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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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

countstack(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, 5 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) UsableRulesReductionPairsProof [EQUIVALENT, 0 ms]
(12) QDP
(13) PisEmptyProof [EQUIVALENT, 0 ms]
(14) YES


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(0)
Obligation:
Clauses:

countstack(empty, 0).
countstack(push(nil, T), X) :- countstack(T, X).
countstack(push(cons(U, V), T), s(X)) :- countstack(push(U, push(V, T)), X).


Query: countstack(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:

countstack_in_2: (b,f)

Transforming Prolog into the following Term Rewriting System:

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

   countstack_in_ga(empty, 0) -> countstack_out_ga(empty, 0)
   countstack_in_ga(push(nil, T), X) -> U1_ga(T, X, countstack_in_ga(T, X))
   countstack_in_ga(push(cons(U, V), T), s(X)) -> U2_ga(U, V, T, X, countstack_in_ga(push(U, push(V, T)), X))
   U2_ga(U, V, T, X, countstack_out_ga(push(U, push(V, T)), X)) -> countstack_out_ga(push(cons(U, V), T), s(X))
   U1_ga(T, X, countstack_out_ga(T, X)) -> countstack_out_ga(push(nil, T), X)

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

empty  =  empty

countstack_out_ga(x1, x2)  =  countstack_out_ga(x2)

push(x1, x2)  =  push(x1, x2)

nil  =  nil

U1_ga(x1, x2, x3)  =  U1_ga(x3)

cons(x1, x2)  =  cons(x1, x2)

U2_ga(x1, x2, x3, x4, x5)  =  U2_ga(x5)

s(x1)  =  s(x1)





Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog



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

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

   countstack_in_ga(empty, 0) -> countstack_out_ga(empty, 0)
   countstack_in_ga(push(nil, T), X) -> U1_ga(T, X, countstack_in_ga(T, X))
   countstack_in_ga(push(cons(U, V), T), s(X)) -> U2_ga(U, V, T, X, countstack_in_ga(push(U, push(V, T)), X))
   U2_ga(U, V, T, X, countstack_out_ga(push(U, push(V, T)), X)) -> countstack_out_ga(push(cons(U, V), T), s(X))
   U1_ga(T, X, countstack_out_ga(T, X)) -> countstack_out_ga(push(nil, T), X)

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

empty  =  empty

countstack_out_ga(x1, x2)  =  countstack_out_ga(x2)

push(x1, x2)  =  push(x1, x2)

nil  =  nil

U1_ga(x1, x2, x3)  =  U1_ga(x3)

cons(x1, x2)  =  cons(x1, x2)

U2_ga(x1, x2, x3, x4, x5)  =  U2_ga(x5)

s(x1)  =  s(x1)



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

(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:

   COUNTSTACK_IN_GA(push(nil, T), X) -> U1_GA(T, X, countstack_in_ga(T, X))
   COUNTSTACK_IN_GA(push(nil, T), X) -> COUNTSTACK_IN_GA(T, X)
   COUNTSTACK_IN_GA(push(cons(U, V), T), s(X)) -> U2_GA(U, V, T, X, countstack_in_ga(push(U, push(V, T)), X))
   COUNTSTACK_IN_GA(push(cons(U, V), T), s(X)) -> COUNTSTACK_IN_GA(push(U, push(V, T)), X)

The TRS R consists of the following rules:

   countstack_in_ga(empty, 0) -> countstack_out_ga(empty, 0)
   countstack_in_ga(push(nil, T), X) -> U1_ga(T, X, countstack_in_ga(T, X))
   countstack_in_ga(push(cons(U, V), T), s(X)) -> U2_ga(U, V, T, X, countstack_in_ga(push(U, push(V, T)), X))
   U2_ga(U, V, T, X, countstack_out_ga(push(U, push(V, T)), X)) -> countstack_out_ga(push(cons(U, V), T), s(X))
   U1_ga(T, X, countstack_out_ga(T, X)) -> countstack_out_ga(push(nil, T), X)

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

empty  =  empty

countstack_out_ga(x1, x2)  =  countstack_out_ga(x2)

push(x1, x2)  =  push(x1, x2)

nil  =  nil

U1_ga(x1, x2, x3)  =  U1_ga(x3)

cons(x1, x2)  =  cons(x1, x2)

U2_ga(x1, x2, x3, x4, x5)  =  U2_ga(x5)

s(x1)  =  s(x1)

COUNTSTACK_IN_GA(x1, x2)  =  COUNTSTACK_IN_GA(x1)

U1_GA(x1, x2, x3)  =  U1_GA(x3)

U2_GA(x1, x2, x3, x4, x5)  =  U2_GA(x5)


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

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

   COUNTSTACK_IN_GA(push(nil, T), X) -> U1_GA(T, X, countstack_in_ga(T, X))
   COUNTSTACK_IN_GA(push(nil, T), X) -> COUNTSTACK_IN_GA(T, X)
   COUNTSTACK_IN_GA(push(cons(U, V), T), s(X)) -> U2_GA(U, V, T, X, countstack_in_ga(push(U, push(V, T)), X))
   COUNTSTACK_IN_GA(push(cons(U, V), T), s(X)) -> COUNTSTACK_IN_GA(push(U, push(V, T)), X)

