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Theorem List for Metamath Proof Explorer - 37901-38000   *Has distinct variable group(s)
TypeLabelDescription
Statement
 
Theorematbiffatnnbalt 37901 If a implies b, then a implies not not b. (Contributed by Jarvin Udandy, 29-Aug-2016.)
 |-  (
 ( ph  ->  ps )  ->  ( ph  ->  -.  -.  ps ) )
 
Theoremabnotbtaxb 37902 Assuming a, not b, there exists a proof a-xor-b.) (Contributed by Jarvin Udandy, 31-Aug-2016.)
 |-  ph   &    |-  -.  ps   =>    |-  ( ph  \/_  ps )
 
Theoremabnotataxb 37903 Assuming not a, b, there exists a proof a-xor-b.) (Contributed by Jarvin Udandy, 31-Aug-2016.)
 |-  -.  ph   &    |-  ps   =>    |-  ( ph  \/_  ps )
 
Theoremconimpf 37904 Assuming a, not b, and a implies b, there exists a proof that a is false.) (Contributed by Jarvin Udandy, 28-Aug-2016.)
 |-  ph   &    |-  -.  ps   &    |-  ( ph  ->  ps )   =>    |-  ( ph  <-> F.  )
 
Theoremconimpfalt 37905 Assuming a, not b, and a implies b, there exists a proof that a is false.) (Contributed by Jarvin Udandy, 29-Aug-2016.)
 |-  ph   &    |-  -.  ps   &    |-  ( ph  ->  ps )   =>    |-  ( ph  <-> F.  )
 
Theoremaistbisfiaxb 37906 Given a is equivalent to T., Given b is equivalent to F. there exists a proof for a-xor-b. (Contributed by Jarvin Udandy, 31-Aug-2016.)
 |-  ( ph 
 <-> T.  )   &    |-  ( ps  <-> F.  )   =>    |-  ( ph  \/_  ps )
 
Theoremaisfbistiaxb 37907 Given a is equivalent to F., Given b is equivalent to T., there exists a proof for a-xor-b. (Contributed by Jarvin Udandy, 31-Aug-2016.)
 |-  ( ph 
 <-> F.  )   &    |-  ( ps  <-> T.  )   =>    |-  ( ph  \/_  ps )
 
Theoremaifftbifffaibif 37908 Given a is equivalent to T., Given b is equivalent to F., there exists a proof for that a implies b is false. (Contributed by Jarvin Udandy, 7-Sep-2020.)
 |-  ( ph 
 <-> T.  )   &    |-  ( ps  <-> F.  )   =>    |-  ( ( ph  ->  ps )  <-> F.  )
 
Theoremaifftbifffaibifff 37909 Given a is equivalent to T., Given b is equivalent to F., there exists a proof for that a iff b is false. (Contributed by Jarvin Udandy, 7-Sep-2020.)
 |-  ( ph 
 <-> T.  )   &    |-  ( ps  <-> F.  )   =>    |-  ( ( ph  <->  ps ) 
 <-> F.  )
 
Theorematnaiana 37910 Given a, it is not the case a implies a self contradiction. (Contributed by Jarvin Udandy, 7-Sep-2020.)
 |-  ph   =>    |- 
 -.  ( ph  ->  (
 ph  /\  -.  ph )
 )
 
Theoremainaiaandna 37911 Given a, a implies it is not the case a implies a self contradiction. (Contributed by Jarvin Udandy, 7-Sep-2020.)
 |-  ph   =>    |-  ( ph  ->  -.  ( ph  ->  ( ph  /\  -.  ph ) ) )
 
Theoremabcdta 37912 Given (((a and b) and c) and d), there exists a proof for a (Contributed by Jarvin Udandy, 3-Sep-2016.)
 |-  (
 ( ( ph  /\  ps )  /\  ch )  /\  th )   =>    |-  ph
 
Theoremabcdtb 37913 Given (((a and b) and c) and d), there exists a proof for b (Contributed by Jarvin Udandy, 3-Sep-2016.)
 |-  (
 ( ( ph  /\  ps )  /\  ch )  /\  th )   =>    |- 
 ps
 
Theoremabcdtc 37914 Given (((a and b) and c) and d), there exists a proof for c (Contributed by Jarvin Udandy, 3-Sep-2016.)
 |-  (
 ( ( ph  /\  ps )  /\  ch )  /\  th )   =>    |- 
 ch
 
Theoremabcdtd 37915 Given (((a and b) and c) and d), there exists a proof for d (Contributed by Jarvin Udandy, 3-Sep-2016.)
 |-  (
 ( ( ph  /\  ps )  /\  ch )  /\  th )   =>    |- 
 th
 
Theoremabciffcbatnabciffncba 37916 Operands in a biconditional expression converted negated. Additionally biconditional converted to show antecedent implies sequent. Closed form. (Contributed by Jarvin Udandy, 7-Sep-2020.)
 |-  ( -.  ( ( ph  /\  ps )  /\  ch )  ->  -.  ( ( ch  /\  ps )  /\  ph )
 )
 
Theoremabciffcbatnabciffncbai 37917 Operands in a biconditional expression converted negated. Additionally biconditional converted to show antecedent implies sequent. (Contributed by Jarvin Udandy, 7-Sep-2020.)
 |-  (
 ( ( ph  /\  ps )  /\  ch )  <->  ( ( ch 
 /\  ps )  /\  ph )
 )   =>    |-  ( -.  ( (
 ph  /\  ps )  /\  ch )  ->  -.  (
 ( ch  /\  ps )  /\  ph ) )
 
