/export/starexec/sandbox/solver/bin/starexec_run_standard /export/starexec/sandbox/benchmark/theBenchmark.pl /export/starexec/sandbox/output/output_files


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YES
proof of /export/starexec/sandbox/benchmark/theBenchmark.pl
# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty


Left Termination of the query pattern

bin_tree(g)

w.r.t. the given Prolog program could successfully be proven:

(0) Prolog
(1) PrologToPiTRSProof [SOUND, 0 ms]
(2) PiTRS
(3) DependencyPairsProof [EQUIVALENT, 0 ms]
(4) PiDP
(5) DependencyGraphProof [EQUIVALENT, 0 ms]
(6) PiDP
(7) PiDPToQDPProof [EQUIVALENT, 9 ms]
(8) QDP
(9) QDPSizeChangeProof [EQUIVALENT, 0 ms]
(10) YES


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(0)
Obligation:
Clauses:

bin_tree(void).
bin_tree(tree(X1, Left, Right)) :- ','(bin_tree(Left), bin_tree(Right)).


Query: bin_tree(g)
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(1) PrologToPiTRSProof (SOUND)
We use the technique of [TOCL09]. With regard to the inferred argument filtering the predicates were used in the following modes:

bin_tree_in_1: (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))

Pi is empty.



Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog



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(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))

Pi is empty.

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(3) DependencyPairsProof (EQUIVALENT)
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))

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

Pi is empty.
We have to consider all (P,R,Pi)-chains
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(5) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 1 less node.
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(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))

Pi is empty.
We have to consider all (P,R,Pi)-chains
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(7) PiDPToQDPProof (EQUIVALENT)
Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.
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(8)
Obligation:
Q 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 set Q consists of the following terms:

   bin_tree_in_g(x0)
   U1_g(x0, x1, x2, x3)
   U2_g(x0, x1, x2, x3)

We have to consider all (P,Q,R)-chains.
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(9) QDPSizeChangeProof (EQUIVALENT)
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(X1, Left, Right, bin_tree_in_g(Left))
The graph contains the following edges 1 > 1, 1 > 2, 1 > 3


*BIN_TREE_IN_G(tree(X1, Left, Right)) -> BIN_TREE_IN_G(Left)
The graph contains the following edges 1 > 1


*U1_G(X1, Left, Right, bin_tree_out_g(Left)) -> BIN_TREE_IN_G(Right)
The graph contains the following edges 3 >= 1


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(10)
YES