Possible bug in Solve function?












7














In 11.3.0 for Microsoft Windows (64-bit) (March 7, 2018) writing:



f[w_, x_, y_, z_] := w*x^2*y^3 - z*(w^2 + x^2 + y^2 - 1)



eqn = {D[f[w, x, y, z], w] == 0, 

       D[f[w, x, y, z], x] == 0, 

       D[f[w, x, y, z], y] == 0, 

       D[f[w, x, y, z], z] == 0};



sol = Solve[eqn];



Table[eqn /. sol[[n]], {n, Length[sol]}]


I get:




{{True, True, True, True},
{True, True, True, True},
{True, True, True, True},
{True, True, True, True},
{True, True, True, True},
{True, True, True, True},
{True, True, True, True},
{True, True, True, True},
{False, True, True, False},
{True, True, True, True},
{True, True, True, True},
{False, True, True, False},
{True, True, True, True},
{True, True, True, True},
{False, True, True, False},
{True, True, True, True},
{True, True, True, True},
{False, True, True, False},
{True, True, True, True},
{True, True, True, True}}




from which there are four wrong solutions.



Am I wrong or is it a Solve bug?






equation-solving bugs







share|improve this question






















  • You could use List@ToRules@Reduce[eqn, {x, y, z, w}] to get all valid solutions. Filter for those that only have numeric values on the RHS of ->.
    – Szabolcs
    3 hours ago










  • Select[sol, And @@ eqn /. # &]
    – Bob Hanlon
    1 hour ago
















7














In 11.3.0 for Microsoft Windows (64-bit) (March 7, 2018) writing:



f[w_, x_, y_, z_] := w*x^2*y^3 - z*(w^2 + x^2 + y^2 - 1)



eqn = {D[f[w, x, y, z], w] == 0, 

       D[f[w, x, y, z], x] == 0, 

       D[f[w, x, y, z], y] == 0, 

       D[f[w, x, y, z], z] == 0};



sol = Solve[eqn];



Table[eqn /. sol[[n]], {n, Length[sol]}]


I get:




{{True, True, True, True},
{True, True, True, True},
{True, True, True, True},
{True, True, True, True},
{True, True, True, True},
{True, True, True, True},
{True, True, True, True},
{True, True, True, True},
{False, True, True, False},
{True, True, True, True},
{True, True, True, True},
{False, True, True, False},
{True, True, True, True},
{True, True, True, True},
{False, True, True, False},
{True, True, True, True},
{True, True, True, True},
{False, True, True, False},
{True, True, True, True},
{True, True, True, True}}




from which there are four wrong solutions.



Am I wrong or is it a Solve bug?






equation-solving bugs







share|improve this question






















  • You could use List@ToRules@Reduce[eqn, {x, y, z, w}] to get all valid solutions. Filter for those that only have numeric values on the RHS of ->.
    – Szabolcs
    3 hours ago










  • Select[sol, And @@ eqn /. # &]
    – Bob Hanlon
    1 hour ago














7












7








7


1





In 11.3.0 for Microsoft Windows (64-bit) (March 7, 2018) writing:



f[w_, x_, y_, z_] := w*x^2*y^3 - z*(w^2 + x^2 + y^2 - 1)



eqn = {D[f[w, x, y, z], w] == 0, 

       D[f[w, x, y, z], x] == 0, 

       D[f[w, x, y, z], y] == 0, 

       D[f[w, x, y, z], z] == 0};



sol = Solve[eqn];



Table[eqn /. sol[[n]], {n, Length[sol]}]


I get:




{{True, True, True, True},
{True, True, True, True},
{True, True, True, True},
{True, True, True, True},
{True, True, True, True},
{True, True, True, True},
{True, True, True, True},
{True, True, True, True},
{False, True, True, False},
{True, True, True, True},
{True, True, True, True},
{False, True, True, False},
{True, True, True, True},
{True, True, True, True},
{False, True, True, False},
{True, True, True, True},
{True, True, True, True},
{False, True, True, False},
{True, True, True, True},
{True, True, True, True}}




from which there are four wrong solutions.



