/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty


Left Termination of the query pattern

suffix(a,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) 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:

suffix(Xs, Ys) :- app(X1, Xs, Ys).
app([], X, X).
app(.(X, Xs), Ys, .(X, Zs)) :- app(Xs, Ys, Zs).


Query: suffix(a,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:

suffix_in_2: (f,b)

app_in_3: (f,f,b)

Transforming Prolog into the following Term Rewriting System:

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

   suffix_in_ag(Xs, Ys) -> U1_ag(Xs, Ys, app_in_aag(X1, Xs, Ys))
   app_in_aag([], X, X) -> app_out_aag([], X, X)
   app_in_aag(.(X, Xs), Ys, .(X, Zs)) -> U2_aag(X, Xs, Ys, Zs, app_in_aag(Xs, Ys, Zs))
   U2_aag(X, Xs, Ys, Zs, app_out_aag(Xs, Ys, Zs)) -> app_out_aag(.(X, Xs), Ys, .(X, Zs))
   U1_ag(Xs, Ys, app_out_aag(X1, Xs, Ys)) -> suffix_out_ag(Xs, Ys)

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

U1_ag(x1, x2, x3)  =  U1_ag(x2, x3)

app_in_aag(x1, x2, x3)  =  app_in_aag(x3)

app_out_aag(x1, x2, x3)  =  app_out_aag(x1, x2, x3)

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

U2_aag(x1, x2, x3, x4, x5)  =  U2_aag(x1, x4, x5)

suffix_out_ag(x1, x2)  =  suffix_out_ag(x1, x2)





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:

   suffix_in_ag(Xs, Ys) -> U1_ag(Xs, Ys, app_in_aag(X1, Xs, Ys))
   app_in_aag([], X, X) -> app_out_aag([], X, X)
   app_in_aag(.(X, Xs), Ys, .(X, Zs)) -> U2_aag(X, Xs, Ys, Zs, app_in_aag(Xs, Ys, Zs))
   U2_aag(X, Xs, Ys, Zs, app_out_aag(Xs, Ys, Zs)) -> app_out_aag(.(X, Xs), Ys, .(X, Zs))
   U1_ag(Xs, Ys, app_out_aag(X1, Xs, Ys)) -> suffix_out_ag(Xs, Ys)

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

U1_ag(x1, x2, x3)  =  U1_ag(x2, x3)

app_in_aag(x1, x2, x3)  =  app_in_aag(x3)

app_out_aag(x1, x2, x3)  =  app_out_aag(x1, x2, x3)

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

U2_aag(x1, x2, x3, x4, x5)  =  U2_aag(x1, x4, x5)

suffix_out_ag(x1, x2)  =  suffix_out_ag(x1, x2)



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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:

   SUFFIX_IN_AG(Xs, Ys) -> U1_AG(Xs, Ys, app_in_aag(X1, Xs, Ys))
   SUFFIX_IN_AG(Xs, Ys) -> APP_IN_AAG(X1, Xs, Ys)
   APP_IN_AAG(.(X, Xs), Ys, .(X, Zs)) -> U2_AAG(X, Xs, Ys, Zs, app_in_aag(Xs, Ys, Zs))
   APP_IN_AAG(.(X, Xs), Ys, .(X, Zs)) -> APP_IN_AAG(Xs, Ys, Zs)

The TRS R consists of the following rules:

   suffix_in_ag(Xs, Ys) -> U1_ag(Xs, Ys, app_in_aag(X1, Xs, Ys))
   app_in_aag([], X, X) -> app_out_aag([], X, X)
   app_in_aag(.(X, Xs), Ys, .(X, Zs)) -> U2_aag(X, Xs, Ys, Zs, app_in_aag(Xs, Ys, Zs))
   U2_aag(X, Xs, Ys, Zs, app_out_aag(Xs, Ys, Zs)) -> app_out_aag(.(X, Xs), Ys, .(X, Zs))
   U1_ag(Xs, Ys, app_out_aag(X1, Xs, Ys)) -> suffix_out_ag(Xs, Ys)

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

U1_ag(x1, x2, x3)  =  U1_ag(x2, x3)

app_in_aag(x1, x2, x3)  =  app_in_aag(x3)

app_out_aag(x1, x2, x3)  =  app_out_aag(x1, x2, x3)

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

U2_aag(x1, x2, x3, x4, x5)  =  U2_aag(x1, x4, x5)

suffix_out_ag(x1, x2)  =  suffix_out_ag(x1, x2)

SUFFIX_IN_AG(x1, x2)  =  SUFFIX_IN_AG(x2)

