/export/starexec/sandbox2/solver/bin/starexec_run_standard /export/starexec/sandbox2/benchmark/theBenchmark.pl /export/starexec/sandbox2/output/output_files


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YES
proof of /export/starexec/sandbox2/benchmark/theBenchmark.pl
# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty


Left Termination of the query pattern

reverse_concatenate(g,g,a)

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) UsableRulesProof [EQUIVALENT, 0 ms]
(8) PiDP
(9) PiDPToQDPProof [SOUND, 0 ms]
(10) QDP
(11) QDPSizeChangeProof [EQUIVALENT, 0 ms]
(12) YES


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(0)
Obligation:
Clauses:

concatenate([], L, L).
concatenate(.(X, L1), L2, .(X, L3)) :- concatenate(L1, L2, L3).
member(X, .(X, L)).
member(X, .(Y, L)) :- member(X, L).
reverse(L, L1) :- reverse_concatenate(L, [], L1).
reverse_concatenate([], L, L).
reverse_concatenate(.(X, L1), L2, L3) :- reverse_concatenate(L1, .(X, L2), L3).


Query: reverse_concatenate(g,g,a)
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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:

reverse_concatenate_in_3: (b,b,f)

Transforming Prolog into the following Term Rewriting System:

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

   reverse_concatenate_in_gga([], L, L) -> reverse_concatenate_out_gga([], L, L)
   reverse_concatenate_in_gga(.(X, L1), L2, L3) -> U4_gga(X, L1, L2, L3, reverse_concatenate_in_gga(L1, .(X, L2), L3))
   U4_gga(X, L1, L2, L3, reverse_concatenate_out_gga(L1, .(X, L2), L3)) -> reverse_concatenate_out_gga(.(X, L1), L2, L3)

The argument filtering Pi contains the following mapping:
reverse_concatenate_in_gga(x1, x2, x3)  =  reverse_concatenate_in_gga(x1, x2)

[]  =  []

reverse_concatenate_out_gga(x1, x2, x3)  =  reverse_concatenate_out_gga(x3)

.(x1, x2)  =  .(x1, x2)

U4_gga(x1, x2, x3, x4, x5)  =  U4_gga(x5)





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:

   reverse_concatenate_in_gga([], L, L) -> reverse_concatenate_out_gga([], L, L)
   reverse_concatenate_in_gga(.(X, L1), L2, L3) -> U4_gga(X, L1, L2, L3, reverse_concatenate_in_gga(L1, .(X, L2), L3))
   U4_gga(X, L1, L2, L3, reverse_concatenate_out_gga(L1, .(X, L2), L3)) -> reverse_concatenate_out_gga(.(X, L1), L2, L3)

The argument filtering Pi contains the following mapping:
reverse_concatenate_in_gga(x1, x2, x3)  =  reverse_concatenate_in_gga(x1, x2)

[]  =  []

reverse_concatenate_out_gga(x1, x2, x3)  =  reverse_concatenate_out_gga(x3)

.(x1, x2)  =  .(x1, x2)

U4_gga(x1, x2, x3, x4, x5)  =  U4_gga(x5)



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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:

   REVERSE_CONCATENATE_IN_GGA(.(X, L1), L2, L3) -> U4_GGA(X, L1, L2, L3, reverse_concatenate_in_gga(L1, .(X, L2), L3))
   REVERSE_CONCATENATE_IN_GGA(.(X, L1), L2, L3) -> REVERSE_CONCATENATE_IN_GGA(L1, .(X, L2), L3)

The TRS R consists of the following rules:

   reverse_concatenate_in_gga([], L, L) -> reverse_concatenate_out_gga([], L, L)
   reverse_concatenate_in_gga(.(X, L1), L2, L3) -> U4_gga(X, L1, L2, L3, reverse_concatenate_in_gga(L1, .(X, L2), L3))
   U4_gga(X, L1, L2, L3, reverse_concatenate_out_gga(L1, .(X, L2), L3)) -> reverse_concatenate_out_gga(.(X, L1), L2, L3)

The argument filtering Pi contains the following mapping:
reverse_concatenate_in_gga(x1, x2, x3)  =  reverse_concatenate_in_gga(x1, x2)

[]  =  []

reverse_concatenate_out_gga(x1, x2, x3)  =  reverse_concatenate_out_gga(x3)

.(x1, x2)  =  .(x1, x2)

