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

0 Prolog
↳1 PrologToPiTRSProof (⇒, 0 ms)
↳2 PiTRS
↳3 DependencyPairsProof (⇔, 4 ms)
↳4 PiDP
↳5 DependencyGraphProof (⇔, 0 ms)
↳6 PiDP
↳7 PiDPToQDPProof (⇔, 21 ms)
↳8 QDP
↳9 QDPOrderProof (⇔, 150 ms)
↳10 QDP
↳11 DependencyGraphProof (⇔, 0 ms)
↳12 QDP
↳13 UsableRulesProof (⇔, 0 ms)
↳14 QDP
↳15 QReductionProof (⇔, 0 ms)
↳16 QDP
↳17 UsableRulesReductionPairsProof (⇔, 13 ms)
↳18 QDP
↳19 PisEmptyProof (⇔, 0 ms)
↳20 YES

(0) Obligation:

Clauses:

p(cons(X, nil)).
p(cons(s(s(X)), cons(Y, Xs))) :- ','(p(cons(X, cons(Y, Xs))), p(cons(s(s(s(s(Y)))), Xs))).
p(cons(0, Xs)) :- p(Xs).

Query: p(g)

(1) PrologToPiTRSProof (SOUND transformation)

We use the technique of [TOCL09]. With regard to the inferred argument filtering the predicates were used in the following modes:
p_in: (b)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

p_in_g(cons(X, nil)) → p_out_g(cons(X, nil))
p_in_g(cons(s(s(X)), cons(Y, Xs))) → U1_g(X, Y, Xs, p_in_g(cons(X, cons(Y, Xs))))
p_in_g(cons(0, Xs)) → U3_g(Xs, p_in_g(Xs))
U3_g(Xs, p_out_g(Xs)) → p_out_g(cons(0, Xs))
U1_g(X, Y, Xs, p_out_g(cons(X, cons(Y, Xs)))) → U2_g(X, Y, Xs, p_in_g(cons(s(s(s(s(Y)))), Xs)))
U2_g(X, Y, Xs, p_out_g(cons(s(s(s(s(Y)))), Xs))) → p_out_g(cons(s(s(X)), cons(Y, Xs)))

Pi is empty.

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog

(2) Obligation:

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

p_in_g(cons(X, nil)) → p_out_g(cons(X, nil))
p_in_g(cons(s(s(X)), cons(Y, Xs))) → U1_g(X, Y, Xs, p_in_g(cons(X, cons(Y, Xs))))
p_in_g(cons(0, Xs)) → U3_g(Xs, p_in_g(Xs))
U3_g(Xs, p_out_g(Xs)) → p_out_g(cons(0, Xs))
U1_g(X, Y, Xs, p_out_g(cons(X, cons(Y, Xs)))) → U2_g(X, Y, Xs, p_in_g(cons(s(s(s(s(Y)))), Xs)))
U2_g(X, Y, Xs, p_out_g(cons(s(s(s(s(Y)))), Xs))) → p_out_g(cons(s(s(X)), cons(Y, