(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).
Queries:
tree_member(a,g).
(1) PrologToDTProblemTransformerProof (SOUND transformation)
Built DT problem from termination graph.
(2) Obligation:
Triples:tree_member1(T53, tree(T14, tree(T50, T51, T52), T16)) :- tree_member1(T53, T51).
tree_member1(T80, tree(T14, tree(T77, T78, T79), T16)) :- tree_member1(T80, T79).
tree_member1(T106, tree(T103, T104, T105)) :- tree_member1(T106, T105).
tree_member1(T157, tree(T118, tree(T154, T155, T156), T120)) :- tree_member1(T157, T155).
tree_member1(T184, tree(T118, tree(T181, T182, T183), T120)) :- tree_member1(T184, T183).
tree_member1(T210, tree(T207, T208, T209)) :- tree_member1(T210, T209).
Clauses:
tree_memberc1(T6, tree(T6, T7, T8)).
tree_memberc1(T30, tree(T14, tree(T30, T31, T32), T16)).
tree_memberc1(T53, tree(T14, tree(T50, T51, T52), T16)) :- tree_memberc1(T53, T51).
tree_memberc1(T80, tree(T14, tree(T77, T78, T79), T16)) :- tree_memberc1(T80, T79).
tree_memberc1(T106, tree(T103, T104, T105)) :- tree_memberc1(T106, T105).
tree_memberc1(T134, tree(T118, tree(T134, T135, T136), T120)).
tree_memberc1(T157, tree(T118, tree(T154, T155, T156), T120)) :- tree_memberc1(T157, T155).
tree_memberc1(T184, tree(T118, tree(T181, T182, T183), T120)) :- tree_memberc1(T184, T183).
tree_memberc1(T210, tree(T207, T208, T209)) :- tree_memberc1(T210, T209).
Afs:
tree_member1(x1, x2) = tree_member1(x2)
(3) TriplesToPiDPProof (SOUND transformation)
We use the technique of [LOPSTR]. With regard to the inferred argument filtering the predicates were used in the following modes:tree_member1_in: (f,b)
Transforming TRIPLES into the following Term Rewriting System:
Pi DP problem:
The TRS P consists of the following rules:
TREE_MEMBER1_IN_AG(T53, tree(T14, tree(T50, T51, T52), T16)) → U1_AG(T53, T14, T50, T51, T52, T16, tree_member1_in_ag(T53, T51))
TREE_MEMBER1_IN_AG(T53, tree(T14, tree(T50, T51, T52), T16)) → TREE_MEMBER1_IN_AG(T53, T51)
TREE_MEMBER1_IN_AG(T80, tree(T14, tree(T77, T78, T79), T16)) → U2_AG(T80, T14, T77, T78, T79, T16, tree_member1_in_ag(T80, T79))
TREE_MEMBER1_IN_AG(T80, tree(T14, tree(T77, T78, T79), T16)) → TREE_MEMBER1_IN_AG(T80, T79)
TREE_MEMBER1_IN_AG(T106, tree(T103, T104, T105)) → U3_AG(T106, T103, T104, T105, tree_member1_in_ag(T106, T105))
TREE_MEMBER1_IN_AG(T106, tree(T103, T104, T105)) → TREE_MEMBER1_IN_AG(T106, T105)
TREE_MEMBER1_IN_AG(T157, tree(T118, tree(T154, T155, T156), T120)) → U4_AG(T157, T118, T154, T155, T156, T120, tree_member1_in_ag(T157, T155))
TREE_MEMBER1_IN_AG(T184, tree(T118, tree(T181, T182, T183), T120)) → U5_AG(T184, T118, T181, T182, T183, T120, tree_member1_in_ag(T184, T183))
TREE_MEMBER1_IN_AG(T210, tree(T207, T208, T209)) → U6_AG(T210, T207, T208, T209, tree_member1_in_ag(T210, T209))
R is empty.
