(0) Obligation:
Clauses:
bin_tree(void).
bin_tree(tree(X1, Left, Right)) :- ','(bin_tree(Left), bin_tree(Right)).
Query: bin_tree(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:
bin_tree_in: (b)
Transforming
Prolog into the following
Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:
bin_tree_in_g(void) → bin_tree_out_g(void)
bin_tree_in_g(tree(X1, Left, Right)) → U1_g(X1, Left, Right, bin_tree_in_g(Left))
U1_g(X1, Left, Right, bin_tree_out_g(Left)) → U2_g(X1, Left, Right, bin_tree_in_g(Right))
U2_g(X1, Left, Right, bin_tree_out_g(Right)) → bin_tree_out_g(tree(X1, Left, Right))
The argument filtering Pi contains the following mapping:
bin_tree_in_g(
x1) =
bin_tree_in_g(
x1)
void =
void
bin_tree_out_g(
x1) =
bin_tree_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
(2) Obligation:
Pi-finite rewrite system:
The TRS R consists of the following rules:
bin_tree_in_g(void) → bin_tree_out_g(void)
bin_tree_in_g(tree(X1, Left, Right)) → U1_g(X1, Left, Right, bin_tree_in_g(Left))
U1_g(X1, Left, Right, bin_tree_out_g(Left)) → U2_g(X1, Left, Right, bin_tree_in_g(Right))
U2_g(X1, Left, Right, bin_tree_out_g(Right)) → bin_tree_out_g(tree(X1, Left, Right))
The argument filtering Pi contains the following mapping:
bin_tree_in_g(
x1) =
bin_tree_in_g(
x1)
void =
void
bin_tree_out_g(
x1) =
bin_tree_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)
(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:
BIN_TREE_IN_G(tree(X1, Left, Right)) → U1_G(X1, Left, Right, bin_tree_in_g(Left))
BIN_TREE_IN_G(tree(X1, Left, Right)) → BIN_TREE_IN_G(Left)
U1_G(X1, Left, Right, bin_tree_out_g(Left)) → U2_G(X1, Left, Right, bin_tree_in_g(Right))
U1_G(X1, Left, Right, bin_tree_out_g(Left)) → BIN_TREE_IN_G(Right)
The TRS R consists of the following rules:
bin_tree_in_g(void) → bin_tree_out_g(void)
bin_tree_in_g(tree(X1, Left, Right)) → U1_g(X1, Left, Right, bin_tree_in_g(Left))
U1_g(X1, Left, Right, bin_tree_out_g(Left)) → U2_g(X1, Left, Right, bin_tree_in_g(Right))
U2_g(X1, Left, Right, bin_tree_out_g(Right)) → bin_tree_out_g(tree(X1, Left, Right))
The argument filtering Pi contains the following mapping:
bin_tree_in_g(
x1) =
bin_tree_in_g(
x1)
void =
void
bin_tree_out_g(
x1) =
bin_tree_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_TREE_IN_G(
x1) =
BIN_TREE_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
(4) Obligation:
Pi DP problem:
The TRS P consists of the following rules:
BIN_TREE_IN_G(tree(X1, Left, Right)) → U1_G(X1, Left, Right, bin_tree_in_g(Left))
BIN_TREE_IN_G(tree(X1, Left, Right)) → BIN_TREE_IN_G(Left)
U1_G(X1, Left, Right, bin_tree_out_g(Left)) → U2_G(X1, Left, Right, bin_tree_in_g(Right))
U1_G(X1, Left, Right, bin_tree_out_g(Left)) → BIN_TREE_IN_G(Right)
The TRS R consists of the following rules:
bin_tree_in_g(void) → bin_tree_out_g(void)
bin_tree_in_g(tree(X1, Left, Right)) → U1_g(X1, Left, Right, bin_tree_in_g(Left))
U1_g(X1, Left, Right, bin_tree_out_g(Left)) → U2_g(X1, Left, Right, bin_tree_in_g(Right))
U2_g(X1, Left, Right, bin_tree_out_g(Right)) → bin_tree_out_g(tree(X1, Left, Right))
The argument filtering Pi contains the following mapping:
bin_tree_in_g(
x1) =
bin_tree_in_g(
x1)
void =
void
bin_tree_out_g(
x1) =
bin_tree_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_TREE_IN_G(
x1) =
BIN_TREE_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
(5) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 1 less node.
(6) Obligation:
Pi DP problem:
The TRS P consists of the following rules:
U1_G(X1, Left, Right, bin_tree_out_g(Left)) → BIN_TREE_IN_G(Right)
BIN_TREE_IN_G(tree(X1, Left, Right)) → U1_G(X1, Left, Right, bin_tree_in_g(Left))
BIN_TREE_IN_G(tree(X1, Left, Right)) → BIN_TREE_IN_G(Left)
The TRS R consists of the following rules:
bin_tree_in_g(void) → bin_tree_out_g(void)
bin_tree_in_g(tree(X1, Left, Right)) → U1_g(X1, Left, Right, bin_tree_in_g(Left))
U1_g(X1, Left, Right, bin_tree_out_g(Left)) → U2_g(X1, Left, Right, bin_tree_in_g(Right))
U2_g(X1, Left, Right, bin_tree_out_g(Right)) → bin_tree_out_g(tree(X1, Left, Right))
The argument filtering Pi contains the following mapping:
bin_tree_in_g(
x1) =
bin_tree_in_g(
x1)
void =
void
bin_tree_out_g(
x1) =
bin_tree_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_TREE_IN_G(
x1) =
BIN_TREE_IN_G(
x1)
U1_G(
x1,
x2,
x3,
x4) =
U1_G(
x3,
x4)
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:
U1_G(Right, bin_tree_out_g) → BIN_TREE_IN_G(Right)
BIN_TREE_IN_G(tree(X1, Left, Right)) → U1_G(Right, bin_tree_in_g(Left))
BIN_TREE_IN_G(tree(X1, Left, Right)) → BIN_TREE_IN_G(Left)
The TRS R consists of the following rules:
bin_tree_in_g(void) → bin_tree_out_g
bin_tree_in_g(tree(X1, Left, Right)) → U1_g(Right, bin_tree_in_g(Left))
U1_g(Right, bin_tree_out_g) → U2_g(bin_tree_in_g(Right))
U2_g(bin_tree_out_g) → bin_tree_out_g
The set Q consists of the following terms:
bin_tree_in_g(x0)
U1_g(x0, x1)
U2_g(x0)
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:
- BIN_TREE_IN_G(tree(X1, Left, Right)) → U1_G(Right, bin_tree_in_g(Left))
The graph contains the following edges 1 > 1
- BIN_TREE_IN_G(tree(X1, Left, Right)) → BIN_TREE_IN_G(Left)
The graph contains the following edges 1 > 1
- U1_G(Right, bin_tree_out_g) → BIN_TREE_IN_G(Right)
The graph contains the following edges 1 >= 1
(10) YES