarvin Udandy, 6-Sep-2016.)
 |-  ( ph 
 <-> F.  )   &    |-  ( ps  <-> T.  )   &    |-  ( ch 
 <-> F.  )   &    |-  ( th  <-> T.  )   &    |-  ( ta 
 <-> F.  )   &    |-  ( et  <-> T.  )   =>    |-  ( ( ( ( ch  <->  ph )  /\  ( th 
 <->  ps ) )  /\  ( ta  <->  ph ) )  /\  ( et  <->  ps ) )
 
Theoremmdandyv11 38419 Given the equivalences set in the hypotheses, there exist a proof where ch, th, ta, et match ph, ps accordingly (Contributed by Jarvin Udandy, 6-Sep-2016.)
 |-  ( ph 
 <-> F.  )   &    |-  ( ps  <-> T.  )   &    |-  ( ch 
 <-> T.  )   &    |-  ( th  <-> T.  )   &    |-  ( ta 
 <-> F.  )   &    |-  ( et  <-> T.  )   =>    |-  ( ( ( ( ch  <->  ps )  /\  ( th 
 <->  ps ) )  /\  ( ta  <->  ph ) )  /\  ( et  <->  ps ) )
 
Theoremmdandyv12 38420 Given the equivalences set in the hypotheses, there exist a proof where ch, th, ta, et match ph, ps accordingly (Contributed by Jarvin Udandy, 6-Sep-2016.)
 |-  ( ph 
 <-> F.  )   &    |-  ( ps  <-> T.  )   &    |-  ( ch 
 <-> F.  )   &    |-  ( th  <-> F.  )   &    |-  ( ta 
 <-> T.  )   &    |-  ( et  <-> T.  )   =>    |-  ( ( ( ( ch  <->  ph )  /\  ( th 
 <-> 
 ph ) )  /\  ( ta  <->  ps ) )  /\  ( et  <->  ps ) )
 
Theoremmdandyv13 38421 Given the equivalences set in the hypotheses, there exist a proof where ch, th, ta, et match ph, ps accordingly (Contributed by Jarvin Udandy, 6-Sep-2016.)
 |-  ( ph 
 <-> F.  )   &    |-  ( ps  <-> T.  )   &    |-  ( ch 
 <-> T.  )   &    |-  ( th  <-> F.  )   &    |-  ( ta 
 <-> T.  )   &    |-  ( et  <-> T.  )   =>    |-  ( ( ( ( ch  <->  ps )  /\  ( th 
 <-> 
 ph ) )  /\  ( ta  <->  ps ) )  /\  ( et  <->  ps ) )
 
Theoremmdandyv14 38422 Given the equivalences set in the hypotheses, there exist a proof where ch, th, ta, et match ph, ps accordingly (Contributed by Jarvin Udandy, 6-Sep-2016.)
 |-  ( ph 
 <-> F.  )   &    |-  ( ps  <-> T.  )   &    |-  ( ch 
 <-> F.  )   &    |-  ( th  <-> T.  )   &    |-  ( ta 
 <-> T.  )   &    |-  ( et  <-> T.  )   =>    |-  ( ( ( ( ch  <->  ph )  /\  ( th 
 <->  ps ) )  /\  ( ta  <->  ps ) )  /\  ( et  <->  ps ) )
 
Theoremmdandyv15 38423 Given the equivalences set in the hypotheses, there exist a proof where ch, th, ta, et match ph, ps accordingly (Contributed by Jarvin Udandy, 6-Sep-2016.)
 |-  ( ph 
 <-> F.  )   &    |-  ( ps  <-> T.  )   &    |-  ( ch 
 <-> T.  )   &    |-  ( th  <-> T.  )   &    |-  ( ta 
 <-> T.  )   &    |-  ( et  <-> T.  )   =>    |-  ( ( ( ( ch  <->  ps )  /\  ( th 
 <->  ps ) )  /\  ( ta  <->  ps ) )  /\  ( et  <->  ps ) )
 
Theoremmdandyvr0 38424 Given the equivalences set in the hypotheses, there exist a proof where ch, th, ta, et match ze, si accordingly (Contributed by Jarvin Udandy, 7-Sep-2016.)
 |-  ( ph 
 <->  ze )   &    |-  ( ps  <->  si )   &    |-  ( ch  <->  ph )   &    |-  ( th  <->  ph )   &    |-  ( ta  <->  ph )   &    |-  ( et  <->  ph )   =>    |-  ( ( ( ( ch  <->  ze )  /\  ( th 
 <->  ze ) )  /\  ( ta  <->  ze ) )  /\  ( et  <->  ze ) )
 
Theoremmdandyvr1 38425 Given the equivalences set in the hypotheses, there exist a proof where ch, th, ta, et match ze, si accordingly (Contributed by Jarvin Udandy, 7-Sep-2016.)
 |-  ( ph 
 <->  ze )   &    |-  ( ps  <->  si )   &    |-  ( ch  <->  ps )   &    |-  ( th  <->  ph )   &    |-  ( ta  <->  ph )   &    |-  ( et  <->  ph )   =>    |-  ( ( ( ( ch  <->  si )  /\  ( th 
 <->  ze ) )  /\  ( ta  <->  ze ) )  /\  ( et  <->  ze ) )
 
Theoremmdandyvr2 38426 Given the equivalences set in the hypotheses, there exist a proof where ch, th, ta, et match ze, si accordingly (Contributed by Jarvin Udandy, 7-Sep-2016.)
 |-  ( ph 
 <->  ze )   &    |-  ( ps  <->  si )   &    |-  ( ch  <->  ph )   &    |-  ( th  <->  ps )   &    |-  ( ta  <->  ph )   &    |-  ( et  <->  ph )   =>    |-  ( ( ( ( ch  <->  ze )  /\  ( th 
 <-> 
 si ) )  /\  ( ta  <->  ze ) )  /\  ( et  <->  ze ) )
 
Theoremmdandyvr3 38427 Given the equivalences set in the hypotheses, there exist a proof where ch, th, ta, et match ze, si accordingly (Contributed by Jarvin Udandy, 7-Sep-2016.)
 |-  ( ph 
 <->  ze )   &    |-  ( ps  <->  si )   &    |-  ( ch  <->  ps )   &    |-  ( th  <->  ps )   &    |-  ( ta  <->  ph )   &    |-  ( et  <->  ph )   =>    |-  ( ( ( ( ch  <->  si )  /\  ( th 
 <-> 
 si ) )  /\  ( ta  <->  ze ) )  /\  ( et  <->  ze ) )
 
Theoremmdandyvr4 38428 Given the equivalences set in the hypotheses, there exist a proof where ch, th, ta, et match ze, si accordingly (Contributed by Jarvin Udandy, 7-Sep-2016.)
 |-  ( ph 
 <->  ze )   &    |-  ( ps  <->  si )   &    |-  ( ch  <->  ph )   &    |-  ( th  <->  ph )   &    |-  ( ta  <->  ps )   &    |-  ( et  <->  ph )   =>    |-  ( ( ( ( ch  <->  ze )  /\  ( th 
 <->  ze ) )  /\  ( ta  <->  si ) )  /\  ( et  <->  ze ) )
 
