(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) PrologToDTProblemTransformerProof (SOUND transformation)
Built DT problem from termination graph DT10.
(2) Obligation:
Triples:
tree_memberA(X1, tree(X2, tree(X3, X4, X5), X6)) :- tree_memberA(X1, X4).
tree_memberA(X1, tree(X2, tree(X3, X4, X5), X6)) :- tree_memberA(X1, X5).
tree_memberA(X1, tree(X2, X3, X4)) :- tree_memberA(X1, X4).
tree_memberA(X1, tree(X2, tree(X3, X4, X5), X6)) :- tree_memberA(X1, X4).
tree_memberA(X1, tree(X2, tree(X3, X4, X5), X6)) :- tree_memberA(X1, X5).
tree_memberA(X1, tree(X2, X3, X4)) :- tree_memberA(X1, X4).
tree_memberA(X1, tree(X2, X3, tree(X4, X5, X6))) :- tree_memberA(X1, X5).
tree_memberA(X1, tree(X2, X3, tree(X4, X5, X6))) :- tree_memberA(X1, X6).
Clauses:
tree_membercA(X1, tree(X1, X2, X3)).
tree_membercA(X1, tree(X2, tree(X1, X3, X4), X5)).
tree_membercA(X1, tree(X2, tree(X3, X4, X5), X6)) :- tree_membercA(X1, X4).
tree_membercA(X1, tree(X2, tree(X3, X4, X5), X6)) :- tree_membercA(X1, X5).
tree_membercA(X1, tree(X2, X3, X4)) :- tree_membercA(X1, X4).
tree_membercA(X1, tree(X2, tree(X1, X3, X4), X5)).
tree_membercA(X1, tree(X2, tree(X3, X4, X5), X6)) :- tree_membercA(X1, X4).
tree_membercA(X1, tree(X2, tree(X3, X4, X5), X6)) :- tree_membercA(X1, X5).
tree_membercA(X1, tree(X2, X3, X4)) :- tree_membercA(X1, X4).
tree_membercA(X1, tree(X2, X3, tree(X1, X4, X5))).
tree_membercA(X1, tree(X2, X3, tree(X4, X5, X6))) :- tree_membercA(X1, X5).
tree_membercA(X1, tree(X2, X3, tree(X4, X5, X6))) :- tree_membercA(X1, X6).
Afs:
tree_memberA(x1, x2) = tree_memberA(x2)
(3) TriplesToPiDPProof (SOUND transformation)
We use the technique of [DT09]. With regard to the inferred argument filtering the predicates were used in the following modes:
tree_memberA_in: (f,b)
Transforming
TRIPLES into the following
Term Rewriting System:
Pi DP problem:
The TRS P consists of the following rules:
TREE_MEMBERA_IN_AG(X1, tree(X2, tree(X3, X4, X5), X6)) → U1_AG(X1, X2, X3, X4, X5, X6, tree_memberA_in_ag(X1, X4))
TREE_MEMBERA_IN_AG(X1, tree(X2, tree(X3, X4, X5), X6)) → TREE_MEMBERA_IN_AG(X1, X4)
TREE_MEMBERA_IN_AG(X1, tree(X2, tree(X3, X4, X5), X6)) → U2_AG(X1, X2, X3, X4, X5, X6, tree_memberA_in_ag(X1, X5))
TREE_MEMBERA_IN_AG(X1, tree(X2, tree(X3, X4, X5), X6)) → TREE_MEMBERA_IN_AG(X1, X5)
TREE_MEMBERA_IN_AG(X1, tree(X2, X3, X4)) → U3_AG(X1, X2, X3, X4, tree_memberA_in_ag(X1, X4))
TREE_MEMBERA_IN_AG(X1, tree(X2, X3, X4)) → TREE_MEMBERA_IN_AG(X1, X4)
TREE_MEMBERA_IN_AG(X1, tree(X2, X3, tree(X4, X5, X6))) → U4_AG(X1, X2, X3, X4, X5, X6, tree_memberA_in_ag(X1, X5))
TREE_MEMBERA_IN_AG(X1, tree(X2, X3, tree(X4, X5, X6))) → TREE_MEMBERA_IN_AG(X1, X5)
TREE_MEMBERA_IN_AG(X1, tree(X2, X3, tree(X4, X5, X6))) → U5_AG(X1, X2, X3, X4, X5, X6, tree_memberA_in_ag(X1, X6))
TREE_MEMBERA_IN_AG(X1, tree(X2, X3, tree(X4, X5, X6))) → TREE_MEMBERA_IN_AG(X1, X6)
R is empty.