Am I wrong or is it a Solve bug?






equation-solving bugs







share|improve this question













In 11.3.0 for Microsoft Windows (64-bit) (March 7, 2018) writing:



f[w_, x_, y_, z_] := w*x^2*y^3 - z*(w^2 + x^2 + y^2 - 1)



eqn = {D[f[w, x, y, z], w] == 0, 

       D[f[w, x, y, z], x] == 0, 

       D[f[w, x, y, z], y] == 0, 

       D[f[w, x, y, z], z] == 0};



sol = Solve[eqn];



Table[eqn /. sol[[n]], {n, Length[sol]}]


I get:




{{True, True, True, True},
{True, True, True, True},
{True, True, True, True},
{True, True, True, True},
{True, True, True, True},
{True, True, True, True},
{True, True, True, True},
{True, True, True, True},
{False, True, True, False},
{True, True, True, True},
{True, True, True, True},
{False, True, True, False},
{True, True, True, True},
{True, True, True, True},
{False, True, True, False},
{True, True, True, True},
{True, True, True, True},
{False, True, True, False},
{True, True, True, True},
{True, True, True, True}}




from which there are four wrong solutions.



Am I wrong or is it a Solve bug?






equation-solving bugs




equation-solving bugs






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked 3 hours ago









TeM

1,938620




1,938620












  • You could use List@ToRules@Reduce[eqn, {x, y, z, w}] to get all valid solutions. Filter for those that only have numeric values on the RHS of ->.
    – Szabolcs
    3 hours ago










  • Select[sol, And @@ eqn /. # &]
    – Bob Hanlon
    1 hour ago


















  • You could use List@ToRules@Reduce[eqn, {x, y, z, w}] to get all valid solutions. Filter for those that only have numeric values on the RHS of ->.
    – Szabolcs
    3 hours ago










  • Select[sol, And @@ eqn /. # &]
    – Bob Hanlon
    1 hour ago
















You could use List@ToRules@Reduce[eqn, {x, y, z, w}] to get all valid solutions. Filter for those that only have numeric values on the RHS of ->.
– Szabolcs
3 hours ago




You could use List@ToRules@Reduce[eqn, {x, y, z, w}] to get all valid solutions. Filter for those that only have numeric values on the RHS of ->.
– Szabolcs
3 hours ago












Select[sol, And @@ eqn /. # &]
– Bob Hanlon
1 hour ago




Select[sol, And @@ eqn /. # &]
– Bob Hanlon
1 hour ago










2 Answers
2






active

oldest

votes


















2














Thanks for asking! In version 9.0 only 16 solutions are returned and they are all valid. In version 10.2 there are 20 solutions, with the extra 4 all being invalid. Contragulations! I think you found a bug. You may want to click "Help", then "Give Feedback...", and then fill out the form in your browser to report.



As another answer notes, you can always try Reduce instead which may give better results in some cases, but Solve is usually what you want.






share|improve this answer























  • Well, it will mean that the next version will be the first calculation I will perform. Thank you! ^_^
    – TeM
    3 hours ago










  • @TeM You should report it to Wolfram first, otherwise there's no chance for it to get fixed.
    – Szabolcs
    3 hours ago










  • @Szabolcs: Could you direct me where I can do it correctly?
    – TeM
    3 hours ago






  • 1




    @TeM wolfram.com/support/contact
    – Szabolcs
    1 hour ago



















1














You can use Reduce



f[w_, x_, y_, z_] := w*x^2*y^3 - z*(w^2 + x^2 + y^2 - 1)

eqn = {D[f[w, x, y, z], w] == 0, D[f[w, x, y, z], x] == 0, 

   D[f[w, x, y, z], y] == 0, D[f[w, x, y, z], z] == 0};

red = Reduce[eqn, Backsubstitution -> True]



$left(z=0land x=0land w=-sqrt{1-y^2}right)lor left(z=0land x=0land w=sqrt{1-y^2}right)lor left(z=0land y=0land
w=-sqrt{1-x^2}right)lor left(z=0land y=0land w=sqrt{1-x^2}right)lor (z=0land y=-1land x=0land w=0)lor (z=0land y=0land x=0land
w=-1)lor (z=0land y=0land x=0land w=1)lor (z=0land y=1land x=0land w=0)lor left(z=-frac{1}{4 sqrt{3}}land y=-frac{1}{sqrt{2}}land
x=-frac{1}{sqrt{3}}land w=frac{1}{sqrt{6}}right)lor left(z=-frac{1}{4 sqrt{3}}land y=-frac{1}{sqrt{2}}land x=frac{1}{sqrt{3}}land
w=frac{1}{sqrt{6}}right)lor left(z=-frac{1}{4 sqrt{3}}land y=frac{1}{sqrt{2}}land x=-frac{1}{sqrt{3}}land
w=-frac{1}{sqrt{6}}right)lor left(z=-frac{1}{4 sqrt{3}}land y=frac{1}{sqrt{2}}land x=frac{1}{sqrt{3}}land
w=-frac{1}{sqrt{6}}right)lor left(z=frac{1}{4 sqrt{3}}land y=-frac{1}{sqrt{2}}land x=-frac{1}{sqrt{3}}land
w=-frac{1}{sqrt{6}}right)lor left(z=frac{1}{4 sqrt{3}}land y=-frac{1}{sqrt{2}}land x=frac{1}{sqrt{3}}land
w=-frac{1}{sqrt{6}}right)lor left(z=frac{1}{4 sqrt{3}}land y=frac{1}{sqrt{2}}land x=-frac{1}{sqrt{3}}land
w=frac{1}{sqrt{6}}right)\
lor left(z=frac{1}{4 sqrt{3}}land y=frac{1}{sqrt{2}}land x=frac{1}{sqrt{3}}land w=frac{1}{sqrt{6}}right)$




First@eqn //. {ToRules[red]}



{True, True, True, True, True, True, True, True, True, True, True,
True, True, True, True, True}







share|improve this answer





















  • Yes, of course, I had already tried. I was almost certain it was a bug from Solve, so I pointed out. Thank you very much anyway, always very kind!
    – TeM
    33 mins ago






  • 1




    I believe Solve use the function Reduce under the hood. when you remove Backsubstitution -> True, you'll find implicit solution, somehow Solve messes up somewhere and it is definitely a bug..
    – Okkes Dulgerci
    30 mins ago











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









2














Thanks for asking! In version 9.0 only 16 solutions are returned and they are all valid. In version 10.2 there are 20 solutions, with the extra 4 all being invalid. Contragulations! I think you found a bug. You may want to click "Help", then "Give Feedback...", and then fill out the form in your browser to report.



As another answer notes, you can always try Reduce instead which may give better results in some cases, but Solve is usually what you want.






share|improve this answer























  • Well, it will mean that the next version will be the first calculation I will perform. Thank you! ^_^
    – TeM
    3 hours ago










  • @TeM You should report it to Wolfram first, otherwise there's no chance for it to get fixed.
    – Szabolcs
    3 hours ago










  • @Szabolcs: Could you direct me where I can do it correctly?
    – TeM
    3 hours ago






  • 1




    @TeM wolfram.com/support/contact
    – Szabolcs
    1 hour ago
















2














Thanks for asking! In version 9.0 only 16 solutions are returned and they are all valid. In version 10.2 there are 20 solutions, with the extra 4 all being invalid. Contragulations! I think you found a bug. You may want to click "Help", then "Give Feedback...", and then fill out the form in your browser to report.



As another answer notes, you can always try Reduce instead which may give better results in some cases, but Solve is usually what you want.






share|improve this answer























  • Well, it will mean that the next version will be the first calculation I will perform. Thank you! ^_^
    – TeM
    3 hours ago










  • @TeM You should report it to Wolfram first, otherwise there's no chance for it to get fixed.
    – Szabolcs
    3 hours ago










  • @Szabolcs: Could you direct me where I can do it correctly?
    – TeM
    3 hours ago






  • 1




    @TeM wolfram.com/support/contact
    – Szabolcs
    1 hour ago














2












2








2






Thanks for asking! In version 9.0 only 16 solutions are returned and they are all valid. In version 10.2 there are 20 solutions, with the extra 4 all being invalid. Contragulations! I think you found a bug. You may want to click "Help", then "Give Feedback...", and then fill out the form in your browser to report.



As another answer notes, you can always try Reduce instead which may give better results in some cases, but Solve is usually what you want.






share|improve this answer














Thanks for asking! In version 9.0 only 16 solutions are returned and they are all valid. In version 10.2 there are 20 solutions, with the extra 4 all being invalid. Contragulations! I think you found a bug. You may want to click "Help", then "Give Feedback...", and then fill out the form in your browser to report.



As another answer notes, you can always try Reduce instead which may give better results in some cases, but Solve is usually what you want.







share|improve this answer














share|improve this answer



share|improve this answer








edited 31 mins ago

























answered 3 hours ago









Somos

2727




2727












  • Well, it will mean that the next version will be the first calculation I will perform. Thank you! ^_^
    – TeM
    3 hours ago










  • @TeM You should report it to Wolfram first, otherwise there's no chance for it to get fixed.
    – Szabolcs
    3 hours ago










  • @Szabolcs: Could you direct me where I can do it correctly?
    – TeM
    3 hours ago






  • 1




    @TeM wolfram.com/support/contact
    – Szabolcs
    1 hour ago


















  • Well, it will mean that the next version will be the first calculation I will perform. Thank you! ^_^
    – TeM
    3 hours ago










  • @TeM You should report it to Wolfram first, otherwise there's no chance for it to get fixed.
    – Szabolcs
    3 hours ago










  • @Szabolcs: Could you direct me where I can do it correctly?
    – TeM
    3 hours ago






  • 1




    @TeM wolfram.com/support/contact
    – Szabolcs
    1 hour ago
















Well, it will mean that the next version will be the first calculation I will perform. Thank you! ^_^
– TeM
3 hours ago




Well, it will mean that the next version will be the first calculation I will perform. Thank you! ^_^
– TeM
3 hours ago












@TeM You should report it to Wolfram first, otherwise there's no chance for it to get fixed.
– Szabolcs
3 hours ago




@TeM You should report it to Wolfram first, otherwise there's no chance for it to get fixed.
– Szabolcs
3 hours ago












@Szabolcs: Could you direct me where I can do it correctly?
– TeM
3 hours ago




@Szabolcs: Could you direct me where I can do it correctly?
– TeM
3 hours ago




1




1




@TeM wolfram.com/support/contact
– Szabolcs
1 hour ago




@TeM wolfram.com/support/contact
– Szabolcs
1 hour ago











1














You can use Reduce



f[w_, x_, y_, z_] := w*x^2*y^3 - z*(w^2 + x^2 + y^2 - 1)

eqn = {D[f[w, x, y, z], w] == 0, D[f[w, x, y, z], x] == 0, 

   D[f[w, x, y, z], y] == 0, D[f[w, x, y, z], z] == 0};

red = Reduce[eqn, Backsubstitution -> True]