Theoremmdandyvr5 38429 Given the equivalences set in the hypotheses, there exist a proof where ch, th, ta, et match ze, si accordingly (Contributed by Jarvin Udandy, 7-Sep-2016.)
 |-  ( ph 
 <->  ze )   &    |-  ( ps  <->  si )   &    |-  ( ch  <->  ps )   &    |-  ( th  <->  ph )   &    |-  ( ta  <->  ps )   &    |-  ( et  <->  ph )   =>    |-  ( ( ( ( ch  <->  si )  /\  ( th 
 <->  ze ) )  /\  ( ta  <->  si ) )  /\  ( et  <->  ze ) )
 
Theoremmdandyvr6 38430 Given the equivalences set in the hypotheses, there exist a proof where ch, th, ta, et match ze, si accordingly (Contributed by Jarvin Udandy, 7-Sep-2016.)
 |-  ( ph 
 <->  ze )   &    |-  ( ps  <->  si )   &    |-  ( ch  <->  ph )   &    |-  ( th  <->  ps )   &    |-  ( ta  <->  ps )   &    |-  ( et  <->  ph )   =>    |-  ( ( ( ( ch  <->  ze )  /\  ( th 
 <-> 
 si ) )  /\  ( ta  <->  si ) )  /\  ( et  <->  ze ) )
 
Theoremmdandyvr7 38431 Given the equivalences set in the hypotheses, there exist a proof where ch, th, ta, et match ze, si accordingly (Contributed by Jarvin Udandy, 7-Sep-2016.)
 |-  ( ph 
 <->  ze )   &    |-  ( ps  <->  si )   &    |-  ( ch  <->  ps )   &    |-  ( th  <->  ps )   &    |-  ( ta  <->  ps )   &    |-  ( et  <->  ph )   =>    |-  ( ( ( ( ch  <->  si )  /\  ( th 
 <-> 
 si ) )  /\  ( ta  <->  si ) )  /\  ( et  <->  ze ) )
 
Theoremmdandyvr8 38432 Given the equivalences set in the hypotheses, there exist a proof where ch, th, ta, et match ze, si accordingly (Contributed by Jarvin Udandy, 7-Sep-2016.)
 |-  ( ph 
 <->  ze )   &    |-  ( ps  <->  si )   &    |-  ( ch  <->  ph )   &    |-  ( th  <->  ph )   &    |-  ( ta  <->  ph )   &    |-  ( et  <->  ps )   =>    |-  ( ( ( ( ch  <->  ze )  /\  ( th 
 <->  ze ) )  /\  ( ta  <->  ze ) )  /\  ( et  <->  si ) )
 
Theoremmdandyvr9 38433 Given the equivalences set in the hypotheses, there exist a proof where ch, th, ta, et match ze, si accordingly (Contributed by Jarvin Udandy, 7-Sep-2016.)
 |-  ( ph 
 <->  ze )   &    |-  ( ps  <->  si )   &    |-  ( ch  <->  ps )   &    |-  ( th  <->  ph )   &    |-  ( ta  <->  ph )   &    |-  ( et  <->  ps )   =>    |-  ( ( ( ( ch  <->  si )  /\  ( th 
 <->  ze ) )  /\  ( ta  <->  ze ) )  /\  ( et  <->  si ) )
 
Theoremmdandyvr10 38434 Given the equivalences set in the hypotheses, there exist a proof where ch, th, ta, et match ze, si accordingly (Contributed by Jarvin Udandy, 7-Sep-2016.)
 |-  ( ph 
 <->  ze )   &    |-  ( ps  <->  si )   &    |-  ( ch  <->  ph )   &    |-  ( th  <->  ps )   &    |-  ( ta  <->  ph )   &    |-  ( et  <->  ps )   =>    |-  ( ( ( ( ch  <->  ze )  /\  ( th 
 <-> 
 si ) )  /\  ( ta  <->  ze ) )  /\  ( et  <->  si ) )
 
Theoremmdandyvr11 38435 Given the equivalences set in the hypotheses, there exist a proof where ch, th, ta, et match ze, si accordingly (Contributed by Jarvin Udandy, 7-Sep-2016.)
 |-  ( ph 
 <->  ze )   &    |-  ( ps  <->  si )   &    |-  ( ch  <->  ps )   &    |-  ( th  <->  ps )   &    |-  ( ta  <->  ph )   &    |-  ( et  <->  ps )   =>    |-  ( ( ( ( ch  <->  si )  /\  ( th 
 <-> 
 si ) )  /\  ( ta  <->  ze ) )  /\  ( et  <->  si ) )
 
Theoremmdandyvr12 38436 Given the equivalences set in the hypotheses, there exist a proof where ch, th, ta, et match ze, si accordingly (Contributed by Jarvin Udandy, 7-Sep-2016.)
 |-  ( ph 
 <->  ze )   &    |-  ( ps  <->  si )   &    |-  ( ch  <->  ph )   &    |-  ( th  <->  ph )   &    |-  ( ta  <->  ps )   &    |-  ( et  <->  ps )   =>    |-  ( ( ( ( ch  <->  ze )  /\  ( th 
 <->  ze ) )  /\  ( ta  <->  si ) )  /\  ( et  <->  si ) )
 
Theoremmdandyvr13 38437 Given the equivalences set in the hypotheses, there exist a proof where ch, th, ta, et match ze, si accordingly (Contributed by Jarvin Udandy, 7-Sep-2016.)
 |-  ( ph 
 <->  ze )   &    |-  ( ps  <->  si )   &    |-  ( ch  <->  ps )   &    |-  ( th  <->  ph )   &    |-  ( ta  <->  ps )   &    |-  ( et  <->  ps )   =>    |-  ( ( ( ( ch  <->  si )  /\  ( th 
 <->  ze ) )  /\  ( ta  <->  si ) )  /\  ( et  <->  si ) )
 
Theoremmdandyvr14 38438 Given the equivalences set in the hypotheses, there exist a proof where ch, th, ta, et match ze, si accordingly (Contributed by Jarvin Udandy, 7-Sep-2016.)
 |-  ( ph 
 <->  ze )   &    |-  ( ps  <->  si )   &    |-  ( ch  <->  ph )   &    |-  ( th  <->  ps )   &    |-  ( ta  <->  ps )   &    |-  ( et  <->  ps )   =>    |-  ( ( ( ( ch  <->  ze )  /\  ( th 
 <-> 
 si ) )  /\  ( ta  <->  si ) )  /\  ( et  <->  si ) )
 
