Left Termination of the query pattern countstack_in_2(g, a) w.r.t. the given Prolog program could successfully be proven:

(0) Obligation:

Clauses:

countstack(empty, X) :- ','(!, eq(X, 0)).
countstack(S, X) :- ','(pop(S, nil), ','(!, ','(popped(S, Pd), countstack(Pd, X)))).
countstack(S, s(X)) :- ','(pop(S, P), ','(head(P, H), ','(tail(P, T), ','(popped(S, Pd), countstack(push(H, push(T, Pd)), X))))).
pop(empty, X1).
pop(push(P, X2), P).
popped(empty, empty).
popped(push(X3, Pd), Pd).
head(nil, X4).
head(cons(H, X5), H).
tail(nil, nil).
tail(cons(X6, T), T).
eq(X, X).

Queries:

countstack(g,a).

(1) PrologToDTProblemTransformerProof (SOUND transformation)

Built DT problem from termination graph.

(2) Obligation:

Triples:

countstack1(push(nil, T24), T13) :- countstack1(T24, T13).
countstack1(push(cons(T71, T72), T73), s(T32)) :- countstack1(push(T71, push(T72, T73)), T32).

Clauses:

countstackc1(empty, 0).
countstackc1(push(nil, T24), T13) :- countstackc1(T24, T13).
countstackc1(push(cons(T71, T72), T73), s(T32)) :- countstackc1(push(T71, push(T72, T73)), T32).

Afs:

countstack1(x1, x2)  =  countstack1(x1)

(3) TriplesToPiDPProof (SOUND transformation)

We use the technique of [LOPSTR]. With regard to the inferred argument filtering the predicates were used in the following modes:
countstack1_in: (b,f)
Transforming TRIPLES into the following Term Rewriting System:
Pi DP problem:
The TRS P consists of the following rules:

COUNTSTACK1_IN_GA(push(nil, T24), T13) → U1_GA(T24, T13, countstack1_in_ga(T24, T13))
COUNTSTACK1_IN_GA(push(nil, T24), T13) → COUNTSTACK1_IN_GA(T24, T13)
COUNTSTACK1_IN_GA(push(cons(T71, T72), T73), s(T32)) → U2_GA(T71, T72, T73, T32, countstack1_in_ga(push(T71, push(T72, T73)), T32))
COUNTSTACK1_IN_GA(push(cons(T71, T72), T73), s(T32)) → COUNTSTACK1_IN_GA(push(T71, push(T72, T73)), T32)

R is empty.
The argument filtering Pi contains the following mapping:
countstack1_in_ga(x1, x2)  =  countstack1_in_ga(x1)
push(x1, x2)  =  push(x1, x2)
nil  =  nil
cons(x1, x2)  =  cons(x1, x2)
s(x1)  =  s(x1)
COUNTSTACK1_IN_GA(x1, x2)  =  COUNTSTACK1_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

Infinitary Constructor Rewriting Termination of PiDP implies Termination of TRIPLES

(4) Obligation:

Pi DP problem:
The TRS P consists of the following rules:

COUNTSTACK1_IN_GA(push(nil, T24), T13) → U1_GA(T24, T13, countstack1_in_ga(T24, T13))
COUNTSTACK1_IN_GA(push(nil, T24), T13) → COUNTSTACK1_IN_GA(T24, T13)
COUNTSTACK1_IN_GA(push(cons(T71, T72), T73), s(T32)) → U2_GA(T71, T72, T73, T32, countstack1_in_ga(push(T71, push(T72, T73)), T32))
COUNTSTACK1_IN_GA(push(cons(T71, T72), T73), s(T32)) → COUNTSTACK1_IN_GA(push(T71, push(T72, T73)), T32)

R is empty.
The argument filtering Pi contains the following mapping:
countstack1_in_ga(x1, x2)  =  countstack1_in_ga(x1)
push(x1, x2)  =  push(x1, x2)
nil  =  nil'
cons(x1, x2)  =  cons(x1, x2)
s(x1)  =  s(x1)
COUNTSTACK1_IN_GA(x1, x2)  =  COUNTSTACK1_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

(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:

COUNTSTACK1_IN_GA(push(cons(T71, T72), T73), s(T32)) → COUNTSTACK1_IN_GA(push(T71, push(T72, T73)), T32)
COUNTSTACK1_IN_GA(push(nil, T24), T13) → COUNTSTACK1_IN_GA(T24, T13)

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)
COUNTSTACK1_IN_GA(x1, x2)  =  COUNTSTACK1_IN_GA(x1)

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

(7) PiDPToQDPProof (SOUND transformation)

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

(8) Obligation:

Q DP problem:
The TRS P consists of the following rules:

COUNTSTACK1_IN_GA(push(cons(T71, T72), T73)) → COUNTSTACK1_IN_GA(push(T71, push(T72, T73)))
COUNTSTACK1_IN_GA(push(nil, T24)) → COUNTSTACK1_IN_GA(T24)

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

(9) 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:

COUNTSTACK1_IN_GA(push(cons(T71, T72), T73)) → COUNTSTACK1_IN_GA(push(T71, push(T72, T73)))
COUNTSTACK1_IN_GA(push(nil, T24)) → COUNTSTACK1_IN_GA(T24)

No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [POLO]:

POL(COUNTSTACK1_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   

(10) Obligation:

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

(11) PisEmptyProof (EQUIVALENT transformation)

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

(12) YES