(0) Obligation:

Clauses:

bin_tree(void).
bin_tree(T) :- ','(no(empty(T)), ','(left(T, L), ','(right(T, R), ','(bin_tree(L), bin_tree(R))))).
left(void, void).
left(tree(X1, L, X2), L).
right(void, void).
right(tree(X3, X4, R), R).
empty(void).
no(X) :- ','(X, ','(!, failure(a))).
no(X5).
failure(b).

Queries:

bin_tree(g).

(1) PrologToPrologProblemTransformerProof (SOUND transformation)

Built Prolog problem from termination graph.

(2) Obligation:

Clauses:

bin_tree1(void).
bin_tree1(tree(T8, T9, T10)) :- bin_tree1(T9).
bin_tree1(tree(T8, T9, T10)) :- ','(bin_tree1(T9), bin_tree1(T10)).

Queries:

bin_tree1(g).

(3) PrologToPiTRSProof (SOUND transformation)

We use the technique of [LOPSTR]. With regard to the inferred argument filtering the predicates were used in the following modes:
bin_tree1_in: (b)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

bin_tree1_in_g(void) → bin_tree1_out_g(void)
bin_tree1_in_g(tree(T8, T9, T10)) → U1_g(T8, T9, T10, bin_tree1_in_g(T9))
U1_g(T8, T9, T10, bin_tree1_out_g(T9)) → bin_tree1_out_g(tree(T8, T9, T10))
U1_g(T8, T9, T10, bin_tree1_out_g(T9)) → U2_g(T8, T9, T10, bin_tree1_in_g(T10))
U2_g(T8, T9, T10, bin_tree1_out_g(T10)) → bin_tree1_out_g(tree(T8, T9, T10))

The argument filtering Pi contains the following mapping:
bin_tree1_in_g(x1)  =  bin_tree1_in_g(x1)
void  =  void
bin_tree1_out_g(x1)  =  bin_tree1_out_g
tree(x1, x2, x3)  =  tree(x1, x2, x3)
U1_g(x1, x2, x3, x4)  =  U1_g(x3, x4)
U2_g(x1, x2, x3, x4)  =  U2_g(x4)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog

(4) Obligation:

Pi-finite rewrite system:
The TRS R consists of the following rules:

bin_tree1_in_g(void) → bin_tree1_out_g(void)
bin_tree1_in_g(tree(T8, T9, T10)) → U1_g(T8, T9, T10, bin_tree1_in_g(T9))
U1_g(T8, T9, T10, bin_tree1_out_g(T9)) → bin_tree1_out_g(tree(T8, T9, T10))
U1_g(T8, T9, T10, bin_tree1_out_g(T9)) → U2_g(T8, T9, T10, bin_tree1_in_g(T10))
U2_g(T8, T9, T10, bin_tree1_out_g(T10)) → bin_tree1_out_g(tree(T8, T9, T10))

The argument filtering Pi contains the following mapping:
bin_tree1_in_g(x1)  =  bin_tree1_in_g(x1)
void  =  void
bin_tree1_out_g(x1)  =  bin_tree1_out_g
tree(x1, x2, x3)  =  tree(x1, x2, x3)
U1_g(x1, x2, x3, x4)  =  U1_g(x3, x4)
U2_g(x1, x2, x3, x4)  =  U2_g(x4)

(5) 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:

BIN_TREE1_IN_G(tree(T8, T9, T10)) → U1_G(T8, T9, T10, bin_tree1_in_g(T9))
BIN_TREE1_IN_G(tree(T8, T9, T10)) → BIN_TREE1_IN_G(T9)
U1_G(T8, T9, T10, bin_tree1_out_g(T9)) → U2_G(T8, T9, T10, bin_tree1_in_g(T10))
U1_G(T8, T9, T10, bin_tree1_out_g(T9)) → BIN_TREE1_IN_G(T10)

The TRS R consists of the following rules:

bin_tree1_in_g(void) → bin_tree1_out_g(void)
bin_tree1_in_g(tree(T8, T9, T10)) → U1_g(T8, T9, T10, bin_tree1_in_g(T9))
U1_g(T8, T9, T10, bin_tree1_out_g(T9)) → bin_tree1_out_g(tree(T8, T9, T10))
U1_g(T8, T9, T10, bin_tree1_out_g(T9)) → U2_g(T8, T9, T10, bin_tree1_in_g(T10))
U2_g(T8, T9, T10, bin_tree1_out_g(T10)) → bin_tree1_out_g(tree(T8, T9, T10))

The argument filtering Pi contains the following mapping:
bin_tree1_in_g(x1)  =  bin_tree1_in_g(x1)
void  =  void
bin_tree1_out_g(x1)  =  bin_tree1_out_g
tree(x1, x2, x3)  =  tree(x1, x2, x3)
U1_g(x1, x2, x3, x4)  =  U1_g(x3, x4)
U2_g(x1, x2, x3, x4)  =  U2_g(x4)
BIN_TREE1_IN_G(x1)  =  BIN_TREE1_IN_G(x1)
U1_G(x1, x2, x3, x4)  =  U1_G(x3, x4)
U2_G(x1, x2, x3, x4)  =  U2_G(x4)