Theoremmdandyvr15 38439 Given the equivalences set in the hypotheses, there exist a proof where ch, th, ta, et match ze, si accordingly (Contributed by Jarvin Udandy, 7-Sep-2016.)
 |-  ( ph 
 <->  ze )   &    |-  ( ps  <->  si )   &    |-  ( ch  <->  ps )   &    |-  ( th  <->  ps )   &    |-  ( ta  <->  ps )   &    |-  ( et  <->  ps )   =>    |-  ( ( ( ( ch  <->  si )  /\  ( th 
 <-> 
 si ) )  /\  ( ta  <->  si ) )  /\  ( et  <->  si ) )
 
Theoremmdandyvrx0 38440 Given the exclusivities set in the hypotheses, there exist a proof where ch, th, ta, et exclude ze, si accordingly (Contributed by Jarvin Udandy, 7-Sep-2016.)
 |-  ( ph  \/_  ze )   &    |-  ( ps  \/_  si )   &    |-  ( ch  <->  ph )   &    |-  ( th  <->  ph )   &    |-  ( ta  <->  ph )   &    |-  ( et  <->  ph )   =>    |-  ( ( ( ( ch  \/_  ze )  /\  ( th  \/_  ze )
 )  /\  ( ta  \/_ 
 ze ) )  /\  ( et  \/_  ze )
 )
 
Theoremmdandyvrx1 38441 Given the exclusivities set in the hypotheses, there exist a proof where ch, th, ta, et exclude ze, si accordingly (Contributed by Jarvin Udandy, 7-Sep-2016.)
 |-  ( ph  \/_  ze )   &    |-  ( ps  \/_  si )   &    |-  ( ch  <->  ps )   &    |-  ( th  <->  ph )   &    |-  ( ta  <->  ph )   &    |-  ( et  <->  ph )   =>    |-  ( ( ( ( ch  \/_  si )  /\  ( th  \/_  ze )
 )  /\  ( ta  \/_ 
 ze ) )  /\  ( et  \/_  ze )
 )
 
Theoremmdandyvrx2 38442 Given the exclusivities set in the hypotheses, there exist a proof where ch, th, ta, et exclude ze, si accordingly (Contributed by Jarvin Udandy, 7-Sep-2016.)
 |-  ( ph  \/_  ze )   &    |-  ( ps  \/_  si )   &    |-  ( ch  <->  ph )   &    |-  ( th  <->  ps )   &    |-  ( ta  <->  ph )   &    |-  ( et  <->  ph )   =>    |-  ( ( ( ( ch  \/_  ze )  /\  ( th  \/_  si )
 )  /\  ( ta  \/_ 
 ze ) )  /\  ( et  \/_  ze )
 )
 
Theoremmdandyvrx3 38443 Given the exclusivities set in the hypotheses, there exist a proof where ch, th, ta, et exclude ze, si accordingly (Contributed by Jarvin Udandy, 7-Sep-2016.)
 |-  ( ph  \/_  ze )   &    |-  ( ps  \/_  si )   &    |-  ( ch  <->  ps )   &    |-  ( th  <->  ps )   &    |-  ( ta  <->  ph )   &    |-  ( et  <->  ph )   =>    |-  ( ( ( ( ch  \/_  si )  /\  ( th  \/_  si ) ) 
 /\  ( ta  \/_  ze ) )  /\  ( et  \/_  ze ) )
 
Theoremmdandyvrx4 38444 Given the exclusivities set in the hypotheses, there exist a proof where ch, th, ta, et exclude ze, si accordingly (Contributed by Jarvin Udandy, 7-Sep-2016.)
 |-  ( ph  \/_  ze )   &    |-  ( ps  \/_  si )   &    |-  ( ch  <->  ph )   &    |-  ( th  <->  ph )   &    |-  ( ta  <->  ps )   &    |-  ( et  <->  ph )   =>    |-  ( ( ( ( ch  \/_  ze )  /\  ( th  \/_  ze )
 )  /\  ( ta  \/_  si ) )  /\  ( et  \/_  ze ) )
 
Theoremmdandyvrx5 38445 Given the exclusivities set in the hypotheses, there exist a proof where ch, th, ta, et exclude ze, si accordingly (Contributed by Jarvin Udandy, 7-Sep-2016.)
 |-  ( ph  \/_  ze )   &    |-  ( ps  \/_  si )   &    |-  ( ch  <->  ps )   &    |-  ( th  <->  ph )   &    |-  ( ta  <->  ps )   &    |-  ( et  <->  ph )   =>    |-  ( ( ( ( ch  \/_  si )  /\  ( th  \/_  ze )
 )  /\  ( ta  \/_  si ) )  /\  ( et  \/_  ze ) )
 
Theoremmdandyvrx6 38446 Given the exclusivities set in the hypotheses, there exist a proof where ch, th, ta, et exclude ze, si accordingly (Contributed by Jarvin Udandy, 7-Sep-2016.)
 |-  ( ph  \/_  ze )   &    |-  ( ps  \/_  si )   &    |-  ( ch  <->  ph )   &    |-  ( th  <->  ps )   &    |-  ( ta  <->  ps )   &    |-  ( et  <->  ph )   =>    |-  ( ( ( ( ch  \/_  ze )  /\  ( th  \/_  si )
 )  /\  ( ta  \/_  si ) )  /\  ( et  \/_  ze ) )
 
Theoremmdandyvrx7 38447 Given the exclusivities set in the hypotheses, there exist a proof where ch, th, ta, et exclude ze, si accordingly (Contributed by Jarvin Udandy, 7-Sep-2016.)
 |-  ( ph  \/_  ze )   &    |-  ( ps  \/_  si )   &    |-  ( ch  <->  ps )   &    |-  ( th  <->  ps )   &    |-  ( ta  <->  ps )   &    |-  ( et  <->  ph )   =>    |-  ( ( ( ( ch  \/_  si )  /\  ( th  \/_  si ) ) 
 /\  ( ta  \/_  si ) )  /\  ( et  \/_  ze ) )
 
Theoremmdandyvrx8 38448 Given the exclusivities set in the hypotheses, there exist a proof where ch, th, ta, et exclude ze, si accordingly (Contributed by Jarvin Udandy, 7-Sep-2016.)
 |-  ( ph  \/_  ze )   &    |-  ( ps  \/_  si )   &    |-  ( ch  <->  ph )   &    |-  ( th  <->  ph )   &    |-  ( ta  <->  ph )   &    |-  ( et  <->  ps )   =>    |-  ( ( ( ( ch  \/_  ze )  /\  ( th  \/_  ze )
 )  /\  ( ta  \/_ 
 ze ) )  /\  ( et  \/_  si )
 )
 
Theoremmdandyvrx9 38449 Given the exclusivities set in the hypotheses, there exist a proof where ch, th, ta, et exclude ze, si accordingly (Contributed by Jarvin Udandy, 7-Sep-2016.)
 |-  ( ph  \/_  ze )   &    |-  ( ps  \/_  si )   &    |-  ( ch  <->  ps )   &    |-  ( th  <->  ph )   &    |-  ( ta  <->  ph )   &    |-  ( et  <->  ps )   =>    |-  ( ( ( ( ch  \/_  si )  /\  ( th  \/_  ze )
 )  /\  ( ta  \/_ 
 ze ) )  /\  ( et  \/_  si )
 )
 
Theoremmdandyvrx10 38450 Given the exclusivities set in the hypotheses, there exist a proof where ch, th, ta, et exclude ze, si accordingly (Contributed by Jarvin Udandy, 7-Sep-2016.)
 |-  ( ph  \/_  ze )   &    |-  ( ps  \/_  si )   &    |-  ( ch  <->  ph )   &    |-  ( th  <->  ps )   &    |-  ( ta  <->  ph )   &    |-  ( et  <->  ps )   =>    |-  ( ( ( ( ch  \/_  ze )  /\  ( th  \/_  si )
 )  /\  ( ta  \/_ 
 ze ) )  /\  ( et  \/_  si )
 )
 
Theoremmdandyvrx11 38451 Given the exclusivities set in the hypotheses, there exist a proof where ch, th, ta, et exclude ze, si accordingly (Contributed by Jarvin Udandy, 7-Sep-2016.)
 |-  ( ph  \/_  ze )   &    |-  ( ps  \/_  si )   &    |-  ( ch  <->  ps )   &    |-  ( th  <->  ps )   &    |-  ( ta  <->  ph )   &    |-  ( et  <->  ps )   =>    |-  ( ( ( ( ch  \/_  si )  /\  ( th  \/_  si ) ) 
 /\  ( ta  \/_  ze ) )  /\  ( et  \/_  si ) )
 
Theoremmdandyvrx12 38452 Given the exclusivities set in the hypotheses, there exist a proof where ch, th, ta, et exclude ze, si accordingly (Contributed by Jarvin Udandy, 7-Sep-2016.)
 |-  ( ph  \/_  ze )   &    |-  ( ps  \/_  si )   &    |-  ( ch  <->  ph )   &    |-  ( th  <->  ph )   &    |-  ( ta  <->  ps )   &    |-  ( et  <->  ps )   =>    |-  ( ( ( ( ch  \/_  ze )  /\  ( th  \/_  ze )
 )  /\  ( ta  \/_  si ) )  /\  ( et  \/_  si ) )
 
Theoremmdandyvrx13 38453 Given the exclusivities set in the hypotheses, there exist a proof where ch, th, ta, et exclude ze, si accordingly (Contributed by Jarvin Udandy, 7-Sep-2016.)
 |-  ( ph  \/_  ze )   &    |-  ( ps  \/_  si )   &    |-  ( ch  <->  ps )   &    |-  ( th  <->  ph )   &    |-  ( ta  <->  ps )   &    |-  ( et  <->  ps )   =>    |-  ( ( ( ( ch  \/_  si )  /\  ( th  \/_  ze )
 )  /\  ( ta  \/_  si ) )  /\  ( et  \/_  si ) )
 
Theoremmdandyvrx14 38454 Given the exclusivities set in the hypotheses, there exist a proof where ch, th, ta, et exclude ze, si accordingly (Contributed by Jarvin Udandy, 7-Sep-2016.)
 |-  ( ph  \/_  ze )   &    |-  ( ps  \/_  si )   &    |-  ( ch  <->  ph )   &    |-  ( th  <->  ps )   &    |-  ( ta  <->  ps )   &    |-  ( et  <->  ps )   =>    |-  ( ( ( ( ch  \/_  ze )  /\  ( th  \/_  si )
 )  /\  ( ta  \/_  si ) )  /\  ( et  \/_  si ) )
 
Theoremmdandyvrx15 38455 Given the exclusivities set in the hypotheses, there exist a proof where ch, th, ta, et exclude ze, si accordingly (Contributed by Jarvin Udandy, 7-Sep-2016.)
 |-  ( ph  \/_  ze )   &    |-  ( ps  \/_  si )   &    |-  ( ch  <->  ps )   &    |-  ( th  <->  ps )   &    |-  ( ta  <->  ps )   &    |-  ( et  <->  ps )   =>    |-  ( ( ( ( ch  \/_  si )  /\  ( th  \/_  si ) ) 
 /\  ( ta  \/_  si ) )  /\  ( et  \/_  si ) )
 
TheoremH15NH16TH15IH16 38456 Given 15 hypotheses and a 16th hypothesis, there exists a proof the 15 imply the 16th. (Contributed by Jarvin Udandy, 8-Sep-2016.)
 |-  ph   &    |-  ps   &    |-  ch   &    |-  th   &    |-  ta   &    |-  et   &    |-  ze   &    |-  si   &    |-  rh   &    |-  mu   &    |-  la   &    |-  ka   &    |- jph   &    |- jps   &    |- jch   &    |- jth   =>    |-  (
 ( ( ( ( ( ( ( ( ( ( ( ( ( ( ph  /\  ps )  /\  ch )  /\  th )  /\  ta )  /\  et )  /\  ze )  /\  si )  /\  rh )  /\  mu )  /\  la )  /\  ka )  /\ jph )  /\ jps
 )  /\ jch ) 
 -> jth )
 
Theoremdandysum2p2e4 38457

CONTRADICTION PROVED AT 1 + 1 = 2 .

Given the right hypotheses we can prove a dandysum of 2+2=4. The qed step is the value '4' in Decimal BEING IMPLIED by the hypotheses.

Note: Values that when added which exceed a 4bit value are not supported.

Note: Digits begin from left (least) to right (greatest). e.g. 1000 would be '1', 0100 would be '2'. 0010 would be '4'.

How to perceive the hypotheses' bits in order: ( th <-> F. ), ( ta <-> F. ) Would be input value X's first bit, and input value Y's first bit.

( et <-> F ), ( ze <-> F. ) would be input value X's second bit, and input value Y's second bit. (Contributed by Jarvin Udandy, 6-Sep-2016.)

 |-  ( ph 
 <->  ( th  /\  ta ) )   &    |-  ( ps  <->  ( et  /\  ze ) )   &    |-  ( ch  <->  ( si  /\  rh ) )   &    |-  ( th  <-> F.  )   &    |-  ( ta 
 <-> F.  )   &    |-  ( et  <-> T.  )   &    |-  ( ze 
 <-> T.  )   &    |-  ( si  <-> F.  )   &    |-  ( rh 
 <-> F.  )   &    |-  ( mu  <-> F.  )   &    |-  ( la 
 <-> F.  )   &    |-  ( ka  <->  ( ( th  \/_ 
 ta )  \/_  ( th  /\  ta ) ) )   &    |-  (jph  <->  (
 ( et  \/_  ze )  \/  ph ) )   &    |-  (jps  <->  ( ( si  \/_  rh )  \/  ps ) )   &    |-  (jch  <->  ( ( mu 
 \/_  la )  \/  ch ) )   =>    |-  ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( (
 ph 
 <->  ( th  /\  ta ) )  /\  ( ps  <->  ( et  /\  ze )
 ) )  /\  ( ch 
 <->  ( si  /\  rh ) ) )  /\  ( th  <-> F.  ) )  /\  ( ta  <-> F.  ) )  /\  ( et  <-> T.  ) )  /\  ( ze  <-> T.  ) )  /\  ( si  <-> F.  ) )  /\  ( rh  <-> F.  ) )  /\  ( mu  <-> F.  ) )  /\  ( la  <-> F.  ) )  /\  ( ka  <->  ( ( th  \/_ 
 ta )  \/_  ( th  /\  ta ) ) ) )  /\  (jph  <->  ( ( et  \/_  ze )  \/  ph ) ) ) 
 /\  (jps  <->  (
 ( si  \/_  rh )  \/  ps ) ) ) 
 /\  (jch  <->  (
 ( mu  \/_  la )  \/  ch ) ) ) 
 ->  ( ( ( ( ka  <-> F.  )  /\  (jph  <-> F.  ) )  /\  (jps  <-> T.  ) )  /\  (jch  <-> F.  ) )
 )
 