The argument filtering Pi contains the following mapping:
tree_memberA_in_ag(
x1,
x2) =
tree_memberA_in_ag(
x2)
tree(
x1,
x2,
x3) =
tree(
x1,
x2,
x3)
TREE_MEMBERA_IN_AG(
x1,
x2) =
TREE_MEMBERA_IN_AG(
x2)
U1_AG(
x1,
x2,
x3,
x4,
x5,
x6,
x7) =
U1_AG(
x2,
x3,
x4,
x5,
x6,
x7)
U2_AG(
x1,
x2,
x3,
x4,
x5,
x6,
x7) =
U2_AG(
x2,
x3,
x4,
x5,
x6,
x7)
U3_AG(
x1,
x2,
x3,
x4,
x5) =
U3_AG(
x2,
x3,
x4,
x5)
U4_AG(
x1,
x2,
x3,
x4,
x5,
x6,
x7) =
U4_AG(
x2,
x3,
x4,
x5,
x6,
x7)
U5_AG(
x1,
x2,
x3,
x4,
x5,
x6,
x7) =
U5_AG(
x2,
x3,
x4,
x5,
x6,
x7)
We have to consider all (P,R,Pi)-chains
Infinitary Constructor Rewriting Termination of PiDP implies Termination of TRIPLES
(4) Obligation:
Pi DP problem:
The TRS P consists of the following rules:
TREE_MEMBERA_IN_AG(X1, tree(X2, tree(X3, X4, X5), X6)) → U1_AG(X1, X2, X3, X4, X5, X6, tree_memberA_in_ag(X1, X4))
TREE_MEMBERA_IN_AG(X1, tree(X2, tree(X3, X4, X5), X6)) → TREE_MEMBERA_IN_AG(X1, X4)
TREE_MEMBERA_IN_AG(X1, tree(X2, tree(X3, X4, X5), X6)) → U2_AG(X1, X2, X3, X4, X5, X6, tree_memberA_in_ag(X1, X5))
TREE_MEMBERA_IN_AG(X1, tree(X2, tree(X3, X4, X5), X6)) → TREE_MEMBERA_IN_AG(X1, X5)
TREE_MEMBERA_IN_AG(X1, tree(X2, X3, X4)) → U3_AG(X1, X2, X3, X4, tree_memberA_in_ag(X1, X4))
TREE_MEMBERA_IN_AG(X1, tree(X2, X3, X4)) → TREE_MEMBERA_IN_AG(X1, X4)
TREE_MEMBERA_IN_AG(X1, tree(X2, X3, tree(X4, X5, X6))) → U4_AG(X1, X2, X3, X4, X5, X6, tree_memberA_in_ag(X1, X5))
TREE_MEMBERA_IN_AG(X1, tree(X2, X3, tree(X4, X5, X6))) → TREE_MEMBERA_IN_AG(X1, X5)
TREE_MEMBERA_IN_AG(X1, tree(X2, X3, tree(X4, X5, X6))) → U5_AG(X1, X2, X3, X4, X5, X6, tree_memberA_in_ag(X1, X6))
TREE_MEMBERA_IN_AG(X1, tree(X2, X3, tree(X4, X5, X6))) → TREE_MEMBERA_IN_AG(X1, X6)
R is empty.
