/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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

con(g)

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

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


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(0)
Obligation:
Clauses:

dis(or(B1, B2)) :- ','(con(B1), dis(B2)).
dis(B) :- con(B).
con(and(B1, B2)) :- ','(dis(B1), con(B2)).
con(B) :- bool(B).
bool(0).
bool(1).


Query: con(g)
----------------------------------------

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

con_in_1: (b)

dis_in_1: (b)

Transforming Prolog into the following Term Rewriting System:

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

   con_in_g(and(B1, B2)) -> U4_g(B1, B2, dis_in_g(B1))
   dis_in_g(or(B1, B2)) -> U1_g(B1, B2, con_in_g(B1))
   con_in_g(B) -> U6_g(B, bool_in_g(B))
   bool_in_g(0) -> bool_out_g(0)
   bool_in_g(1) -> bool_out_g(1)
   U6_g(B, bool_out_g(B)) -> con_out_g(B)
   U1_g(B1, B2, con_out_g(B1)) -> U2_g(B1, B2, dis_in_g(B2))
   dis_in_g(B) -> U3_g(B, con_in_g(B))
   U3_g(B, con_out_g(B)) -> dis_out_g(B)
   U2_g(B1, B2, dis_out_g(B2)) -> dis_out_g(or(B1, B2))
   U4_g(B1, B2, dis_out_g(B1)) -> U5_g(B1, B2, con_in_g(B2))
   U5_g(B1, B2, con_out_g(B2)) -> con_out_g(and(B1, B2))

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:

   con_in_g(and(B1, B2)) -> U4_g(B1, B2, dis_in_g(B1))
   dis_in_g(or(B1, B2)) -> U1_g(B1, B2, con_in_g(B1))
   con_in_g(B) -> U6_g(B, bool_in_g(B))
   bool_in_g(0) -> bool_out_g(0)
   bool_in_g(1) -> bool_out_g(1)
   U6_g(B, bool_out_g(B)) -> con_out_g(B)
   U1_g(B1, B2, con_out_g(B1)) -> U2_g(B1, B2, dis_in_g(B2))
   dis_in_g(B) -> U3_g(B, con_in_g(B))
   U3_g(B, con_out_g(B)) -> dis_out_g(B)
   U2_g(B1, B2, dis_out_g(B2)) -> dis_out_g(or(B1, B2))
   U4_g(B1, B2, dis_out_g(B1)) -> U5_g(B1, B2, con_in_g(B2))
   U5_g(B1, B2, con_out_g(B2)) -> con_out_g(and(B1, B2))

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:

   CON_IN_G(and(B1, B2)) -> U4_G(B1, B2, dis_in_g(B1))
   CON_IN_G(and(B1, B2)) -> DIS_IN_G(B1)
   DIS_IN_G(or(B1, B2)) -> U1_G(B1, B2, con_in_g(B1))
   DIS_IN_G(or(B1, B2)) -> CON_IN_G(B1)
   CON_IN_G(B) -> U6_G(B, bool_in_g(B))
   CON_IN_G(B) -> BOOL_IN_G(B)
   U1_G(B1, B2, con_out_g(B1)) -> U2_G(B1, B2, dis_in_g(B2))
   U1_G(B1, B2, con_out_g(B1)) -> DIS_IN_G(B2)
   DIS_IN_G(B) -> U3_G(B, con_in_g(B))
   DIS_IN_G(B) -> CON_IN_G(B)
   U4_G(B1, B2, dis_out_g(B1)) -> U5_G(B1, B2, con_in_g(B2))
   U4_G(B1, B2, dis_out_g(B1)) -> CON_IN_G(B2)