Theoremmdandysum2p2e4 38458 CONTRADICTION PROVED AT 1 + 1 = 2 . Luckily Mario Carneiro did a successful version of his own.

See Mario's Relevant Work: 1.3.14 Half adder and full adder in propositional calculus.

Given the right hypotheses we can prove a dandysum of 2+2=4. The qed step is the value '4' in Decimal BEING IMPLIED by the hypotheses.

Note: Values that when added which exceed a 4bit value are not supported.

Note: Digits begin from left (least) to right (greatest). e.g. 1000 would be '1', 0100 would be '2'. 0010 would be '4'.

How to perceive the hypotheses' bits in order: ( th <-> F. ), ( ta <-> F. ) Would be input value X's first bit, and input value Y's first bit.

( et <-> F. ), ( ze <-> F. ) would be input value X's second bit, and input value Y's second bit.

In mdandysum2p2e4, one might imagine what jth or jta could be then do the math with their truths. Also limited to the restriction jth, jta are having opposite truths equivalent to the stated truth constants.

(Contributed by Jarvin Udandy, 6-Sep-2016.)

 |-  (jth  <-> F.  )   &    |-  (jta  <-> T.  )   &    |-  ( ph  <->  ( th  /\  ta ) )   &    |-  ( ps  <->  ( et  /\  ze ) )   &    |-  ( ch  <->  ( si  /\  rh ) )   &    |-  ( th  <-> jth )   &    |-  ( ta 
 <-> jth
 )   &    |-  ( et  <-> jta )   &    |-  ( ze 
 <-> jta
 )   &    |-  ( si  <-> jth )   &    |-  ( rh 
 <-> jth
 )   &    |-  ( mu  <-> jth )   &    |-  ( la 
 <-> jth
 )   &    |-  ( ka  <->  ( ( th  \/_ 
 ta )  \/_  ( th  /\  ta ) ) )   &    |-  (jph  <->  (
 ( et  \/_  ze )  \/  ph ) )   &    |-  (jps  <->  ( ( si  \/_  rh )  \/  ps ) )   &    |-  (jch  <->  ( ( mu 
 \/_  la )  \/  ch ) )   =>    |-  ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( (
 ph 
 <->  ( th  /\  ta ) )  /\  ( ps  <->  ( et  /\  ze )
 ) )  /\  ( ch 
 <->  ( si  /\  rh ) ) )  /\  ( th  <-> F.  ) )  /\  ( ta  <-> F.  ) )  /\  ( et  <-> T.  ) )  /\  ( ze  <-> T.  ) )  /\  ( si  <-> F.  ) )  /\  ( rh  <-> F.  ) )  /\  ( mu  <-> F.  ) )  /\  ( la  <-> F.  ) )  /\  ( ka  <->  ( ( th  \/_ 
 ta )  \/_  ( th  /\  ta ) ) ) )  /\  (jph  <->  ( ( et  \/_  ze )  \/  ph ) ) ) 
 /\  (jps  <->  (
 ( si  \/_  rh )  \/  ps ) ) ) 
 /\  (jch  <->  (
 ( mu  \/_  la )  \/  ch ) ) ) 
 ->  ( ( ( ( ka  <-> F.  )  /\  (jph  <-> F.  ) )  /\  (jps  <-> T.  ) )  /\  (jch  <-> F.  ) )
 )
 
21.33  Mathbox for Alexander van der Vekens
 
21.33.1  Double restricted existential uniqueness
 
21.33.1.1  Restricted quantification (extension)
 
Theoremr19.32 38459 Theorem 19.32 of [Margaris] p. 90 with restricted quantifiers, analogous to r19.32v 2971. (Contributed by Alexander van der Vekens, 29-Jun-2017.)
 |-  F/ x ph   =>    |-  ( A. x  e.  A  ( ph  \/  ps )  <->  ( ph  \/  A. x  e.  A  ps ) )
 
Theoremrexsb 38460* An equivalent expression for restricted existence, analogous to exsb 2267. (Contributed by Alexander van der Vekens, 1-Jul-2017.)
 |-  ( E. x  e.  A  ph  <->  E. y  e.  A  A. x ( x  =  y  ->  ph ) )
 
Theoremrexrsb 38461* An equivalent expression for restricted existence, analogous to exsb 2267. (Contributed by Alexander van der Vekens, 1-Jul-2017.)
 |-  ( E. x  e.  A  ph  <->  E. y  e.  A  A. x  e.  A  ( x  =  y  ->  ph ) )
 
Theorem2rexsb 38462* An equivalent expression for double restricted existence, analogous to rexsb 38460. (Contributed by Alexander van der Vekens, 1-Jul-2017.)
 |-  ( E. x  e.  A  E. y  e.  B  ph  <->  E. z  e.  A  E. w  e.  B  A. x A. y ( ( x  =  z  /\  y  =  w )  ->  ph )
 )
 
Theorem2rexrsb 38463* An equivalent expression for double restricted existence, analogous to 2exsb 2268. (Contributed by Alexander van der Vekens, 1-Jul-2017.)
 |-  ( E. x  e.  A  E. y  e.  B  ph  <->  E. z  e.  A  E. w  e.  B  A. x  e.  A  A. y  e.  B  ( ( x  =  z  /\  y  =  w )  ->  ph )
 )
 
Theoremcbvral2 38464* Change bound variables of double restricted universal quantification, using implicit substitution, analogous to cbvral2v 3062. (Contributed by Alexander van der Vekens, 2-Jul-2017.)
 |-  F/ z ph   &    |-  F/ x ch   &    |-  F/ w ch   &    |-  F/ y ps   &    |-  ( x  =  z  ->  ( ph  <->  ch ) )   &    |-  (
 y  =  w  ->  ( ch  <->  ps ) )   =>    |-  ( A. x  e.  A  A. y  e.  B  ph  <->  A. z  e.  A  A. w  e.  B  ps )
 