$left(z=0land x=0land w=-sqrt{1-y^2}right)lor left(z=0land x=0land w=sqrt{1-y^2}right)lor left(z=0land y=0land
w=-sqrt{1-x^2}right)lor left(z=0land y=0land w=sqrt{1-x^2}right)lor (z=0land y=-1land x=0land w=0)lor (z=0land y=0land x=0land
w=-1)lor (z=0land y=0land x=0land w=1)lor (z=0land y=1land x=0land w=0)lor left(z=-frac{1}{4 sqrt{3}}land y=-frac{1}{sqrt{2}}land
x=-frac{1}{sqrt{3}}land w=frac{1}{sqrt{6}}right)lor left(z=-frac{1}{4 sqrt{3}}land y=-frac{1}{sqrt{2}}land x=frac{1}{sqrt{3}}land
w=frac{1}{sqrt{6}}right)lor left(z=-frac{1}{4 sqrt{3}}land y=frac{1}{sqrt{2}}land x=-frac{1}{sqrt{3}}land
w=-frac{1}{sqrt{6}}right)lor left(z=-frac{1}{4 sqrt{3}}land y=frac{1}{sqrt{2}}land x=frac{1}{sqrt{3}}land
w=-frac{1}{sqrt{6}}right)lor left(z=frac{1}{4 sqrt{3}}land y=-frac{1}{sqrt{2}}land x=-frac{1}{sqrt{3}}land
w=-frac{1}{sqrt{6}}right)lor left(z=frac{1}{4 sqrt{3}}land y=-frac{1}{sqrt{2}}land x=frac{1}{sqrt{3}}land
w=-frac{1}{sqrt{6}}right)lor left(z=frac{1}{4 sqrt{3}}land y=frac{1}{sqrt{2}}land x=-frac{1}{sqrt{3}}land
w=frac{1}{sqrt{6}}right)\
lor left(z=frac{1}{4 sqrt{3}}land y=frac{1}{sqrt{2}}land x=frac{1}{sqrt{3}}land w=frac{1}{sqrt{6}}right)$




First@eqn //. {ToRules[red]}



{True, True, True, True, True, True, True, True, True, True, True,
True, True, True, True, True}







share|improve this answer





















  • Yes, of course, I had already tried. I was almost certain it was a bug from Solve, so I pointed out. Thank you very much anyway, always very kind!
    – TeM
    33 mins ago






  • 1




    I believe Solve use the function Reduce under the hood. when you remove Backsubstitution -> True, you'll find implicit solution, somehow Solve messes up somewhere and it is definitely a bug..
    – Okkes Dulgerci
    30 mins ago
















1














You can use Reduce



f[w_, x_, y_, z_] := w*x^2*y^3 - z*(w^2 + x^2 + y^2 - 1)

eqn = {D[f[w, x, y, z], w] == 0, D[f[w, x, y, z], x] == 0, 

   D[f[w, x, y, z], y] == 0, D[f[w, x, y, z], z] == 0};

red = Reduce[eqn, Backsubstitution -> True]



$left(z=0land x=0land w=-sqrt{1-y^2}right)lor left(z=0land x=0land w=sqrt{1-y^2}right)lor left(z=0land y=0land
w=-sqrt{1-x^2}right)lor left(z=0land y=0land w=sqrt{1-x^2}right)lor (z=0land y=-1land x=0land w=0)lor (z=0land y=0land x=0land
w=-1)lor (z=0land y=0land x=0land w=1)lor (z=0land y=1land x=0land w=0)lor left(z=-frac{1}{4 sqrt{3}}land y=-frac{1}{sqrt{2}}land
x=-frac{1}{sqrt{3}}land w=frac{1}{sqrt{6}}right)lor left(z=-frac{1}{4 sqrt{3}}land y=-frac{1}{sqrt{2}}land x=frac{1}{sqrt{3}}land
w=frac{1}{sqrt{6}}right)lor left(z=-frac{1}{4 sqrt{3}}land y=frac{1}{sqrt{2}}land x=-frac{1}{sqrt{3}}land
w=-frac{1}{sqrt{6}}right)lor left(z=-frac{1}{4 sqrt{3}}land y=frac{1}{sqrt{2}}land x=frac{1}{sqrt{3}}land
w=-frac{1}{sqrt{6}}right)lor left(z=frac{1}{4 sqrt{3}}land y=-frac{1}{sqrt{2}}land x=-frac{1}{sqrt{3}}land
w=-frac{1}{sqrt{6}}right)lor left(z=frac{1}{4 sqrt{3}}land y=-frac{1}{sqrt{2}}land x=frac{1}{sqrt{3}}land
w=-frac{1}{sqrt{6}}right)lor left(z=frac{1}{4 sqrt{3}}land y=frac{1}{sqrt{2}}land x=-frac{1}{sqrt{3}}land
w=frac{1}{sqrt{6}}right)\
lor left(z=frac{1}{4 sqrt{3}}land y=frac{1}{sqrt{2}}land x=frac{1}{sqrt{3}}land w=frac{1}{sqrt{6}}right)$




