(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).

Queries:

countstack(g,a).

(1) PrologToPiTRSProof (SOUND transformation)

We use the technique of [LOPSTR]. With regard to the inferred argument filtering the predicates were used in the following modes:
countstack_in: (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 transformation)

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 transformation)

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 transformation)

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 transformation)

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 transformation)

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(x1)) = 2·x1   
POL(cons(x1, x2)) = 2 + 2·x1 + 2·x2   
POL(nil) = 0   
POL(push(x1, x2)) = 1 + 2·x1 + x2   

(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 transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(14) TRUE

(15) PrologToPiTRSProof (SOUND transformation)

We use the technique of [LOPSTR]. With regard to the inferred argument filtering the predicates were used in the following modes:
countstack_in: (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(x1, x2)
push(x1, x2)  =  push(x1, x2)
nil  =  nil
U1_ga(x1, x2, x3)  =  U1_ga(x1, x3)
cons(x1, x2)  =  cons(x1, x2)
U2_ga(x1, x2, x3, x4, x5)  =  U2_ga(x1, x2, x3, x5)
s(x1)  =  s(x1)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog

(16) 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(x1, x2)
push(x1, x2)  =  push(x1, x2)
nil  =  nil
U1_ga(x1, x2, x3)  =  U1_ga(x1, x3)
cons(x1, x2)  =  cons(x1, x2)
U2_ga(x1, x2, x3, x4, x5)  =  U2_ga(x1, x2, x3, x5)
s(x1)  =  s(x1)

(17) DependencyPairsProof (EQUIVALENT transformation)

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(x1, x2)
push(x1, x2)  =  push(x1, x2)
nil  =  nil
U1_ga(x1, x2, x3)  =  U1_ga(x1, x3)
cons(x1, x2)  =  cons(x1, x2)
U2_ga(x1, x2, x3, x4, x5)  =  U2_ga(x1, x2, x3, x5)
s(x1)  =  s(x1)
COUNTSTACK_IN_GA(x1, x2)  =  COUNTSTACK_IN_GA(x1)
U1_GA(x1, x2, x3)  =  U1_GA(x1, x3)
U2_GA(x1, x2, x3, x4, x5)  =  U2_GA(x1, x2, x3, x5)

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

(18) 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(x1, x2)
push(x1, x2)  =  push(x1, x2)
nil  =  nil
U1_ga(x1, x2, x3)  =  U1_ga(x1, x3)
cons(x1, x2)  =  cons(x1, x2)
U2_ga(x1, x2, x3, x4, x5)  =  U2_ga(x1, x2, x3, x5)
s(x1)  =  s(x1)
COUNTSTACK_IN_GA(x1, x2)  =  COUNTSTACK_IN_GA(x1)
U1_GA(x1, x2, x3)  =  U1_GA(x1, x3)
U2_GA(x1, x2, x3, x4, x5)  =  U2_GA(x1, x2, x3, x5)

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

(19) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 2 less nodes.

(20) 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(x1, x2)
push(x1, x2)  =  push(x1, x2)
nil  =  nil
U1_ga(x1, x2, x3)  =  U1_ga(x1, x3)
cons(x1, x2)  =  cons(x1, x2)
U2_ga(x1, x2, x3, x4, x5)  =  U2_ga(x1, x2, x3, x5)
s(x1)  =  s(x1)
COUNTSTACK_IN_GA(x1, x2)  =  COUNTSTACK_IN_GA(x1)

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

(21) UsableRulesProof (EQUIVALENT transformation)

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

(22) 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

(23) PiDPToQDPProof (SOUND transformation)

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

(24) 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.