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



↳ Prolog
  ↳ PrologToPiTRSProof

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).

We use the technique of [30].Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

countstack_in(push(cons(U, V), T), s(X)) → U2(U, V, T, X, countstack_in(push(U, push(V, T)), X))
countstack_in(push(nil, T), X) → U1(T, X, countstack_in(T, X))
countstack_in(empty, 0) → countstack_out(empty, 0)
U1(T, X, countstack_out(T, X)) → countstack_out(push(nil, T), X)
U2(U, V, T, X, countstack_out(push(U, push(V, T)), X)) → countstack_out(push(cons(U, V), T), s(X))

The argument filtering Pi contains the following mapping:
countstack_in(x1, x2)  =  countstack_in(x1)
push(x1, x2)  =  push(x1, x2)
cons(x1, x2)  =  cons(x1, x2)
s(x1)  =  s(x1)
U2(x1, x2, x3, x4, x5)  =  U2(x5)
nil  =  nil
U1(x1, x2, x3)  =  U1(x3)
empty  =  empty
0  =  0
countstack_out(x1, x2)  =  countstack_out(x2)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog



↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof

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

countstack_in(push(cons(U, V), T), s(X)) → U2(U, V, T, X, countstack_in(push(U, push(V, T)), X))
countstack_in(push(nil, T), X) → U1(T, X, countstack_in(T, X))
countstack_in(empty, 0) → countstack_out(empty, 0)
U1(T, X, countstack_out(T, X)) → countstack_out(push(nil, T), X)
U2(U, V, T, X, countstack_out(push(U, push(V, T)), X)) → countstack_out(push(cons(U, V), T), s(X))

The argument filtering Pi contains the following mapping:
countstack_in(x1, x2)  =  countstack_in(x1)
push(x1, x2)  =  push(x1, x2)
cons(x1, x2)  =  cons(x1, x2)
s(x1)  =  s(x1)
U2(x1, x2, x3, x4, x5)  =  U2(x5)
nil  =  nil
U1(x1, x2, x3)  =  U1(x3)
empty  =  empty
0  =  0
countstack_out(x1, x2)  =  countstack_out(x2)


Using Dependency Pairs [1,30] we result in the following initial DP problem:
Pi DP problem:
The TRS P consists of the following rules:

COUNTSTACK_IN(push(cons(U, V), T), s(X)) → U21(U, V, T, X, countstack_in(push(U, push(V, T)), X))
COUNTSTACK_IN(push(cons(U, V), T), s(X)) → COUNTSTACK_IN(push(U, push(V, T)), X)
COUNTSTACK_IN(push(nil, T), X) → U11(T, X, countstack_in(T, X))
COUNTSTACK_IN(push(nil, T), X) → COUNTSTACK_IN(T, X)

The TRS R consists of the following rules:

countstack_in(push(cons(U, V), T), s(X)) → U2(U, V, T, X, countstack_in(push(U, push(V, T)), X))
countstack_in(push(nil, T), X) → U1(T, X, countstack_in(T, X))
countstack_in(empty, 0) → countstack_out(empty, 0)
U1(T, X, countstack_out(T, X)) → countstack_out(push(nil, T), X)
U2(U, V, T, X, countstack_out(push(U, push(V, T)), X)) → countstack_out(push(cons(U, V), T), s(X))

The argument filtering Pi contains the following mapping:
countstack_in(x1, x2)  =  countstack_in(x1)
push(x1, x2)  =  push(x1, x2)
cons(x1, x2)  =  cons(x1, x2)
s(x1)  =  s(x1)
U2(x1, x2, x3, x4, x5)  =  U2(x5)
nil  =  nil
U1(x1, x2, x3)  =  U1(x3)
empty  =  empty
0  =  0
countstack_out(x1, x2)  =  countstack_out(x2)
U21(x1, x2, x3, x4, x5)  =  U21(x5)
U11(x1, x2, x3)  =  U11(x3)
COUNTSTACK_IN(x1, x2)  =  COUNTSTACK_IN(x1)

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

↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof

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

COUNTSTACK_IN(push(cons(U, V), T), s(X)) → U21(U, V, T, X, countstack_in(push(U, push(V, T)), X))
COUNTSTACK_IN(push(cons(U, V), T), s(X)) → COUNTSTACK_IN(push(U, push(V, T)), X)
COUNTSTACK_IN(push(nil, T), X) → U11(T, X, countstack_in(T, X))
COUNTSTACK_IN(push(nil, T), X) → COUNTSTACK_IN(T, X)

The TRS R consists of the following rules:

countstack_in(push(cons(U, V), T), s(X)) → U2(U, V, T, X, countstack_in(push(U, push(V, T)), X))
countstack_in(push(nil, T), X) → U1(T, X, countstack_in(T, X))
countstack_in(empty, 0) → countstack_out(empty, 0)
U1(T, X, countstack_out(T, X)) → countstack_out(push(nil, T), X)
U2(U, V, T, X, countstack_out(push(U, push(V, T)), X)) → countstack_out(push(cons(U, V), T), s(X))

The argument filtering Pi contains the following mapping:
countstack_in(x1, x2)  =  countstack_in(x1)
push(x1, x2)  =  push(x1, x2)
cons(x1, x2)  =  cons(x1, x2)
s(x1)  =  s(x1)
U2(x1, x2, x3, x4, x5)  =  U2(x5)
nil  =  nil
U1(x1, x2, x3)  =  U1(x3)
empty  =  empty
0  =  0
countstack_out(x1, x2)  =  countstack_out(x2)
U21(x1, x2, x3, x4, x5)  =  U21(x5)
U11(x1, x2, x3)  =  U11(x3)
COUNTSTACK_IN(x1, x2)  =  COUNTSTACK_IN(x1)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph [30] contains 1 SCC with 2 less nodes.

↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ PiDP
              ↳ UsableRulesProof

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

COUNTSTACK_IN(push(nil, T), X) → COUNTSTACK_IN(T, X)
COUNTSTACK_IN(push(cons(U, V), T), s(X)) → COUNTSTACK_IN(push(U, push(V, T)), X)

The TRS R consists of the following rules:

countstack_in(push(cons(U, V), T), s(X)) → U2(U, V, T, X, countstack_in(push(U, push(V, T)), X))
countstack_in(push(nil, T), X) → U1(T, X, countstack_in(T, X))
countstack_in(empty, 0) → countstack_out(empty, 0)
U1(T, X, countstack_out(T, X)) → countstack_out(push(nil, T), X)
U2(U, V, T, X, countstack_out(push(U, push(V, T)), X)) → countstack_out(push(cons(U, V), T), s(X))

The argument filtering Pi contains the following mapping:
countstack_in(x1, x2)  =  countstack_in(x1)
push(x1, x2)  =  push(x1, x2)
cons(x1, x2)  =  cons(x1, x2)
s(x1)  =  s(x1)
U2(x1, x2, x3, x4, x5)  =  U2(x5)
nil  =  nil
U1(x1, x2, x3)  =  U1(x3)
empty  =  empty
0  =  0
countstack_out(x1, x2)  =  countstack_out(x2)
COUNTSTACK_IN(x1, x2)  =  COUNTSTACK_IN(x1)

We have to consider all (P,R,Pi)-chains
For (infinitary) constructor rewriting [30] we can delete all non-usable rules from R.

↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ PiDP
              ↳ UsableRulesProof
                ↳ PiDP
                  ↳ PiDPToQDPProof

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

COUNTSTACK_IN(push(nil, T), X) → COUNTSTACK_IN(T, X)
COUNTSTACK_IN(push(cons(U, V), T), s(X)) → COUNTSTACK_IN(push(U, push(V, T)), X)

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

We have to consider all (P,R,Pi)-chains
Transforming (infinitary) constructor rewriting Pi-DP problem [30] into ordinary QDP problem [15] by application of Pi.

↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ PiDP
              ↳ UsableRulesProof
                ↳ PiDP
                  ↳ PiDPToQDPProof
                    ↳ QDP
                      ↳ UsableRulesReductionPairsProof

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

COUNTSTACK_IN(push(cons(U, V), T)) → COUNTSTACK_IN(push(U, push(V, T)))
COUNTSTACK_IN(push(nil, T)) → COUNTSTACK_IN(T)

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
By using the usable rules with reduction pair processor [15] with a polynomial ordering [25], all dependency pairs and the corresponding usable rules [17] 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(push(cons(U, V), T)) → COUNTSTACK_IN(push(U, push(V, T)))
COUNTSTACK_IN(push(nil, T)) → COUNTSTACK_IN(T)
No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [25]:

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



↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ PiDP
              ↳ UsableRulesProof
                ↳ PiDP
                  ↳ PiDPToQDPProof
                    ↳ QDP
                      ↳ UsableRulesReductionPairsProof
                        ↳ QDP
                          ↳ PisEmptyProof

Q DP problem:
P is empty.
R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.