Theoremcbvrex2 38465* Change bound variables of double restricted universal quantification, using implicit substitution, analogous to cbvrex2v 3063. (Contributed by Alexander van der Vekens, 2-Jul-2017.)
 |-  F/ z ph   &    |-  F/ x ch   &    |-  F/ w ch   &    |-  F/ y ps   &    |-  ( x  =  z  ->  ( ph  <->  ch ) )   &    |-  (
 y  =  w  ->  ( ch  <->  ps ) )   =>    |-  ( E. x  e.  A  E. y  e.  B  ph  <->  E. z  e.  A  E. w  e.  B  ps )
 
Theorem2ralbiim 38466 Split a biconditional and distribute 2 quantifiers, analogous to 2albiim 1747 and ralbiim 2957. (Contributed by Alexander van der Vekens, 2-Jul-2017.)
 |-  ( A. x  e.  A  A. y  e.  B  (
 ph 
 <->  ps )  <->  ( A. x  e.  A  A. y  e.  B  ( ph  ->  ps )  /\  A. x  e.  A  A. y  e.  B  ( ps  ->  ph ) ) )
 
21.33.1.2  The empty set (extension)
 
Theoremraaan2 38467* Rearrange restricted quantifiers with two different restricting classes, analogous to raaan 3907. It is necessary that either both restricting classes are empty or both are not empty. (Contributed by Alexander van der Vekens, 29-Jun-2017.)
 |-  F/ y ph   &    |-  F/ x ps   =>    |-  (
 ( A  =  (/)  <->  B  =  (/) )  ->  ( A. x  e.  A  A. y  e.  B  (
 ph  /\  ps )  <->  (
 A. x  e.  A  ph 
 /\  A. y  e.  B  ps ) ) )
 
21.33.1.3  Restricted uniqueness and "at most one" quantification
 
Theoremrmoimi 38468 Restricted "at most one" is preserved through implication (note wff reversal). (Contributed by Alexander van der Vekens, 17-Jun-2017.)
 |-  ( ph  ->  ps )   =>    |-  ( E* x  e.  A  ps  ->  E* x  e.  A  ph )
 
Theorem2reu5a 38469 Double restricted existential uniqueness in terms of restricted existence and restricted "at most one." (Contributed by Alexander van der Vekens, 17-Jun-2017.)
 |-  ( E! x  e.  A  E! y  e.  B  ph  <->  ( E. x  e.  A  ( E. y  e.  B  ph 
 /\  E* y  e.  B  ph )  /\  E* x  e.  A  ( E. y  e.  B  ph  /\  E* y  e.  B  ph ) ) )
 
Theoremreuimrmo 38470 Restricted uniqueness implies restricted "at most one" through implication, analogous to euimmo 2321. (Contributed by Alexander van der Vekens, 25-Jun-2017.)
 |-  ( A. x  e.  A  ( ph  ->  ps )  ->  ( E! x  e.  A  ps  ->  E* x  e.  A  ph ) )
 
Theoremrmoanim 38471* Introduction of a conjunct into restricted "at most one" quantifier, analogous to moanim 2328. (Contributed by Alexander van der Vekens, 25-Jun-2017.)
 |-  F/ x ph   =>    |-  ( E* x  e.  A  ( ph  /\  ps ) 
 <->  ( ph  ->  E* x  e.  A  ps ) )
 
Theoremreuan 38472* Introduction of a conjunct into restricted uniqueness quantifier, analogous to euan 2329. (Contributed by Alexander van der Vekens, 2-Jul-2017.)
 |-  F/ x ph   =>    |-  ( E! x  e.  A  ( ph  /\  ps ) 
 <->  ( ph  /\  E! x  e.  A  ps ) )
 
21.33.1.4  Analogs to Existential uniqueness (double quantification)
 
Theorem2reurex 38473* Double restricted quantification with existential uniqueness, analogous to 2euex 2343. (Contributed by Alexander van der Vekens, 24-Jun-2017.)
 |-  ( E! x  e.  A  E. y  e.  B  ph 
 ->  E. y  e.  B  E! x  e.  A  ph )
 
Theorem2reurmo 38474* Double restricted quantification with restricted existential uniqueness and restricted "at most one.", analogous to 2eumo 2344. (Contributed by Alexander van der Vekens, 24-Jun-2017.)
 |-  ( E! x  e.  A  E* y  e.  B  ph 
 ->  E* x  e.  A  E! y  e.  B  ph )
 
Theorem2reu2rex 38475* Double restricted existential uniqueness, analogous to 2eu2ex 2345. (Contributed by Alexander van der Vekens, 25-Jun-2017.)
 |-  ( E! x  e.  A  E! y  e.  B  ph 
 ->  E. x  e.  A  E. y  e.  B  ph )
 
Theorem2rmoswap 38476* A condition allowing swap of restricted "at most one" and restricted existential quantifiers, analogous to 2moswap 2346. (Contributed by Alexander van der Vekens, 25-Jun-2017.)
 |-  ( A. x  e.  A  E* y  e.  B  ph 
 ->  ( E* x  e.  A  E. y  e.  B  ph  ->  E* y  e.  B  E. x  e.  A  ph ) )
 
Theorem2rexreu 38477* Double restricted existential uniqueness implies double restricted uniqueness quantification, analogous to 2exeu 2348. (Contributed by Alexander van der Vekens, 25-Jun-2017.)
 |-  (
 ( E! x  e.  A  E. y  e.  B  ph  /\  E! y  e.  B  E. x  e.  A  ph )  ->  E! x  e.  A  E! y  e.  B  ph )
 
Theorem2reu1 38478* Double restricted existential uniqueness. This theorem shows a condition under which a "naive" definition matches the correct one, analogous to 2eu1 2352. (Contributed by Alexander van der Vekens, 25-Jun-2017.)
 |-  ( A. x  e.  A  E* y  e.  B  ph 
 ->  ( E! x  e.  A  E! y  e.  B  ph  <->  ( E! x  e.  A  E. y  e.  B  ph  /\  E! y  e.  B  E. x  e.  A  ph ) ) )
 
Theorem2reu2 38479* Double restricted existential uniqueness, analogous to 2eu2 2353. (Contributed by Alexander van der Vekens, 29-Jun-2017.)
 |-  ( E! y  e.  B  E. x  e.  A  ph 
 ->  ( E! x  e.  A  E! y  e.  B  ph  <->  E! x  e.  A  E. y  e.  B  ph ) )
 
Theorem2reu3 38480* Double restricted existential uniqueness, analogous to 2eu3 2354. (Contributed by Alexander van der Vekens, 29-Jun-2017.)
 |-  ( A. x  e.  A  A. y  e.  B  ( E* x  e.  A  ph 
 \/  E* y  e.  B  ph )  ->  ( ( E! x  e.  A  E! y  e.  B  ph 
 /\  E! y  e.  B  E! x  e.  A  ph )  <->  ( E! x  e.  A  E. y  e.  B  ph  /\  E! y  e.  B  E. x  e.  A  ph ) ) )
 