The TRS R consists of the following rules:

   con_in_g(and(B1, B2)) -> U4_g(B1, B2, dis_in_g(B1))
   dis_in_g(or(B1, B2)) -> U1_g(B1, B2, con_in_g(B1))
   con_in_g(B) -> U6_g(B, bool_in_g(B))
   bool_in_g(0) -> bool_out_g(0)
   bool_in_g(1) -> bool_out_g(1)
   U6_g(B, bool_out_g(B)) -> con_out_g(B)
   U1_g(B1, B2, con_out_g(B1)) -> U2_g(B1, B2, dis_in_g(B2))
   dis_in_g(B) -> U3_g(B, con_in_g(B))
   U3_g(B, con_out_g(B)) -> dis_out_g(B)
   U2_g(B1, B2, dis_out_g(B2)) -> dis_out_g(or(B1, B2))
   U4_g(B1, B2, dis_out_g(B1)) -> U5_g(B1, B2, con_in_g(B2))
   U5_g(B1, B2, con_out_g(B2)) -> con_out_g(and(B1, B2))

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:

   CON_IN_G(and(B1, B2)) -> U4_G(B1, B2, dis_in_g(B1))
   CON_IN_G(and(B1, B2)) -> DIS_IN_G(B1)
   DIS_IN_G(or(B1, B2)) -> U1_G(B1, B2, con_in_g(B1))
   DIS_IN_G(or(B1, B2)) -> CON_IN_G(B1)
   CON_IN_G(B) -> U6_G(B, bool_in_g(B))
   CON_IN_G(B) -> BOOL_IN_G(B)
   U1_G(B1, B2, con_out_g(B1)) -> U2_G(B1, B2, dis_in_g(B2))
   U1_G(B1, B2, con_out_g(B1)) -> DIS_IN_G(B2)
   DIS_IN_G(B) -> U3_G(B, con_in_g(B))
   DIS_IN_G(B) -> CON_IN_G(B)
   U4_G(B1, B2, dis_out_g(B1)) -> U5_G(B1, B2, con_in_g(B2))
   U4_G(B1, B2, dis_out_g(B1)) -> CON_IN_G(B2)

The TRS R consists of the following rules:

   con_in_g(and(B1, B2)) -> U4_g(B1, B2, dis_in_g(B1))
   dis_in_g(or(B1, B2)) -> U1_g(B1, B2, con_in_g(B1))
   con_in_g(B) -> U6_g(B, bool_in_g(B))
   bool_in_g(0) -> bool_out_g(0)
   bool_in_g(1) -> bool_out_g(1)
   U6_g(B, bool_out_g(B)) -> con_out_g(B)
   U1_g(B1, B2, con_out_g(B1)) -> U2_g(B1, B2, dis_in_g(B2))
   dis_in_g(B) -> U3_g(B, con_in_g(B))
   U3_g(B, con_out_g(B)) -> dis_out_g(B)
   U2_g(B1, B2, dis_out_g(B2)) -> dis_out_g(or(B1, B2))
   U4_g(B1, B2, dis_out_g(B1)) -> U5_g(B1, B2, con_in_g(B2))
   U5_g(B1, B2, con_out_g(B2)) -> con_out_g(and(B1, B2))

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 5 less nodes.
----------------------------------------

(6)
Obligation:
Pi DP problem:
The TRS P consists of the following rules:

   U4_G(B1, B2, dis_out_g(B1)) -> CON_IN_G(B2)
   CON_IN_G(and(B1, B2)) -> U4_G(B1, B2, dis_in_g(B1))
   CON_IN_G(and(B1, B2)) -> DIS_IN_G(B1)
   DIS_IN_G(or(B1, B2)) -> U1_G(B1, B2, con_in_g(B1))
   U1_G(B1, B2, con_out_g(B1)) -> DIS_IN_G(B2)
   DIS_IN_G(or(B1, B2)) -> CON_IN_G(B1)
   DIS_IN_G(B) -> CON_IN_G(B)