The TRS R consists of the following rules:

   countstack_in_ga(empty, 0) -> countstack_out_ga(empty, 0)
   countstack_in_ga(push(nil, T), X) -> U1_ga(T, X, countstack_in_ga(T, X))
   countstack_in_ga(push(cons(U, V), T), s(X)) -> U2_ga(U, V, T, X, countstack_in_ga(push(U, push(V, T)), X))
   U2_ga(U, V, T, X, countstack_out_ga(push(U, push(V, T)), X)) -> countstack_out_ga(push(cons(U, V), T), s(X))
   U1_ga(T, X, countstack_out_ga(T, X)) -> countstack_out_ga(push(nil, T), X)

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

empty  =  empty

countstack_out_ga(x1, x2)  =  countstack_out_ga(x2)

push(x1, x2)  =  push(x1, x2)

nil  =  nil

U1_ga(x1, x2, x3)  =  U1_ga(x3)

cons(x1, x2)  =  cons(x1, x2)

U2_ga(x1, x2, x3, x4, x5)  =  U2_ga(x5)

s(x1)  =  s(x1)

COUNTSTACK_IN_GA(x1, x2)  =  COUNTSTACK_IN_GA(x1)

U1_GA(x1, x2, x3)  =  U1_GA(x3)

U2_GA(x1, x2, x3, x4, x5)  =  U2_GA(x5)


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

(5) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 2 less nodes.
----------------------------------------

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

   COUNTSTACK_IN_GA(push(cons(U, V), T), s(X)) -> COUNTSTACK_IN_GA(push(U, push(V, T)), X)
   COUNTSTACK_IN_GA(push(nil, T), X) -> COUNTSTACK_IN_GA(T, X)

The TRS R consists of the following rules:

   countstack_in_ga(empty, 0) -> countstack_out_ga(empty, 0)
   countstack_in_ga(push(nil, T), X) -> U1_ga(T, X, countstack_in_ga(T, X))
   countstack_in_ga(push(cons(U, V), T), s(X)) -> U2_ga(U, V, T, X, countstack_in_ga(push(U, push(V, T)), X))
   U2_ga(U, V, T, X, countstack_out_ga(push(U, push(V, T)), X)) -> countstack_out_ga(push(cons(U, V), T), s(X))
   U1_ga(T, X, countstack_out_ga(T, X)) -> countstack_out_ga(push(nil, T), X)

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

empty  =  empty

countstack_out_ga(x1, x2)  =  countstack_out_ga(x2)

push(x1, x2)  =  push(x1, x2)

nil  =  nil

U1_ga(x1, x2, x3)  =  U1_ga(x3)

cons(x1, x2)  =  cons(x1, x2)

U2_ga(x1, x2, x3, x4, x5)  =  U2_ga(x5)

s(x1)  =  s(x1)

COUNTSTACK_IN_GA(x1, x2)  =  COUNTSTACK_IN_GA(x1)


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

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

(8)
Obligation:
Pi DP problem:
The TRS P consists of the following rules:

   COUNTSTACK_IN_GA(push(cons(U, V), T), s(X)) -> COUNTSTACK_IN_GA(push(U, push(V, T)), X)
   COUNTSTACK_IN_GA(push(nil, T), X) -> COUNTSTACK_IN_GA(T, X)

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

nil  =  nil

cons(x1, x2)  =  cons(x1, x2)

s(x1)  =  s(x1)

COUNTSTACK_IN_GA(x1, x2)  =  COUNTSTACK_IN_GA(x1)


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

(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:

   COUNTSTACK_IN_GA(push(cons(U, V), T)) -> COUNTSTACK_IN_GA(push(U, push(V, T)))
   COUNTSTACK_IN_GA(push(nil, T)) -> COUNTSTACK_IN_GA(T)

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

(11) UsableRulesReductionPairsProof (EQUIVALENT)
By using the usable rules with reduction pair processor [LPAR04] with a polynomial ordering [POLO], all dependency pairs and the corresponding usable rules [FROCOS05] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.

The following dependency pairs can be deleted:

   COUNTSTACK_IN_GA(push(cons(U, V), T)) -> COUNTSTACK_IN_GA(push(U, push(V, T)))
   COUNTSTACK_IN_GA(push(nil, T)) -> COUNTSTACK_IN_GA(T)
No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [POLO]:

   POL(COUNTSTACK_IN_GA(x_1)) = 2*x_1
   POL(cons(x_1, x_2)) = 1 + x_1 + 2*x_2
   POL(nil) = 0
   POL(push(x_1, x_2)) = 1 + 2*x_1 + x_2


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(12)
Obligation:
Q DP problem:
P is empty.
R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
----------------------------------------

(13) PisEmptyProof (EQUIVALENT)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
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(14)
YES