Theorem2reu4a 38481* Definition of double restricted existential uniqueness ("exactly one  x and exactly one  y"), analogous to 2eu4 2355 with the additional requirement that the restricting classes are not empty (which is not necessary as shown in 2reu4 38482). (Contributed by Alexander van der Vekens, 1-Jul-2017.)
 |-  (
 ( A  =/=  (/)  /\  B  =/= 
 (/) )  ->  (
 ( E! x  e.  A  E. y  e.  B  ph  /\  E! y  e.  B  E. x  e.  A  ph )  <->  ( E. x  e.  A  E. y  e.  B  ph  /\  E. z  e.  A  E. w  e.  B  A. x  e.  A  A. y  e.  B  ( ph  ->  ( x  =  z  /\  y  =  w )
 ) ) ) )
 
Theorem2reu4 38482* Definition of double restricted existential uniqueness ("exactly one  x and exactly one  y"), analogous to 2eu4 2355. (Contributed by Alexander van der Vekens, 1-Jul-2017.)
 |-  (
 ( E! x  e.  A  E. y  e.  B  ph  /\  E! y  e.  B  E. x  e.  A  ph )  <->  ( E. x  e.  A  E. y  e.  B  ph  /\  E. z  e.  A  E. w  e.  B  A. x  e.  A  A. y  e.  B  ( ph  ->  ( x  =  z  /\  y  =  w )
 ) ) )
 
Theorem2reu7 38483* Two equivalent expressions for double restricted existential uniqueness, analogous to 2eu7 2358. (Contributed by Alexander van der Vekens, 2-Jul-2017.)
 |-  (
 ( E! x  e.  A  E. y  e.  B  ph  /\  E! y  e.  B  E. x  e.  A  ph )  <->  E! x  e.  A  E! y  e.  B  ( E. x  e.  A  ph 
 /\  E. y  e.  B  ph ) )
 
Theorem2reu8 38484* Two equivalent expressions for double restricted existential uniqueness, analogous to 2eu8 2359. Curiously, we can put  E! on either of the internal conjuncts but not both. We can also commute  E! x  e.  A E! y  e.  B using 2reu7 38483. (Contributed by Alexander van der Vekens, 2-Jul-2017.)
 |-  ( E! x  e.  A  E! y  e.  B  ( E. x  e.  A  ph 
 /\  E. y  e.  B  ph )  <->  E! x  e.  A  E! y  e.  B  ( E! x  e.  A  ph 
 /\  E. y  e.  B  ph ) )
 
21.33.2  Alternative definitions of function's and operation's values

The current definition of the value 
( F `  A
) of a function  F for an argument  A (see df-fv 5609) assures that this value is always a set, see fex 6153. This is because this definition can be applied to any classes  F and  A, and evaluates to the empty set when it is not meaningful (as shown by ndmfv 5905 and fvprc 5875).

Although it is very convenient for many theorems on functions and their proofs, there are some cases in which from  ( F `  A
)  =  (/) alone it cannot be decided/derived if  ( F `  A ) is meaningful ( F is actually a function which is defined for  A and really has the function value  (/)) or not. Therefore, additional assumptions are required, such as  (/)  e/  ran  F,  (/)  e.  ran  F or 
Fun  F  /\  A  e. 
dom  F (see, for example, ndmfvrcl 5906).

To avoid such an ambiguity, an alternative definition  ( F''' A ) (see df-afv 38489) would be possible which evaluates to the universal class ( ( F''' A )  =  _V) if it is not meaningful (see afvnfundmuv 38511, ndmafv 38512, afvprc 38516 and nfunsnafv 38514), and which corresponds to the current definition ( ( F `  A )  =  ( F''' A )) if it is (see afvfundmfveq 38510). That means  ( F''' A )  =  _V  ->  ( F `  A )  =  (/) (see afvpcfv0 38518), but  ( F `  A )  =  (/)  ->  ( F''' A )  =  _V is not generally valid.

In the theory of partial functions, it is a common case that  F is not defined at  A, which also would result in  ( F''' A )  =  _V. In this context we say  ( F''' A ) "is not defined" instead of "is not meaningful".

With this definition the following intuitive equivalence holds:  ( F''' A )  e.  _V <-> " ( F''' A ) is meaningful/defined".

An interesting question would be if 
( F `  A
) could be replaced by  ( F''' A ) in most of the theorems based on function's values. If we look at the (currently 19) proofs using the definition df-fv 5609 of 
( F `  A
), we see that analogons for the following 8 theorems can be proven using the alternative definition: fveq1 5880-> afveq1 38506, fveq2 5881-> afveq2 38507, nffv 5888-> nfafv 38508, csbfv12 5916-> csbafv12g , fvres 5895-> afvres 38544, rlimdm 13614-> rlimdmafv 38549, tz6.12-1 5897-> tz6.12-1-afv 38546, fveu 5873-> afveu 38525.

3 theorems proved by directly using df-fv 5609 are within a mathbox (fvsb 36775) or not used (isumclim3 13819, avril1 25898).

However, the remaining 8 theorems proved by directly using df-fv 5609 are used more or less often:

* fvex 5891: used in about 1750 proofs.

* tz6.12-1 5897: root theorem of many theorems which have not a strict analogon, and which are used many times: fvprc 5875 (used in about 127 proofs), tz6.12i 5901 (used - indirectly via fvbr0 5902 and fvrn0 5903- in 18 proofs, and in fvclss 6162 used in fvclex 6779 used in fvresex 6780, which is not used!), dcomex 8884 (used in 4 proofs), ndmfv 5905 (used in 86 proofs) and nfunsn 5912 (used by dffv2 5954 which is not used).

* fv2 5876: only used by elfv 5879, which is only used by fv3 5894, which is not used.

* dffv3 5877: used by dffv4 5878 (the previous "df-fv"), which now is only used in deprecated (usage discouraged) theorems or within mathboxes (csbfv12gALTOLD 37186, csbfv12gALTVD 37269), by shftval 13137 (itself used in 9 proofs), by dffv5 30696 (mathbox) and by fvco2 5956, which has the analogon afvco2 38548.

* fvopab5 5989: used only by ajval 26501 (not used) and by adjval 27541 ( used - indirectly - in 9 proofs).

* zsum 13783: used (via isum 13784, sum0 13786 and fsumsers 13793) in more than 90 proofs.

* isumshft 13896: used in pserdv2 23383 and (via logtayl 23603) 4 other proofs.

* ovtpos 6999: used in 14 proofs.

As a result of this analysis we can say that the current definition of a function's value is crucial for Metamath and cannot be exchanged easily with an alternative definition. While fv2 5876, dffv3 5877, fvopab5 5989, zsum 13783, isumshft 13896 and ovtpos 6999 are not critical or are, hopefully, also valid for the alternative definition, fvex 5891 and tz6.12-1 5897 (and the theorems based on them) are essential for the current definition of function values.