First@eqn //. {ToRules[red]}



{True, True, True, True, True, True, True, True, True, True, True,
True, True, True, True, True}







share|improve this answer





















  • Yes, of course, I had already tried. I was almost certain it was a bug from Solve, so I pointed out. Thank you very much anyway, always very kind!
    – TeM
    33 mins ago






  • 1




    I believe Solve use the function Reduce under the hood. when you remove Backsubstitution -> True, you'll find implicit solution, somehow Solve messes up somewhere and it is definitely a bug..
    – Okkes Dulgerci
    30 mins ago














1












1








1






You can use Reduce



f[w_, x_, y_, z_] := w*x^2*y^3 - z*(w^2 + x^2 + y^2 - 1)

eqn = {D[f[w, x, y, z], w] == 0, D[f[w, x, y, z], x] == 0, 

   D[f[w, x, y, z], y] == 0, D[f[w, x, y, z], z] == 0};

red = Reduce[eqn, Backsubstitution -> True]



$left(z=0land x=0land w=-sqrt{1-y^2}right)lor left(z=0land x=0land w=sqrt{1-y^2}right)lor left(z=0land y=0land
w=-sqrt{1-x^2}right)lor left(z=0land y=0land w=sqrt{1-x^2}right)lor (z=0land y=-1land x=0land w=0)lor (z=0land y=0land x=0land
w=-1)lor (z=0land y=0land x=0land w=1)lor (z=0land y=1land x=0land w=0)lor left(z=-frac{1}{4 sqrt{3}}land y=-frac{1}{sqrt{2}}land
x=-frac{1}{sqrt{3}}land w=frac{1}{sqrt{6}}right)lor left(z=-frac{1}{4 sqrt{3}}land y=-frac{1}{sqrt{2}}land x=frac{1}{sqrt{3}}land
w=frac{1}{sqrt{6}}right)lor left(z=-frac{1}{4 sqrt{3}}land y=frac{1}{sqrt{2}}land x=-frac{1}{sqrt{3}}land
w=-frac{1}{sqrt{6}}right)lor left(z=-frac{1}{4 sqrt{3}}land y=frac{1}{sqrt{2}}land x=frac{1}{sqrt{3}}land
w=-frac{1}{sqrt{6}}right)lor left(z=frac{1}{4 sqrt{3}}land y=-frac{1}{sqrt{2}}land x=-frac{1}{sqrt{3}}land
w=-frac{1}{sqrt{6}}right)lor left(z=frac{1}{4 sqrt{3}}land y=-frac{1}{sqrt{2}}land x=frac{1}{sqrt{3}}land
w=-frac{1}{sqrt{6}}right)lor left(z=frac{1}{4 sqrt{3}}land y=frac{1}{sqrt{2}}land x=-frac{1}{sqrt{3}}land
w=frac{1}{sqrt{6}}right)\
lor left(z=frac{1}{4 sqrt{3}}land y=frac{1}{sqrt{2}}land x=frac{1}{sqrt{3}}land w=frac{1}{sqrt{6}}right)$




First@eqn //. {ToRules[red]}



{True, True, True, True, True, True, True, True, True, True, True,
True, True, True, True, True}







share|improve this answer












You can use Reduce



f[w_, x_, y_, z_] := w*x^2*y^3 - z*(w^2 + x^2 + y^2 - 1)

eqn = {D[f[w, x, y, z], w] == 0, D[f[w, x, y, z], x] == 0, 

   D[f[w, x, y, z], y] == 0, D[f[w, x, y, z], z] == 0};

red = Reduce[eqn, Backsubstitution -> True]



