ProveDetails Command
- ProveDetails( <Boolean Expression> )
-
Returns some details of the result of the automated proof.
Normally, GeoGebra decides whether a boolean expression is true or not by using numerical computations. However, the ProveDetails command uses symbolic methods to determine whether a statement is true or false in general. This command works like the Prove command, but also returns some details of the result as a list:
-
An empty list {} if GeoGebra cannot determine the answer.
-
A list with one element: {false}, if the statement is not true in general.
-
A list with one element: {true}, if the statement is always true (in all cases when the diagram can be constructed).
-
A list with more elements, containing the boolean value true and another list for some so-called non-degeneracy conditions, if the statement is true under certain conditions, e.g. {true, {"AreCollinear(A,B,C),AreEqual(C,D)"}}. This means that if none of the conditions are true (and the diagram can be constructed), then the statement will be true.
-
A list {true,{"…"}}, if the statement is true under certain conditions, but these conditions cannot be translated to human readable form for some reasons.
Let us define a triangle with vertices A, B and C, and define D=MidPoint(B,C)
, E=MidPoint(A,C)
,
p=Line(A,B)
, q=Line(D,E)
. Now ProveDetails(p∥q)
returns {true}, which means that if the diagram can
be constructed, then the midline DE of the triangle is parallel to the side AB.
Let AB be the segment a, and define C=MidPoint(A,B)
, b=PerpendicularBisector(A,B)
,
D=Intersect(a,b)
. Now ProveDetails(C==D)
returns {true,{"AreEqual(A,B)"}}: it means that if the points A
and B differ, then the points C and D will coincide.
Let AB be the segment a, and define l=Line(A,B)
. Let C be an arbitrary point on line l, moreover let
b=Segment(B,C)
, c=Segment(A,C)
. Now ProveDetails(a==b+c)
returns {true,{"a+b==c", "b==a+c"}}: it
means that if neither \(a+b=c\), nor \(b=a+c\), then \(a=b+c\).
It is possible that the list of the non-degeneracy conditions is not the simplest possible set. For the above example, the simplest set would be the empty set.
See also Prove command, Boolean values, GeoGebra Automated Reasoning Tools: A Tutorial and technical details of the algorithms. |