(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