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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty


Left Termination of the query pattern

reverse(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, 9 ms]
(4) PiDP
(5) DependencyGraphProof [EQUIVALENT, 0 ms]
(6) AND
    (7) PiDP
        (8) UsableRulesProof [EQUIVALENT, 0 ms]
        (9) PiDP
        (10) PiDPToQDPProof [SOUND, 0 ms]
        (11) QDP
        (12) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (13) YES
    (14) PiDP
        (15) UsableRulesProof [EQUIVALENT, 0 ms]
        (16) PiDP
        (17) PiDPToQDPProof [SOUND, 0 ms]
        (18) QDP
        (19) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (20) YES


----------------------------------------

(0)
Obligation:
Clauses:

app(.(X, Xs), Ys, .(X, Zs)) :- app(Xs, Ys, Zs).
app([], Ys, Ys).
reverse(.(X, Xs), Ys) :- ','(reverse(Xs, Zs), app(Zs, .(X, []), Ys)).
reverse([], []).


Query: reverse(g,a)
----------------------------------------

(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_in_2: (b,f)

app_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_in_ga(.(X, Xs), Ys) -> U2_ga(X, Xs, Ys, reverse_in_ga(Xs, Zs))
   reverse_in_ga([], []) -> reverse_out_ga([], [])
   U2_ga(X, Xs, Ys, reverse_out_ga(Xs, Zs)) -> U3_ga(X, Xs, Ys, app_in_gga(Zs, .(X, []), Ys))
   app_in_gga(.(X, Xs), Ys, .(X, Zs)) -> U1_gga(X, Xs, Ys, Zs, app_in_gga(Xs, Ys, Zs))
   app_in_gga([], Ys, Ys) -> app_out_gga([], Ys, Ys)
   U1_gga(X, Xs, Ys, Zs, app_out_gga(Xs, Ys, Zs)) -> app_out_gga(.(X, Xs), Ys, .(X, Zs))
   U3_ga(X, Xs, Ys, app_out_gga(Zs, .(X, []), Ys)) -> reverse_out_ga(.(X, Xs), Ys)

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

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

U2_ga(x1, x2, x3, x4)  =  U2_ga(x1, x4)

[]  =  []

reverse_out_ga(x1, x2)  =  reverse_out_ga(x2)

U3_ga(x1, x2, x3, x4)  =  U3_ga(x4)

app_in_gga(x1, x2, x3)  =  app_in_gga(x1, x2)

U1_gga(x1, x2, x3, x4, x5)  =  U1_gga(x1, x5)

app_out_gga(x1, x2, x3)  =  app_out_gga(x3)





Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog



----------------------------------------

(2)
Obligation:
Pi-finite rewrite system:
The TRS R consists of the following rules:

   reverse_in_ga(.(X, Xs), Ys) -> U2_ga(X, Xs, Ys, reverse_in_ga(Xs, Zs))
   reverse_in_ga([], []) -> reverse_out_ga([], [])
   U2_ga(X, Xs, Ys, reverse_out_ga(Xs, Zs)) -> U3_ga(X, Xs, Ys, app_in_gga(Zs, .(X, []), Ys))
   app_in_gga(.(X, Xs), Ys, .(X, Zs)) -> U1_gga(X, Xs, Ys, Zs, app_in_gga(Xs, Ys, Zs))
   app_in_gga([], Ys, Ys) -> app_out_gga([], Ys, Ys)
   U1_gga(X, Xs, Ys, Zs, app_out_gga(Xs, Ys, Zs)) -> app_out_gga(.(X, Xs), Ys, .(X, Zs))
   U3_ga(X, Xs, Ys, app_out_gga(Zs, .(X, []), Ys)) -> reverse_out_ga(.(X, Xs), Ys)

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

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

U2_ga(x1, x2, x3, x4)  =  U2_ga(x1, x4)

[]  =  []

reverse_out_ga(x1, x2)  =  reverse_out_ga(x2)

U3_ga(x1, x2, x3, x4)  =  U3_ga(x4)

app_in_gga(x1, x2, x3)  =  app_in_gga(x1, x2)

U1_gga(x1, x2, x3, x4, x5)  =  U1_gga(x1, x5)

app_out_gga(x1, x2, x3)  =  app_out_gga(x3)



----------------------------------------

(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_IN_GA(.(X, Xs), Ys) -> U2_GA(X, Xs, Ys, reverse_in_ga(Xs, Zs))
   REVERSE_IN_GA(.(X, Xs), Ys) -> REVERSE_IN_GA(Xs, Zs)
   U2_GA(X, Xs, Ys, reverse_out_ga(Xs, Zs)) -> U3_GA(X, Xs, Ys, app_in_gga(Zs, .(X, []), Ys))
   U2_GA(X, Xs, Ys, reverse_out_ga(Xs, Zs)) -> APP_IN_GGA(Zs, .(X, []), Ys)
   APP_IN_GGA(.(X, Xs), Ys, .(X, Zs)) -> U1_GGA(X, Xs, Ys, Zs, app_in_gga(Xs, Ys, Zs))
   APP_IN_GGA(.(X, Xs), Ys, .(X, Zs)) -> APP_IN_GGA(Xs, Ys, Zs)