$left(z=0land x=0land w=-sqrt{1-y^2}right)lor left(z=0land x=0land w=sqrt{1-y^2}right)lor left(z=0land y=0land
w=-sqrt{1-x^2}right)lor left(z=0land y=0land w=sqrt{1-x^2}right)lor (z=0land y=-1land x=0land w=0)lor (z=0land y=0land x=0land
w=-1)lor (z=0land y=0land x=0land w=1)lor (z=0land y=1land x=0land w=0)lor left(z=-frac{1}{4 sqrt{3}}land y=-frac{1}{sqrt{2}}land
x=-frac{1}{sqrt{3}}land w=frac{1}{sqrt{6}}right)lor left(z=-frac{1}{4 sqrt{3}}land y=-frac{1}{sqrt{2}}land x=frac{1}{sqrt{3}}land
w=frac{1}{sqrt{6}}right)lor left(z=-frac{1}{4 sqrt{3}}land y=frac{1}{sqrt{2}}land x=-frac{1}{sqrt{3}}land
w=-frac{1}{sqrt{6}}right)lor left(z=-frac{1}{4 sqrt{3}}land y=frac{1}{sqrt{2}}land x=frac{1}{sqrt{3}}land
w=-frac{1}{sqrt{6}}right)lor left(z=frac{1}{4 sqrt{3}}land y=-frac{1}{sqrt{2}}land x=-frac{1}{sqrt{3}}land
w=-frac{1}{sqrt{6}}right)lor left(z=frac{1}{4 sqrt{3}}land y=-frac{1}{sqrt{2}}land x=frac{1}{sqrt{3}}land
w=-frac{1}{sqrt{6}}right)lor left(z=frac{1}{4 sqrt{3}}land y=frac{1}{sqrt{2}}land x=-frac{1}{sqrt{3}}land
w=frac{1}{sqrt{6}}right)\
lor left(z=frac{1}{4 sqrt{3}}land y=frac{1}{sqrt{2}}land x=frac{1}{sqrt{3}}land w=frac{1}{sqrt{6}}right)$




First@eqn //. {ToRules[red]}



{True, True, True, True, True, True, True, True, True, True, True,
True, True, True, True, True}








share|improve this answer












share|improve this answer



share|improve this answer










answered 38 mins ago









Okkes Dulgerci

4,1551816




4,1551816












  • Yes, of course, I had already tried. I was almost certain it was a bug from Solve, so I pointed out. Thank you very much anyway, always very kind!
    – TeM
    33 mins ago






  • 1




    I believe Solve use the function Reduce under the hood. when you remove Backsubstitution -> True, you'll find implicit solution, somehow Solve messes up somewhere and it is definitely a bug..
    – Okkes Dulgerci
    30 mins ago


















  • Yes, of course, I had already tried. I was almost certain it was a bug from Solve, so I pointed out. Thank you very much anyway, always very kind!
    – TeM
    33 mins ago






  • 1




    I believe Solve use the function Reduce under the hood. when you remove Backsubstitution -> True, you'll find implicit solution, somehow Solve messes up somewhere and it is definitely a bug..
    – Okkes Dulgerci
    30 mins ago
















Yes, of course, I had already tried. I was almost certain it was a bug from Solve, so I pointed out. Thank you very much anyway, always very kind!
– TeM
33 mins ago




Yes, of course, I had already tried. I was almost certain it was a bug from Solve, so I pointed out. Thank you very much anyway, always very kind!
– TeM
33 mins ago




1




1




I believe Solve use the function Reduce under the hood. when you remove Backsubstitution -> True, you'll find implicit solution, somehow Solve messes up somewhere and it is definitely a bug..
– Okkes Dulgerci
30 mins ago




I believe Solve use the function Reduce under the hood. when you remove Backsubstitution -> True, you'll find implicit solution, somehow Solve messes up somewhere and it is definitely a bug..
– Okkes Dulgerci
30 mins ago


















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