Theoremnabctnabc 37918 not ( a -> ( b /\ c ) ) we can show: not a implies ( b /\ c ) (Contributed by Jarvin Udandy, 7-Sep-2020.)
 |-  -.  ( ph  ->  ( ps  /\ 
 ch ) )   =>    |-  ( -.  ph  ->  ( ps  /\  ch ) )
 
Theoremconfun 37919 Given the hypotheses there exists a proof for (c implies ( d iff a ) ) (Contributed by Jarvin Udandy, 6-Sep-2020.)
 |-  ph   &    |-  ( ch  ->  ps )   &    |-  ( ch  ->  th )   &    |-  ( ph  ->  (
 ph  ->  ps ) )   =>    |-  ( ch  ->  ( th  <->  ph ) )
 
Theoremconfun2 37920 Confun simplified to two propositions. (Contributed by Jarvin Udandy, 6-Sep-2020.)
 |-  ( ps  ->  ph )   &    |-  ( ps  ->  -.  ( ps  ->  ( ps  /\  -.  ps )
 ) )   &    |-  ( ( ps 
 ->  ph )  ->  (
 ( ps  ->  ph )  -> 
 ph ) )   =>    |-  ( ps  ->  ( -.  ( ps  ->  ( ps  /\  -.  ps ) )  <->  ( ps  ->  ph ) ) )
 
Theoremconfun3 37921 Confun's more complex form where both a,d have been "defined". (Contributed by Jarvin Udandy, 6-Sep-2020.)
 |-  ( ph 
 <->  ( ch  ->  ps )
 )   &    |-  ( th  <->  -.  ( ch  ->  ( ch  /\  -.  ch ) ) )   &    |-  ( ch  ->  ps )   &    |-  ( ch  ->  -.  ( ch  ->  ( ch  /\  -.  ch )
 ) )   &    |-  ( ( ch 
 ->  ps )  ->  (
 ( ch  ->  ps )  ->  ps ) )   =>    |-  ( ch  ->  ( -.  ( ch  ->  ( ch  /\  -.  ch ) )  <->  ( ch  ->  ps ) ) )
 
Theoremconfun4 37922 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 37923 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 37924 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 37925 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 37926 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
 
Theoremmdandyv0 37927 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 37928 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 37929 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 37930 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 37931 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 37932 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 37933 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 37934 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 37935 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 37936 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 37937 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 37938 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 37939 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 37940 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 37941 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 37942 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 37943 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 37944 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 37945 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 37946 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 37947 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 37948 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 37949 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 37950 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 37951 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 37952 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 37953 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 37954 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 37955 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 37956 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 37957 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 37958 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 37959 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 37960 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 37961 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 37962 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 37963 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 37964 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 37965 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 37966 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 37967 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 37968 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 37969 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 37970 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 37971 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 37972 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 37973 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 37974 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 37975 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 37976

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 37977 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 37978 Theorem 19.32 of [Margaris] p. 90 with restricted quantifiers, analogous to r19.32v 2981. (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 37979* An equivalent expression for restricted existence, analogous to exsb 2264. (Contributed by Alexander van der Vekens, 1-Jul-2017.)
 |-  ( E. x  e.  A  ph  <->  E. y  e.  A  A. x ( x  =  y  ->  ph ) )
 
Theoremrexrsb 37980* An equivalent expression for restricted existence, analogous to exsb 2264. (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 37981* An equivalent expression for double restricted existence, analogous to rexsb 37979. (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 37982* An equivalent expression for double restricted existence, analogous to 2exsb 2265. (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 37983* Change bound variables of double restricted universal quantification, using implicit substitution, analogous to cbvral2v 3070. (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 37984* Change bound variables of double restricted universal quantification, using implicit substitution, analogous to cbvrex2v 3071. (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 37985 Split a biconditional and distribute 2 quantifiers, analogous to 2albiim 1747 and ralbiim 2967. (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 37986* Rearrange restricted quantifiers with two different restricting classes, analogous to raaan 3911. 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 37987 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 37988 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 37989 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 37990* 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 37991* 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 37992* Double restricted quantification with existential uniqueness, analogous to 2euex 2344. (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 37993* Double restricted quantification with restricted existential uniqueness and restricted "at most one.", analogous to 2eumo 2345. (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 37994* Double restricted existential uniqueness, analogous to 2eu2ex 2346. (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 37995* A condition allowing swap of restricted "at most one" and restricted existential quantifiers, analogous to 2moswap 2347. (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 37996* Double restricted existential uniqueness implies double restricted uniqueness quantification, analogous to 2exeu 2349. (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 37997* Double restricted existential uniqueness. This theorem shows a condition under which a "naive" definition matches the correct one, analogous to 2eu1 2354. (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 37998* Double restricted existential uniqueness, analogous to 2eu2 2356. (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 37999* Double restricted existential uniqueness, analogous to 2eu3 2357. (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 38000* Definition of double restricted existential uniqueness ("exactly one  x and exactly one  y"), analogous to 2eu4 2358 with the additional requirement that the restricting classes are not empty (which is not necessary as shown in 2reu4 38001). (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 )
 ) ) ) )
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