U4_gga(x1, x2, x3, x4, x5)  =  U4_gga(x5)

REVERSE_CONCATENATE_IN_GGA(x1, x2, x3)  =  REVERSE_CONCATENATE_IN_GGA(x1, x2)

U4_GGA(x1, x2, x3, x4, x5)  =  U4_GGA(x5)


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:

   REVERSE_CONCATENATE_IN_GGA(.(X, L1), L2, L3) -> U4_GGA(X, L1, L2, L3, reverse_concatenate_in_gga(L1, .(X, L2), L3))
   REVERSE_CONCATENATE_IN_GGA(.(X, L1), L2, L3) -> REVERSE_CONCATENATE_IN_GGA(L1, .(X, L2), L3)

The TRS R consists of the following rules:

   reverse_concatenate_in_gga([], L, L) -> reverse_concatenate_out_gga([], L, L)
   reverse_concatenate_in_gga(.(X, L1), L2, L3) -> U4_gga(X, L1, L2, L3, reverse_concatenate_in_gga(L1, .(X, L2), L3))
   U4_gga(X, L1, L2, L3, reverse_concatenate_out_gga(L1, .(X, L2), L3)) -> reverse_concatenate_out_gga(.(X, L1), L2, L3)

The argument filtering Pi contains the following mapping:
reverse_concatenate_in_gga(x1, x2, x3)  =  reverse_concatenate_in_gga(x1, x2)

[]  =  []

reverse_concatenate_out_gga(x1, x2, x3)  =  reverse_concatenate_out_gga(x3)

.(x1, x2)  =  .(x1, x2)

U4_gga(x1, x2, x3, x4, x5)  =  U4_gga(x5)

REVERSE_CONCATENATE_IN_GGA(x1, x2, x3)  =  REVERSE_CONCATENATE_IN_GGA(x1, x2)

U4_GGA(x1, x2, x3, x4, x5)  =  U4_GGA(x5)


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:

   REVERSE_CONCATENATE_IN_GGA(.(X, L1), L2, L3) -> REVERSE_CONCATENATE_IN_GGA(L1, .(X, L2), L3)

The TRS R consists of the following rules:

   reverse_concatenate_in_gga([], L, L) -> reverse_concatenate_out_gga([], L, L)
   reverse_concatenate_in_gga(.(X, L1), L2, L3) -> U4_gga(X, L1, L2, L3, reverse_concatenate_in_gga(L1, .(X, L2), L3))
   U4_gga(X, L1, L2, L3, reverse_concatenate_out_gga(L1, .(X, L2), L3)) -> reverse_concatenate_out_gga(.(X, L1), L2, L3)

The argument filtering Pi contains the following mapping:
reverse_concatenate_in_gga(x1, x2, x3)  =  reverse_concatenate_in_gga(x1, x2)

[]  =  []

reverse_concatenate_out_gga(x1, x2, x3)  =  reverse_concatenate_out_gga(x3)

.(x1, x2)  =  .(x1, x2)

U4_gga(x1, x2, x3, x4, x5)  =  U4_gga(x5)

REVERSE_CONCATENATE_IN_GGA(x1, x2, x3)  =  REVERSE_CONCATENATE_IN_GGA(x1, x2)


We have to consider all (P,R,Pi)-chains
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(7) UsableRulesProof (EQUIVALENT)
For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R.
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(8)
Obligation:
Pi DP problem:
The TRS P consists of the following rules:

   REVERSE_CONCATENATE_IN_GGA(.(X, L1), L2, L3) -> REVERSE_CONCATENATE_IN_GGA(L1, .(X, L2), L3)

R is empty.
The argument filtering Pi contains the following mapping:
.(x1, x2)  =  .(x1, x2)

REVERSE_CONCATENATE_IN_GGA(x1, x2, x3)  =  REVERSE_CONCATENATE_IN_GGA(x1, x2)


We have to consider all (P,R,Pi)-chains
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(9) PiDPToQDPProof (SOUND)
Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.
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(10)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   REVERSE_CONCATENATE_IN_GGA(.(X, L1), L2) -> REVERSE_CONCATENATE_IN_GGA(L1, .(X, L2))

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
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(11) 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:
*REVERSE_CONCATENATE_IN_GGA(.(X, L1), L2) -> REVERSE_CONCATENATE_IN_GGA(L1, .(X, L2))
The graph contains the following edges 1 > 1


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(12)
YES