The argument filtering Pi contains the following mapping:
tree_memberA_in_ag(
x1,
x2) =
tree_memberA_in_ag(
x2)
tree(
x1,
x2,
x3) =
tree(
x1,
x2,
x3)
TREE_MEMBERA_IN_AG(
x1,
x2) =
TREE_MEMBERA_IN_AG(
x2)
U1_AG(
x1,
x2,
x3,
x4,
x5,
x6,
x7) =
U1_AG(
x2,
x3,
x4,
x5,
x6,
x7)
U2_AG(
x1,
x2,
x3,
x4,
x5,
x6,
x7) =
U2_AG(
x2,
x3,
x4,
x5,
x6,
x7)
U3_AG(
x1,
x2,
x3,
x4,
x5) =
U3_AG(
x2,
x3,
x4,
x5)
U4_AG(
x1,
x2,
x3,
x4,
x5,
x6,
x7) =
U4_AG(
x2,
x3,
x4,
x5,
x6,
x7)
U5_AG(
x1,
x2,
x3,
x4,
x5,
x6,
x7) =
U5_AG(
x2,
x3,
x4,
x5,
x6,
x7)
We have to consider all (P,R,Pi)-chains
(5) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 5 less nodes.
(6) Obligation:
Pi DP problem:
The TRS P consists of the following rules:
TREE_MEMBERA_IN_AG(X1, tree(X2, tree(X3, X4, X5), X6)) → TREE_MEMBERA_IN_AG(X1, X5)
TREE_MEMBERA_IN_AG(X1, tree(X2, tree(X3, X4, X5), X6)) → TREE_MEMBERA_IN_AG(X1, X4)
TREE_MEMBERA_IN_AG(X1, tree(X2, X3, X4)) → TREE_MEMBERA_IN_AG(X1, X4)
TREE_MEMBERA_IN_AG(X1, tree(X2, X3, tree(X4, X5, X6))) → TREE_MEMBERA_IN_AG(X1, X5)
TREE_MEMBERA_IN_AG(X1, tree(X2, X3, tree(X4, X5, X6))) → TREE_MEMBERA_IN_AG(X1, X6)
R is empty.
The argument filtering Pi contains the following mapping:
tree(
x1,
x2,
x3) =
tree(
x1,
x2,
x3)
TREE_MEMBERA_IN_AG(
x1,
x2) =
TREE_MEMBERA_IN_AG(
x2)
We have to consider all (P,R,Pi)-chains
(7) PiDPToQDPProof (SOUND transformation)
Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.
(8) Obligation:
Q DP problem:
The TRS P consists of the following rules:
TREE_MEMBERA_IN_AG(tree(X2, tree(X3, X4, X5), X6)) → TREE_MEMBERA_IN_AG(X5)
TREE_MEMBERA_IN_AG(tree(X2, tree(X3, X4, X5), X6)) → TREE_MEMBERA_IN_AG(X4)
TREE_MEMBERA_IN_AG(tree(X2, X3, X4)) → TREE_MEMBERA_IN_AG(X4)
TREE_MEMBERA_IN_AG(tree(X2, X3, tree(X4, X5, X6))) → TREE_MEMBERA_IN_AG(X5)
TREE_MEMBERA_IN_AG(tree(X2, X3, tree(X4, X5, X6))) → TREE_MEMBERA_IN_AG(X6)
R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
(9) 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_MEMBERA_IN_AG(tree(X2, tree(X3, X4, X5), X6)) → TREE_MEMBERA_IN_AG(X5)
The graph contains the following edges 1 > 1
- TREE_MEMBERA_IN_AG(tree(X2, tree(X3, X4, X5), X6)) → TREE_MEMBERA_IN_AG(X4)
The graph contains the following edges 1 > 1
- TREE_MEMBERA_IN_AG(tree(X2, X3, X4)) → TREE_MEMBERA_IN_AG(X4)
The graph contains the following edges 1 > 1
- TREE_MEMBERA_IN_AG(tree(X2, X3, tree(X4, X5, X6))) → TREE_MEMBERA_IN_AG(X5)
The graph contains the following edges 1 > 1
- TREE_MEMBERA_IN_AG(tree(X2, X3, tree(X4, X5, X6))) → TREE_MEMBERA_IN_AG(X6)
The graph contains the following edges 1 > 1
(10) YES