The TRS R consists of the following rules:

   reverse_in_ga(.(X, Xs), Ys) -> U2_ga(X, Xs, Ys, reverse_in_ga(Xs, Zs))
   reverse_in_ga([], []) -> reverse_out_ga([], [])
   U2_ga(X, Xs, Ys, reverse_out_ga(Xs, Zs)) -> U3_ga(X, Xs, Ys, app_in_gga(Zs, .(X, []), Ys))
   app_in_gga(.(X, Xs), Ys, .(X, Zs)) -> U1_gga(X, Xs, Ys, Zs, app_in_gga(Xs, Ys, Zs))
   app_in_gga([], Ys, Ys) -> app_out_gga([], Ys, Ys)
   U1_gga(X, Xs, Ys, Zs, app_out_gga(Xs, Ys, Zs)) -> app_out_gga(.(X, Xs), Ys, .(X, Zs))
   U3_ga(X, Xs, Ys, app_out_gga(Zs, .(X, []), Ys)) -> reverse_out_ga(.(X, Xs), Ys)

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

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

U2_ga(x1, x2, x3, x4)  =  U2_ga(x1, x4)

[]  =  []

reverse_out_ga(x1, x2)  =  reverse_out_ga(x2)

U3_ga(x1, x2, x3, x4)  =  U3_ga(x4)

app_in_gga(x1, x2, x3)  =  app_in_gga(x1, x2)

U1_gga(x1, x2, x3, x4, x5)  =  U1_gga(x1, x5)

app_out_gga(x1, x2, x3)  =  app_out_gga(x3)

REVERSE_IN_GA(x1, x2)  =  REVERSE_IN_GA(x1)

U2_GA(x1, x2, x3, x4)  =  U2_GA(x1, x4)

U3_GA(x1, x2, x3, x4)  =  U3_GA(x4)

APP_IN_GGA(x1, x2, x3)  =  APP_IN_GGA(x1, x2)

U1_GGA(x1, x2, x3, x4, x5)  =  U1_GGA(x1, x5)


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

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

   REVERSE_IN_GA(.(X, Xs), Ys) -> U2_GA(X, Xs, Ys, reverse_in_ga(Xs, Zs))
   REVERSE_IN_GA(.(X, Xs), Ys) -> REVERSE_IN_GA(Xs, Zs)
   U2_GA(X, Xs, Ys, reverse_out_ga(Xs, Zs)) -> U3_GA(X, Xs, Ys, app_in_gga(Zs, .(X, []), Ys))
   U2_GA(X, Xs, Ys, reverse_out_ga(Xs, Zs)) -> APP_IN_GGA(Zs, .(X, []), Ys)
   APP_IN_GGA(.(X, Xs), Ys, .(X, Zs)) -> U1_GGA(X, Xs, Ys, Zs, app_in_gga(Xs, Ys, Zs))
   APP_IN_GGA(.(X, Xs), Ys, .(X, Zs)) -> APP_IN_GGA(Xs, Ys, Zs)

The TRS R consists of the following rules:

   reverse_in_ga(.(X, Xs), Ys) -> U2_ga(X, Xs, Ys, reverse_in_ga(Xs, Zs))
   reverse_in_ga([], []) -> reverse_out_ga([], [])
   U2_ga(X, Xs, Ys, reverse_out_ga(Xs, Zs)) -> U3_ga(X, Xs, Ys, app_in_gga(Zs, .(X, []), Ys))
   app_in_gga(.(X, Xs), Ys, .(X, Zs)) -> U1_gga(X, Xs, Ys, Zs, app_in_gga(Xs, Ys, Zs))
   app_in_gga([], Ys, Ys) -> app_out_gga([], Ys, Ys)
   U1_gga(X, Xs, Ys, Zs, app_out_gga(Xs, Ys, Zs)) -> app_out_gga(.(X, Xs), Ys, .(X, Zs))
   U3_ga(X, Xs, Ys, app_out_gga(Zs, .(X, []), Ys)) -> reverse_out_ga(.(X, Xs), Ys)

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

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

U2_ga(x1, x2, x3, x4)  =  U2_ga(x1, x4)

[]  =  []

reverse_out_ga(x1, x2)  =  reverse_out_ga(x2)

U3_ga(x1, x2, x3, x4)  =  U3_ga(x4)

app_in_gga(x1, x2, x3)  =  app_in_gga(x1, x2)

U1_gga(x1, x2, x3, x4, x5)  =  U1_gga(x1, x5)

app_out_gga(x1, x2, x3)  =  app_out_gga(x3)

REVERSE_IN_GA(x1, x2)  =  REVERSE_IN_GA(x1)

U2_GA(x1, x2, x3, x4)  =  U2_GA(x1, x4)

U3_GA(x1, x2, x3, x4)  =  U3_GA(x4)

APP_IN_GGA(x1, x2, x3)  =  APP_IN_GGA(x1, x2)

U1_GGA(x1, x2, x3, x4, x5)  =  U1_GGA(x1, x5)