The TRS R consists of the following rules:

   con_in_g(and(B1, B2)) -> U4_g(B1, B2, dis_in_g(B1))
   dis_in_g(or(B1, B2)) -> U1_g(B1, B2, con_in_g(B1))
   con_in_g(B) -> U6_g(B, bool_in_g(B))
   bool_in_g(0) -> bool_out_g(0)
   bool_in_g(1) -> bool_out_g(1)
   U6_g(B, bool_out_g(B)) -> con_out_g(B)
   U1_g(B1, B2, con_out_g(B1)) -> U2_g(B1, B2, dis_in_g(B2))
   dis_in_g(B) -> U3_g(B, con_in_g(B))
   U3_g(B, con_out_g(B)) -> dis_out_g(B)
   U2_g(B1, B2, dis_out_g(B2)) -> dis_out_g(or(B1, B2))
   U4_g(B1, B2, dis_out_g(B1)) -> U5_g(B1, B2, con_in_g(B2))
   U5_g(B1, B2, con_out_g(B2)) -> con_out_g(and(B1, B2))

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:

   U4_G(B1, B2, dis_out_g(B1)) -> CON_IN_G(B2)
   CON_IN_G(and(B1, B2)) -> U4_G(B1, B2, dis_in_g(B1))
   CON_IN_G(and(B1, B2)) -> DIS_IN_G(B1)
   DIS_IN_G(or(B1, B2)) -> U1_G(B1, B2, con_in_g(B1))
   U1_G(B1, B2, con_out_g(B1)) -> DIS_IN_G(B2)
   DIS_IN_G(or(B1, B2)) -> CON_IN_G(B1)
   DIS_IN_G(B) -> CON_IN_G(B)

The TRS R consists of the following rules:

   con_in_g(and(B1, B2)) -> U4_g(B1, B2, dis_in_g(B1))
   dis_in_g(or(B1, B2)) -> U1_g(B1, B2, con_in_g(B1))
   con_in_g(B) -> U6_g(B, bool_in_g(B))
   bool_in_g(0) -> bool_out_g(0)
   bool_in_g(1) -> bool_out_g(1)
   U6_g(B, bool_out_g(B)) -> con_out_g(B)
   U1_g(B1, B2, con_out_g(B1)) -> U2_g(B1, B2, dis_in_g(B2))
   dis_in_g(B) -> U3_g(B, con_in_g(B))
   U3_g(B, con_out_g(B)) -> dis_out_g(B)
   U2_g(B1, B2, dis_out_g(B2)) -> dis_out_g(or(B1, B2))
   U4_g(B1, B2, dis_out_g(B1)) -> U5_g(B1, B2, con_in_g(B2))
   U5_g(B1, B2, con_out_g(B2)) -> con_out_g(and(B1, B2))

The set Q consists of the following terms:

   con_in_g(x0)
   dis_in_g(x0)
   bool_in_g(x0)
   U6_g(x0, x1)
   U1_g(x0, x1, x2)
   U3_g(x0, x1)
   U2_g(x0, x1, x2)
   U4_g(x0, x1, x2)
   U5_g(x0, x1, x2)

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:
*CON_IN_G(and(B1, B2)) -> U4_G(B1, B2, dis_in_g(B1))
The graph contains the following edges 1 > 1, 1 > 2


*CON_IN_G(and(B1, B2)) -> DIS_IN_G(B1)
The graph contains the following edges 1 > 1


*U4_G(B1, B2, dis_out_g(B1)) -> CON_IN_G(B2)
The graph contains the following edges 2 >= 1


*DIS_IN_G(or(B1, B2)) -> U1_G(B1, B2, con_in_g(B1))
The graph contains the following edges 1 > 1, 1 > 2


*U1_G(B1, B2, con_out_g(B1)) -> DIS_IN_G(B2)
The graph contains the following edges 2 >= 1


*DIS_IN_G(or(B1, B2)) -> CON_IN_G(B1)
The graph contains the following edges 1 > 1


*DIS_IN_G(B) -> CON_IN_G(B)
The graph contains the following edges 1 >= 1


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(10)
YES