The argument filtering Pi contains the following mapping:
tree_member1_in_ag(x1, x2) = tree_member1_in_ag(x2)
tree(x1, x2, x3) = tree(x1, x2, x3)
TREE_MEMBER1_IN_AG(x1, x2) = TREE_MEMBER1_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)
U6_AG(x1, x2, x3, x4, x5) = U6_AG(x2, x3, x4, x5)
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_MEMBER1_IN_AG(T53, tree(T14, tree(T50, T51, T52), T16)) → U1_AG(T53, T14, T50, T51, T52, T16, tree_member1_in_ag(T53, T51))
TREE_MEMBER1_IN_AG(T53, tree(T14, tree(T50, T51, T52), T16)) → TREE_MEMBER1_IN_AG(T53, T51)
TREE_MEMBER1_IN_AG(T80, tree(T14, tree(T77, T78, T79), T16)) → U2_AG(T80, T14, T77, T78, T79, T16, tree_member1_in_ag(T80, T79))
TREE_MEMBER1_IN_AG(T80, tree(T14, tree(T77, T78, T79), T16)) → TREE_MEMBER1_IN_AG(T80, T79)
TREE_MEMBER1_IN_AG(T106, tree(T103, T104, T105)) → U3_AG(T106, T103, T104, T105, tree_member1_in_ag(T106, T105))
TREE_MEMBER1_IN_AG(T106, tree(T103, T104, T105)) → TREE_MEMBER1_IN_AG(T106, T105)
TREE_MEMBER1_IN_AG(T157, tree(T118, tree(T154, T155, T156), T120)) → U4_AG(T157, T118, T154, T155, T156, T120, tree_member1_in_ag(T157, T155))
TREE_MEMBER1_IN_AG(T184, tree(T118, tree(T181, T182, T183), T120)) → U5_AG(T184, T118, T181, T182, T183, T120, tree_member1_in_ag(T184, T183))
TREE_MEMBER1_IN_AG(T210, tree(T207, T208, T209)) → U6_AG(T210, T207, T208, T209, tree_member1_in_ag(T210, T209))
R is empty.
The argument filtering Pi contains the following mapping:
tree_member1_in_ag(x1, x2) = tree_member1_in_ag(x2)
tree(x1, x2, x3) = tree(x1, x2, x3)
TREE_MEMBER1_IN_AG(x1, x2) = TREE_MEMBER1_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)
U6_AG(x1, x2, x3, x4, x5) = U6_AG(x2, x3, x4, 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 6 less nodes.(6) Obligation:
Pi DP problem:The TRS P consists of the following rules:
TREE_MEMBER1_IN_AG(T80, tree(T14, tree(T77, T78, T79), T16)) → TREE_MEMBER1_IN_AG(T80, T79)
TREE_MEMBER1_IN_AG(T53, tree(T14, tree(T50, T51, T52), T16)) → TREE_MEMBER1_IN_AG(T53, T51)
TREE_MEMBER1_IN_AG(T106, tree(T103, T104, T105)) → TREE_MEMBER1_IN_AG(T106, T105)
R is empty.
The argument filtering Pi contains the following mapping:
tree(x1, x2, x3) = tree(x1, x2, x3)
TREE_MEMBER1_IN_AG(x1, x2) = TREE_MEMBER1_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_MEMBER1_IN_AG(tree(T14, tree(T77, T78, T79), T16)) → TREE_MEMBER1_IN_AG(T79)
TREE_MEMBER1_IN_AG(tree(T14, tree(T50, T51, T52), T16)) → TREE_MEMBER1_IN_AG(T51)
TREE_MEMBER1_IN_AG(tree(T103, T104, T105)) → TREE_MEMBER1_IN_AG(T105)
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_MEMBER1_IN_AG(tree(T14, tree(T77, T78, T79), T16)) → TREE_MEMBER1_IN_AG(T79)
The graph contains the following edges 1 > 1 - TREE_MEMBER1_IN_AG(tree(T14, tree(T50, T51, T52), T16)) → TREE_MEMBER1_IN_AG(T51)
The graph contains the following edges 1 > 1 - TREE_MEMBER1_IN_AG(tree(T103, T104, T105)) → TREE_MEMBER1_IN_AG(T105)
The graph contains the following edges 1 > 1