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

(5) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LOPSTR] contains 2 SCCs with 4 less nodes.
----------------------------------------

(6)
Complex Obligation (AND)

----------------------------------------

(7)
Obligation:
Pi DP problem:
The TRS P consists of the following rules:

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

The TRS R consists of the following rules:

   reverse_in_ga(.(X, Xs), Ys) -> U2_ga(X, Xs, Ys, reverse_in_ga(Xs, Zs))
   reverse_in_ga([], []) -> reverse_out_ga([], [])
   U2_ga(X, Xs, Ys, reverse_out_ga(Xs, Zs)) -> U3_ga(X, Xs, Ys, app_in_gga(Zs, .(X, []), Ys))
   app_in_gga(.(X, Xs), Ys, .(X, Zs)) -> U1_gga(X, Xs, Ys, Zs, app_in_gga(Xs, Ys, Zs))
   app_in_gga([], Ys, Ys) -> app_out_gga([], Ys, Ys)
   U1_gga(X, Xs, Ys, Zs, app_out_gga(Xs, Ys, Zs)) -> app_out_gga(.(X, Xs), Ys, .(X, Zs))
   U3_ga(X, Xs, Ys, app_out_gga(Zs, .(X, []), Ys)) -> reverse_out_ga(.(X, Xs), Ys)

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

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

U2_ga(x1, x2, x3, x4)  =  U2_ga(x1, x4)

[]  =  []

reverse_out_ga(x1, x2)  =  reverse_out_ga(x2)

U3_ga(x1, x2, x3, x4)  =  U3_ga(x4)

app_in_gga(x1, x2, x3)  =  app_in_gga(x1, x2)

U1_gga(x1, x2, x3, x4, x5)  =  U1_gga(x1, x5)

app_out_gga(x1, x2, x3)  =  app_out_gga(x3)

APP_IN_GGA(x1, x2, x3)  =  APP_IN_GGA(x1, x2)


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

(8) UsableRulesProof (EQUIVALENT)
For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R.
----------------------------------------

(9)
Obligation:
Pi DP problem:
The TRS P consists of the following rules:

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

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

APP_IN_GGA(x1, x2, x3)  =  APP_IN_GGA(x1, x2)


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

(10) PiDPToQDPProof (SOUND)
Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.
----------------------------------------

(11)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   APP_IN_GGA(.(X, Xs), Ys) -> APP_IN_GGA(Xs, Ys)

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

(12) 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_GGA(.(X, Xs), Ys) -> APP_IN_GGA(Xs, Ys)
The graph contains the following edges 1 > 1, 2 >= 2


----------------------------------------

(13)
YES

----------------------------------------

(14)
Obligation:
Pi DP problem:
The TRS P consists of the following rules:

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

The TRS R consists of the following rules:

   reverse_in_ga(.(X, Xs), Ys) -> U2_ga(X, Xs, Ys, reverse_in_ga(Xs, Zs))
   reverse_in_ga([], []) -> reverse_out_ga([], [])
   U2_ga(X, Xs, Ys, reverse_out_ga(Xs, Zs)) -> U3_ga(X, Xs, Ys, app_in_gga(Zs, .(X, []), Ys))
   app_in_gga(.(X, Xs), Ys, .(X, Zs)) -> U1_gga(X, Xs, Ys, Zs, app_in_gga(Xs, Ys, Zs))
   app_in_gga([], Ys, Ys) -> app_out_gga([], Ys, Ys)
   U1_gga(X, Xs, Ys, Zs, app_out_gga(Xs, Ys, Zs)) -> app_out_gga(.(X, Xs), Ys, .(X, Zs))
   U3_ga(X, Xs, Ys, app_out_gga(Zs, .(X, []), Ys)) -> reverse_out_ga(.(X, Xs), Ys)

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

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

U2_ga(x1, x2, x3, x4)  =  U2_ga(x1, x4)

[]  =  []

reverse_out_ga(x1, x2)  =  reverse_out_ga(x2)

U3_ga(x1, x2, x3, x4)  =  U3_ga(x4)

app_in_gga(x1, x2, x3)  =  app_in_gga(x1, x2)

U1_gga(x1, x2, x3, x4, x5)  =  U1_gga(x1, x5)

app_out_gga(x1, x2, x3)  =  app_out_gga(x3)

REVERSE_IN_GA(x1, x2)  =  REVERSE_IN_GA(x1)


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

(15) UsableRulesProof (EQUIVALENT)
For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R.
----------------------------------------

(16)
Obligation:
Pi DP problem:
The TRS P consists of the following rules:

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

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

REVERSE_IN_GA(x1, x2)  =  REVERSE_IN_GA(x1)


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

(17) PiDPToQDPProof (SOUND)
Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.
----------------------------------------

(18)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   REVERSE_IN_GA(.(X, Xs)) -> REVERSE_IN_GA(Xs)

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

(19) 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_IN_GA(.(X, Xs)) -> REVERSE_IN_GA(Xs)
The graph contains the following edges 1 > 1


----------------------------------------

(20)
YES