With the same arguments, an alternatvie definition of operation's values (( A O B)) could be meaningful to avoid ambiguities, see df-aov 38490.

For additional discussions/material see https://groups.google.com/forum/#!topic/metamath/cteNUppB6A4.

 
Syntaxwdfat 38485 Extend the definition of a wff to include the "defined at" predicate. (Read: (The Function)  F is defined at (the argument)  A). In a previous version, the token "def@" was used. However, since the @ is used (informally) as a replacement for $ in commented out sections that may be deleted some day. While there is no violation of any standard to use the @ in a token, it could make the search for such commented-out sections slightly more difficult. (See remark of Norman Megill at https://groups.google.com/forum/#!topic/metamath/cteNUppB6A4).
 wff  F defAt  A
 
Syntaxcafv 38486 Extend the definition of a class to include the value of a function. (Read: The value of  F at  A, or " F of  A."). In a previous version, the symbol " ' " was used. However, since the similarity with the symbol 
` used for the current definition of a function's value (see df-fv 5609), which, by the way, was intended to visualize that in many cases  ` and " ' " are exchangeable, makes reading the theorems, especially those which uses both definitions as dfafv2 38504, very difficult, 3 apostrophes ''' are used now so that it's easier to distinguish from df-fv 5609 and df-ima 4866. And not three backticks ( three times  ` ) since that would be annoying to escape in a comment. (See remark of Norman Megill and Gerard Lang at https://groups.google.com/forum/#!topic/metamath/cteNUppB6A4).
 class  ( F''' A )
 
Syntaxcaov 38487 Extend class notation to include the value of an operation  F (such as  +) for two arguments  A and  B. Note that the syntax is simply three class symbols in a row surrounded by a pair of parentheses in contrast to the current definition, see df-ov 6308.
 class (( A F B))
 
Definitiondf-dfat 38488 Definition of the predicate that determines if some class  F is defined as function for an argument  A or, in other words, if the function value for some class  F for an argument  A is defined. We say that  F is defined at  A if a  F is a function restricted to the member  A of its domain. (Contributed by Alexander van der Vekens, 25-May-2017.)
 |-  ( F defAt  A  <->  ( A  e.  dom 
 F  /\  Fun  ( F  |`  { A } )
 ) )
 
Definitiondf-afv 38489* Alternative definition of the value of a function,  ( F''' A ), also known as function application. In contrast to  ( F `  A )  =  (/) (see df-fv 5609 and ndmfv 5905),  ( F''' A )  =  _V if F is not defined for A! (Contributed by Alexander van der Vekens, 25-May-2017.)
 |-  ( F''' A )  =  if ( F defAt  A ,  ( iota x A F x ) ,  _V )
 
Definitiondf-aov 38490 Define the value of an operation. In contrast to df-ov 6308, the alternative definition for a function value (see df-afv 38489) is used. By this, the value of the operation applied to two arguments is the universal class if the operation is not defined for these two arguments. There are still no restrictions of any kind on what those class expressions may be, although only certain kinds of class expressions - a binary operation  F and its arguments  A and  B- will be useful for proving meaningful theorems. (Contributed by Alexander van der Vekens, 26-May-2017.)
 |- (( A F B))  =  ( F'''
 <. A ,  B >. )
 
21.33.2.1  Restricted quantification (extension)
 
Theoremralbinrald 38491* Elemination of a restricted universal quantification under certain conditions. (Contributed by Alexander van der Vekens, 2-Aug-2017.)
 |-  ( ph  ->  X  e.  A )   &    |-  ( x  e.  A  ->  x  =  X )   &    |-  ( x  =  X  ->  ( ps  <->  th ) )   =>    |-  ( ph  ->  (
 A. x  e.  A  ps 
 <-> 
 th ) )
 
21.33.2.2  The universal class (extension)
 
Theoremnvelim 38492 If a class is the universal class it doesn't belong to any class, generalisation of nvel 4563. (Contributed by Alexander van der Vekens, 26-May-2017.)
 |-  ( A  =  _V  ->  -.  A  e.  B )
 
21.33.2.3  Introduce the Axiom of Power Sets (extension)
 
Theoremalneu 38493 If a statement holds for all sets, there is not a unique set for which the statement holds. (Contributed by Alexander van der Vekens, 28-Nov-2017.)
 |-  ( A. x ph  ->  -.  E! x ph )
 
Theoremeu2ndop1stv 38494* If there is a unique second component in an ordered pair contained in a given set, the first component must be a set. (Contributed by Alexander van der Vekens, 29-Nov-2017.)
 |-  ( E! y <. A ,  y >.  e.  V  ->  A  e.  _V )
 
21.33.2.4  Relations (extension)
 
Theoremeldmressn 38495 Element of the domain of a restriction to a singleton. (Contributed by Alexander van der Vekens, 2-Jul-2017.)
 |-  ( B  e.  dom  ( F  |`  { A } )  ->  B  =  A )
 
21.33.2.5  Functions (extension)
 
Theoremfveqvfvv 38496 If a function's value at an argument is the universal class (which can never be the case because of fvex 5891), the function's value at this argument is any set (especially the empty set). In short "If a function's value is a proper class, it is a set", which sounds strange/contradictory, but which is a consequence of that a contradiction implies anything (see pm2.21i 134). (Contributed by Alexander van der Vekens, 26-May-2017.)
 |-  (
 ( F `  A )  =  _V  ->  ( F `  A )  =  B )
 
Theoremfunresfunco 38497 Composition of two functions, generalization of funco 5639. (Contributed by Alexander van der Vekens, 25-Jul-2017.)
 |-  (
 ( Fun  ( F  |` 
 ran  G )  /\  Fun  G )  ->  Fun  ( F  o.  G ) )
 
Theoremfnresfnco 38498 Composition of two functions, similar to fnco 5702. (Contributed by Alexander van der Vekens, 25-Jul-2017.)
 |-  (
 ( ( F  |`  ran  G )  Fn  ran  G  /\  G  Fn  B )  ->  ( F  o.  G )  Fn  B )
 
Theoremfuncoressn 38499 A composition restricted to a singleton is a function under certain conditions. (Contributed by Alexander van der Vekens, 25-Jul-2017.)
 |-  (
 ( ( ( G `
  X )  e. 
 dom  F  /\  Fun  ( F  |`  { ( G `
  X ) }
 ) )  /\  ( G  Fn  A  /\  X  e.  A ) )  ->  Fun  ( ( F  o.  G )  |`  { X } ) )
 
Theoremfunressnfv 38500 A restriction to a singleton with a function value is a function under certain conditions. (Contributed by Alexander van der Vekens, 25-Jul-2017.)
 |-  (
 ( ( X  e.  dom  ( F  o.  G )  /\  Fun  ( ( F  o.  G )  |`  { X } ) ) 
 /\  ( G  Fn  A  /\  X  e.  A ) )  ->  Fun  ( F  |`  { ( G `
  X ) }
 ) )
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