(0) Obligation:
Clauses:
tree_member(X, tree(X, X1, X2)).
tree_member(X, tree(X3, Left, X4)) :- tree_member(X, Left).
tree_member(X, tree(X5, X6, Right)) :- tree_member(X, Right).
Query: tree_member(a,g)
(1) PrologToPiTRSProof (SOUND transformation)
We use the technique of [TOCL09]. With regard to the inferred argument filtering the predicates were used in the following modes:
tree_member_in: (f,b)
Transforming
Prolog into the following
Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:
tree_member_in_ag(X, tree(X, X1, X2)) → tree_member_out_ag(X, tree(X, X1, X2))
tree_member_in_ag(X, tree(X3, Left, X4)) → U1_ag(X, X3, Left, X4, tree_member_in_ag(X, Left))
tree_member_in_ag(X, tree(X5, X6, Right)) → U2_ag(X, X5, X6, Right, tree_member_in_ag(X, Right))
U2_ag(X, X5, X6, Right, tree_member_out_ag(X, Right)) → tree_member_out_ag(X, tree(X5, X6, Right))
U1_ag(X, X3, Left, X4, tree_member_out_ag(X, Left)) → tree_member_out_ag(X, tree(X3, Left, X4))
The argument filtering Pi contains the following mapping:
tree_member_in_ag(
x1,
x2) =
tree_member_in_ag(
x2)
tree(
x1,
x2,
x3) =
tree(
x1,
x2,
x3)
tree_member_out_ag(
x1,
x2) =
tree_member_out_ag(
x1)
U1_ag(
x1,
x2,
x3,
x4,
x5) =
U1_ag(
x5)
U2_ag(
x1,
x2,
x3,
x4,
x5) =
U2_ag(
x5)
Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog
(2) Obligation:
Pi-finite rewrite system:
The TRS R consists of the following rules:
tree_member_in_ag(X, tree(X, X1, X2)) → tree_member_out_ag(X, tree(X, X1, X2))
tree_member_in_ag(X, tree(X3, Left, X4)) → U1_ag(X, X3, Left, X4, tree_member_in_ag(X, Left))
tree_member_in_ag(X, tree(X5, X6, Right)) → U2_ag(X, X5, X6, Right, tree_member_in_ag(X, Right))
U2_ag(X, X5, X6, Right, tree_member_out_ag(X, Right)) → tree_member_out_ag(X, tree(X5, X6, Right))
U1_ag(X, X3, Left, X4, tree_member_out_ag(X, Left)) → tree_member_out_ag(X, tree(X3, Left, X4))
The argument filtering Pi contains the following mapping:
tree_member_in_ag(
x1,
x2) =
tree_member_in_ag(
x2)
tree(
x1,
x2,
x3) =
tree(
x1,
x2,
x3)
tree_member_out_ag(
x1,
x2) =
tree_member_out_ag(
x1)
U1_ag(
x1,
x2,
x3,
x4,
x5) =
U1_ag(
x5)
U2_ag(
x1,
x2,
x3,
x4,
x5) =
U2_ag(
x5)
(3) DependencyPairsProof (EQUIVALENT transformation)
Using Dependency Pairs [AG00,LOPSTR] we result in the following initial DP problem:
Pi DP problem:
The TRS P consists of the following rules:
TREE_MEMBER_IN_AG(X, tree(X3, Left, X4)) → U1_AG(X, X3, Left, X4, tree_member_in_ag(X, Left))
TREE_MEMBER_IN_AG(X, tree(X3, Left, X4)) → TREE_MEMBER_IN_AG(X, Left)
TREE_MEMBER_IN_AG(X, tree(X5, X6, Right)) → U2_AG(X, X5, X6, Right, tree_member_in_ag(X, Right))
TREE_MEMBER_IN_AG(X, tree(X5, X6, Right)) → TREE_MEMBER_IN_AG(X, Right)
The TRS R consists of the following rules:
tree_member_in_ag(X, tree(X, X1, X2)) → tree_member_out_ag(X, tree(X, X1, X2))
tree_member_in_ag(X, tree(X3, Left, X4)) → U1_ag(X, X3, Left, X4, tree_member_in_ag(X, Left))
tree_member_in_ag(X, tree(X5, X6, Right)) → U2_ag(X, X5, X6, Right, tree_member_in_ag(X, Right))
U2_ag(X, X5, X6, Right, tree_member_out_ag(X, Right)) → tree_member_out_ag(X, tree(X5, X6, Right))
U1_ag(X, X3, Left, X4, tree_member_out_ag(X, Left)) → tree_member_out_ag(X, tree(X3, Left, X4))
The argument filtering Pi contains the following mapping:
tree_member_in_ag(
x1,
x2) =
tree_member_in_ag(
x2)
tree(
x1,
x2,
x3) =
tree(
x1,
x2,
x3)
tree_member_out_ag(
x1,
x2) =
tree_member_out_ag(
x1)
U1_ag(
x1,
x2,
x3,
x4,
x5) =
U1_ag(
x5)
U2_ag(
x1,
x2,
x3,
x4,
x5) =
U2_ag(
x5)
TREE_MEMBER_IN_AG(
x1,
x2) =
TREE_MEMBER_IN_AG(
x2)
U1_AG(
x1,
x2,
x3,
x4,
x5) =
U1_AG(
x5)
U2_AG(
x1,
x2,
x3,
x4,
x5) =
U2_AG(
x5)
We have to consider all (P,R,Pi)-chains
(4) Obligation:
Pi DP problem:
The TRS P consists of the following rules:
TREE_MEMBER_IN_AG(X, tree(X3, Left, X4)) → U1_AG(X, X3, Left, X4, tree_member_in_ag(X, Left))
TREE_MEMBER_IN_AG(X, tree(X3, Left, X4)) → TREE_MEMBER_IN_AG(X, Left)
TREE_MEMBER_IN_AG(X, tree(X5, X6, Right)) → U2_AG(X, X5, X6, Right, tree_member_in_ag(X, Right))
TREE_MEMBER_IN_AG(X, tree(X5, X6, Right)) → TREE_MEMBER_IN_AG(X, Right)
The TRS R consists of the following rules:
tree_member_in_ag(X, tree(X, X1, X2)) → tree_member_out_ag(X, tree(X, X1, X2))
tree_member_in_ag(X, tree(X3, Left, X4)) → U1_ag(X, X3, Left, X4, tree_member_in_ag(X, Left))
tree_member_in_ag(X, tree(X5, X6, Right)) → U2_ag(X, X5, X6, Right, tree_member_in_ag(X, Right))
U2_ag(X, X5, X6, Right, tree_member_out_ag(X, Right)) → tree_member_out_ag(X, tree(X5, X6, Right))
U1_ag(X, X3, Left, X4, tree_member_out_ag(X, Left)) → tree_member_out_ag(X, tree(X3, Left, X4))
The argument filtering Pi contains the following mapping:
tree_member_in_ag(
x1,
x2) =
tree_member_in_ag(
x2)
tree(
x1,
x2,
x3) =
tree(
x1,
x2,
x3)
tree_member_out_ag(
x1,
x2) =
tree_member_out_ag(
x1)
U1_ag(
x1,
x2,
x3,
x4,
x5) =
U1_ag(
x5)
U2_ag(
x1,
x2,
x3,
x4,
x5) =
U2_ag(
x5)
TREE_MEMBER_IN_AG(
x1,
x2) =
TREE_MEMBER_IN_AG(
x2)
U1_AG(
x1,
x2,
x3,
x4,
x5) =
U1_AG(
x5)
U2_AG(
x1,
x2,
x3,
x4,
x5) =
U2_AG(
x5)
We have to consider all (P,R,Pi)-chains
(5) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 2 less nodes.
(6) Obligation:
Pi DP problem:
The TRS P consists of the following rules:
TREE_MEMBER_IN_AG(X, tree(X5, X6, Right)) → TREE_MEMBER_IN_AG(X, Right)
TREE_MEMBER_IN_AG(X, tree(X3, Left, X4)) → TREE_MEMBER_IN_AG(X, Left)
The TRS R consists of the following rules:
tree_member_in_ag(X, tree(X, X1, X2)) → tree_member_out_ag(X, tree(X, X1, X2))
tree_member_in_ag(X, tree(X3, Left, X4)) → U1_ag(X, X3, Left, X4, tree_member_in_ag(X, Left))
tree_member_in_ag(X, tree(X5, X6, Right)) → U2_ag(X, X5, X6, Right, tree_member_in_ag(X, Right))
U2_ag(X, X5, X6, Right, tree_member_out_ag(X, Right)) → tree_member_out_ag(X, tree(X5, X6, Right))
U1_ag(X, X3, Left, X4, tree_member_out_ag(X, Left)) → tree_member_out_ag(X, tree(X3, Left, X4))
The argument filtering Pi contains the following mapping:
tree_member_in_ag(
x1,
x2) =
tree_member_in_ag(
x2)
tree(
x1,
x2,
x3) =
tree(
x1,
x2,
x3)
tree_member_out_ag(
x1,
x2) =
tree_member_out_ag(
x1)
U1_ag(
x1,
x2,
x3,
x4,
x5) =
U1_ag(
x5)
U2_ag(
x1,
x2,
x3,
x4,
x5) =
U2_ag(
x5)
TREE_MEMBER_IN_AG(
x1,
x2) =
TREE_MEMBER_IN_AG(
x2)
We have to consider all (P,R,Pi)-chains
(7) UsableRulesProof (EQUIVALENT transformation)
For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R.
(8) Obligation:
Pi DP problem:
The TRS P consists of the following rules:
TREE_MEMBER_IN_AG(X, tree(X5, X6, Right)) → TREE_MEMBER_IN_AG(X, Right)
TREE_MEMBER_IN_AG(X, tree(X3, Left, X4)) → TREE_MEMBER_IN_AG(X, Left)
R is empty.
The argument filtering Pi contains the following mapping:
tree(
x1,
x2,
x3) =
tree(
x1,
x2,
x3)
TREE_MEMBER_IN_AG(
x1,
x2) =
TREE_MEMBER_IN_AG(
x2)
We have to consider all (P,R,Pi)-chains
(9) PiDPToQDPProof (SOUND transformation)
Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.
(10) Obligation:
Q DP problem:
The TRS P consists of the following rules:
TREE_MEMBER_IN_AG(tree(X5, X6, Right)) → TREE_MEMBER_IN_AG(Right)
TREE_MEMBER_IN_AG(tree(X3, Left, X4)) → TREE_MEMBER_IN_AG(Left)
R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
(11) QDPSizeChangeProof (EQUIVALENT transformation)
By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.
From the DPs we obtained the following set of size-change graphs:
- TREE_MEMBER_IN_AG(tree(X5, X6, Right)) → TREE_MEMBER_IN_AG(Right)
The graph contains the following edges 1 > 1
- TREE_MEMBER_IN_AG(tree(X3, Left, X4)) → TREE_MEMBER_IN_AG(Left)
The graph contains the following edges 1 > 1
(12) YES