U1_AG(x1, x2, x3)  =  U1_AG(x2, x3)

APP_IN_AAG(x1, x2, x3)  =  APP_IN_AAG(x3)

U2_AAG(x1, x2, x3, x4, x5)  =  U2_AAG(x1, x4, x5)


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

(4)
Obligation:
Pi DP problem:
The TRS P consists of the following rules:

   SUFFIX_IN_AG(Xs, Ys) -> U1_AG(Xs, Ys, app_in_aag(X1, Xs, Ys))
   SUFFIX_IN_AG(Xs, Ys) -> APP_IN_AAG(X1, Xs, Ys)
   APP_IN_AAG(.(X, Xs), Ys, .(X, Zs)) -> U2_AAG(X, Xs, Ys, Zs, app_in_aag(Xs, Ys, Zs))
   APP_IN_AAG(.(X, Xs), Ys, .(X, Zs)) -> APP_IN_AAG(Xs, Ys, Zs)

The TRS R consists of the following rules:

   suffix_in_ag(Xs, Ys) -> U1_ag(Xs, Ys, app_in_aag(X1, Xs, Ys))
   app_in_aag([], X, X) -> app_out_aag([], X, X)
   app_in_aag(.(X, Xs), Ys, .(X, Zs)) -> U2_aag(X, Xs, Ys, Zs, app_in_aag(Xs, Ys, Zs))
   U2_aag(X, Xs, Ys, Zs, app_out_aag(Xs, Ys, Zs)) -> app_out_aag(.(X, Xs), Ys, .(X, Zs))
   U1_ag(Xs, Ys, app_out_aag(X1, Xs, Ys)) -> suffix_out_ag(Xs, Ys)

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

U1_ag(x1, x2, x3)  =  U1_ag(x2, x3)

app_in_aag(x1, x2, x3)  =  app_in_aag(x3)

app_out_aag(x1, x2, x3)  =  app_out_aag(x1, x2, x3)

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

U2_aag(x1, x2, x3, x4, x5)  =  U2_aag(x1, x4, x5)

suffix_out_ag(x1, x2)  =  suffix_out_ag(x1, x2)

SUFFIX_IN_AG(x1, x2)  =  SUFFIX_IN_AG(x2)

U1_AG(x1, x2, x3)  =  U1_AG(x2, x3)

APP_IN_AAG(x1, x2, x3)  =  APP_IN_AAG(x3)

U2_AAG(x1, x2, x3, x4, x5)  =  U2_AAG(x1, x4, 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 3 less nodes.
----------------------------------------

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

   APP_IN_AAG(.(X, Xs), Ys, .(X, Zs)) -> APP_IN_AAG(Xs, Ys, Zs)

The TRS R consists of the following rules:

   suffix_in_ag(Xs, Ys) -> U1_ag(Xs, Ys, app_in_aag(X1, Xs, Ys))
   app_in_aag([], X, X) -> app_out_aag([], X, X)
   app_in_aag(.(X, Xs), Ys, .(X, Zs)) -> U2_aag(X, Xs, Ys, Zs, app_in_aag(Xs, Ys, Zs))
   U2_aag(X, Xs, Ys, Zs, app_out_aag(Xs, Ys, Zs)) -> app_out_aag(.(X, Xs), Ys, .(X, Zs))
   U1_ag(Xs, Ys, app_out_aag(X1, Xs, Ys)) -> suffix_out_ag(Xs, Ys)

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

U1_ag(x1, x2, x3)  =  U1_ag(x2, x3)

app_in_aag(x1, x2, x3)  =  app_in_aag(x3)

app_out_aag(x1, x2, x3)  =  app_out_aag(x1, x2, x3)

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

U2_aag(x1, x2, x3, x4, x5)  =  U2_aag(x1, x4, x5)

suffix_out_ag(x1, x2)  =  suffix_out_ag(x1, x2)

APP_IN_AAG(x1, x2, x3)  =  APP_IN_AAG(x3)


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:

   APP_IN_AAG(.(X, Xs), Ys, .(X, Zs)) -> APP_IN_AAG(Xs, Ys, Zs)

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

APP_IN_AAG(x1, x2, x3)  =  APP_IN_AAG(x3)


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.
----------------------------------------

(10)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   APP_IN_AAG(.(X, Zs)) -> APP_IN_AAG(Zs)

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:
*APP_IN_AAG(.(X, Zs)) -> APP_IN_AAG(Zs)
The graph contains the following edges 1 > 1


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(12)
YES