We have to consider all (P,R,Pi)-chains

(6) Obligation:

Pi DP problem:
The TRS P consists of the following rules:

BIN_TREE1_IN_G(tree(T8, T9, T10)) → U1_G(T8, T9, T10, bin_tree1_in_g(T9))
BIN_TREE1_IN_G(tree(T8, T9, T10)) → BIN_TREE1_IN_G(T9)
U1_G(T8, T9, T10, bin_tree1_out_g(T9)) → U2_G(T8, T9, T10, bin_tree1_in_g(T10))
U1_G(T8, T9, T10, bin_tree1_out_g(T9)) → BIN_TREE1_IN_G(T10)

The TRS R consists of the following rules:

bin_tree1_in_g(void) → bin_tree1_out_g(void)
bin_tree1_in_g(tree(T8, T9, T10)) → U1_g(T8, T9, T10, bin_tree1_in_g(T9))
U1_g(T8, T9, T10, bin_tree1_out_g(T9)) → bin_tree1_out_g(tree(T8, T9, T10))
U1_g(T8, T9, T10, bin_tree1_out_g(T9)) → U2_g(T8, T9, T10, bin_tree1_in_g(T10))
U2_g(T8, T9, T10, bin_tree1_out_g(T10)) → bin_tree1_out_g(tree(T8, T9, T10))

The argument filtering Pi contains the following mapping:
bin_tree1_in_g(x1)  =  bin_tree1_in_g(x1)
void  =  void
bin_tree1_out_g(x1)  =  bin_tree1_out_g
tree(x1, x2, x3)  =  tree(x1, x2, x3)
U1_g(x1, x2, x3, x4)  =  U1_g(x3, x4)
U2_g(x1, x2, x3, x4)  =  U2_g(x4)
BIN_TREE1_IN_G(x1)  =  BIN_TREE1_IN_G(x1)
U1_G(x1, x2, x3, x4)  =  U1_G(x3, x4)
U2_G(x1, x2, x3, x4)  =  U2_G(x4)

We have to consider all (P,R,Pi)-chains

(7) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 1 less node.

(8) Obligation:

Pi DP problem:
The TRS P consists of the following rules:

U1_G(T8, T9, T10, bin_tree1_out_g(T9)) → BIN_TREE1_IN_G(T10)
BIN_TREE1_IN_G(tree(T8, T9, T10)) → U1_G(T8, T9, T10, bin_tree1_in_g(T9))
BIN_TREE1_IN_G(tree(T8, T9, T10)) → BIN_TREE1_IN_G(T9)

The TRS R consists of the following rules:

bin_tree1_in_g(void) → bin_tree1_out_g(void)
bin_tree1_in_g(tree(T8, T9, T10)) → U1_g(T8, T9, T10, bin_tree1_in_g(T9))
U1_g(T8, T9, T10, bin_tree1_out_g(T9)) → bin_tree1_out_g(tree(T8, T9, T10))
U1_g(T8, T9, T10, bin_tree1_out_g(T9)) → U2_g(T8, T9, T10, bin_tree1_in_g(T10))
U2_g(T8, T9, T10, bin_tree1_out_g(T10)) → bin_tree1_out_g(tree(T8, T9, T10))

The argument filtering Pi contains the following mapping:
bin_tree1_in_g(x1)  =  bin_tree1_in_g(x1)
void  =  void
bin_tree1_out_g(x1)  =  bin_tree1_out_g
tree(x1, x2, x3)  =  tree(x1, x2, x3)
U1_g(x1, x2, x3, x4)  =  U1_g(x3, x4)
U2_g(x1, x2, x3, x4)  =  U2_g(x4)
BIN_TREE1_IN_G(x1)  =  BIN_TREE1_IN_G(x1)
U1_G(x1, x2, x3, x4)  =  U1_G(x3, x4)

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:

U1_G(T10, bin_tree1_out_g) → BIN_TREE1_IN_G(T10)
BIN_TREE1_IN_G(tree(T8, T9, T10)) → U1_G(T10, bin_tree1_in_g(T9))
BIN_TREE1_IN_G(tree(T8, T9, T10)) → BIN_TREE1_IN_G(T9)

The TRS R consists of the following rules:

bin_tree1_in_g(void) → bin_tree1_out_g
bin_tree1_in_g(tree(T8, T9, T10)) → U1_g(T10, bin_tree1_in_g(T9))
U1_g(T10, bin_tree1_out_g) → bin_tree1_out_g
U1_g(T10, bin_tree1_out_g) → U2_g(bin_tree1_in_g(T10))
U2_g(bin_tree1_out_g) → bin_tree1_out_g

The set Q consists of the following terms:

bin_tree1_in_g(x0)
U1_g(x0, x1)
U2_g(x0)

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:

  • BIN_TREE1_IN_G(tree(T8, T9, T10)) → U1_G(T10, bin_tree1_in_g(T9))
    The graph contains the following edges 1 > 1

  • BIN_TREE1_IN_G(tree(T8, T9, T10)) → BIN_TREE1_IN_G(T9)
    The graph contains the following edges 1 > 1

  • U1_G(T10, bin_tree1_out_g) → BIN_TREE1_IN_G(T10)
    The graph contains the following edges 1 >= 1

(12) TRUE