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

0 Prolog
↳1 PrologToDTProblemTransformerProof (⇐)
↳2 TRIPLES
↳3 TriplesToPiDPProof (⇐)
↳4 PiDP
↳5 DependencyGraphProof (⇔)
↳6 PiDP
↳7 PiDPToQDPProof (⇐)
↳8 QDP
↳9 UsableRulesReductionPairsProof (⇔)
↳10 QDP
↳11 PisEmptyProof (⇔)
↳12 YES

(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) PrologToDTProblemTransformerProof (SOUND transformation)

Built DT problem from termination graph.

(2) Obligation:

Triples:

countstack1(push(nil, push(nil, T16)), T18) :- countstack1(T16, T18).
countstack1(push(nil, push(cons(T35, T36), T37)), s(T39)) :- countstack1(push(T35, push(T36, T37)), T39).
countstack1(push(cons(nil, T64), T65), s(T67)) :- countstack1(push(T64, T65), T67).
countstack1(push(cons(cons(T80, T81), T82), T83), s(s(T85))) :- countstack1(push(T80, push(T81, push(T82, T83))), T85).

Clauses:

countstackc1(empty, 0).
countstackc1(push(nil, empty), 0).
countstackc1(push(nil, push(nil, T16)), T18) :- countstackc1(T16, T18).
countstackc1(push(nil, push(cons(T35, T36), T37)), s(T39)) :- countstackc1(push(T35, push(T36, T37)), T39).
countstackc1(push(cons(nil, T64), T65), s(T67)) :- countstackc1(push(T64, T65), T67).
countstackc1(push(cons(cons(T80, T81), T82), T83), s(s(T85))) :- countstackc1(push(T80, push(T81, push(T82, T83))), T85).

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, push(nil, T16)), T18) → U1_GA(T16, T18, countstack1_in_ga(T16, T18))
COUNTSTACK1_IN_GA(push(nil, push(nil, T16)), T18) → COUNTSTACK1_IN_GA(T16, T18)
COUNTSTACK1_IN_GA(push(nil, push(cons(T35, T36), T37)), s(T39)) → U2_GA(T35, T36, T37, T39, countstack1_in_ga(push(T35, push(T36, T37)), T39))
COUNTSTACK1_IN_GA(push(nil, push(cons(T35, T36), T37)), s(T39)) → COUNTSTACK1_IN_GA(push(T35, push(T36, T37)), T39)
COUNTSTACK1_IN_GA(push(cons(nil, T64), T65), s(T67)) → U3_GA(T64, T65, T67, countstack1_in_ga(push(T64, T65), T67))
COUNTSTACK1_IN_GA(push(cons(nil, T64), T65), s(T67)) → COUNTSTACK1_IN_GA(push(T64, T65), T67)
COUNTSTACK1_IN_GA(push(cons(cons(T80, T81), T82), T83), s(s(T85))) → U4_GA(T80, T81, T82, T83, T85, countstack1_in_ga(push(T80, push(T81, push(T82, T83))), T85))
COUNTSTACK1_IN_GA(push(cons(cons(T80, T81), T82), T83), s(s(T85))) → COUNTSTACK1_IN_GA(push(T80, push(T81, push(T82, T83))), T85)

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)
U3_GA(x1, x2, x3, x4)  =  U3_GA(x1, x2, x4)
U4_GA(x1, x2, x3, x4, x5, x6)  =  U4_GA(x1, x2, x3, x4, x6)

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, push(nil, T16)), T18) → U1_GA(T16, T18, countstack1_in_ga(T16, T18))
COUNTSTACK1_IN_GA(push(nil, push(nil, T16)), T18) → COUNTSTACK1_IN_GA(T16, T18)
COUNTSTACK1_IN_GA(push(nil, push(cons(T35, T36), T37)), s(T39)) → U2_GA(T35, T36, T37, T39, countstack1_in_ga(push(T35, push(T36, T37)), T39))
COUNTSTACK1_IN_GA(push(nil, push(cons(T35, T36), T37)), s(T39)) → COUNTSTACK1_IN_GA(push(T35, push(T36, T37)), T39)
COUNTSTACK1_IN_GA(push(cons(nil, T64), T65), s(T67)) → U3_GA(T64, T65, T67, countstack1_in_ga(push(T64, T65), T67))
COUNTSTACK1_IN_GA(push(cons(nil, T64), T65), s(T67)) → COUNTSTACK1_IN_GA(push(T64, T65), T67)
COUNTSTACK1_IN_GA(push(cons(cons(T80, T81), T82), T83), s(s(T85))) → U4_GA(T80, T81, T82, T83, T85, countstack1_in_ga(push(T80, push(T81, push(T82, T83))), T85))
COUNTSTACK1_IN_GA(push(cons(cons(T80, T81), T82), T83), s(s(T85))) → COUNTSTACK1_IN_GA(push(T80, push(T81, push(T82, T83))), T85)

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)
U3_GA(x1, x2, x3, x4)  =  U3_GA(x1, x2, x4)
U4_GA(x1, x2, x3, x4, x5, x6)  =  U4_GA(x1, x2, x3, x4, x6)

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 4 less nodes.

(6) Obligation:

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

COUNTSTACK1_IN_GA(push(nil, push(cons(T35, T36), T37)), s(T39)) → COUNTSTACK1_IN_GA(push(T35, push(T36, T37)), T39)
COUNTSTACK1_IN_GA(push(nil, push(nil, T16)), T18) → COUNTSTACK1_IN_GA(T16, T18)
COUNTSTACK1_IN_GA(push(cons(nil, T64), T65), s(T67)) → COUNTSTACK1_IN_GA(push(T64, T65), T67)
COUNTSTACK1_IN_GA(push(cons(cons(T80, T81), T82), T83), s(s(T85))) → COUNTSTACK1_IN_GA(push(T80, push(T81, push(T82, T83))), T85)

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(nil, push(cons(T35, T36), T37))) → COUNTSTACK1_IN_GA(push(T35, push(T36, T37)))
COUNTSTACK1_IN_GA(push(nil, push(nil, T16))) → COUNTSTACK1_IN_GA(T16)
COUNTSTACK1_IN_GA(push(cons(nil, T64), T65)) → COUNTSTACK1_IN_GA(push(T64, T65))
COUNTSTACK1_IN_GA(push(cons(cons(T80, T81), T82), T83)) → COUNTSTACK1_IN_GA(push(T80, push(T81, push(T82, T83))))

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(nil, push(cons(T35, T36), T37))) → COUNTSTACK1_IN_GA(push(T35, push(T36, T37)))
COUNTSTACK1_IN_GA(push(nil, push(nil, T16))) → COUNTSTACK1_IN_GA(T16)
COUNTSTACK1_IN_GA(push(cons(nil, T64), T65)) → COUNTSTACK1_IN_GA(push(T64, T65))
COUNTSTACK1_IN_GA(push(cons(cons(T80, T81), T82), T83)) → COUNTSTACK1_IN_GA(push(T80, push(T81, push(T82, T83))))
No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [POLO]:

POL(COUNTSTACK1_IN_GA(x1)) = 2·x1   
POL(cons(x1, x2)) = x1 + x2   
POL(nil) = 0   
POL(push(x1, x2)) = 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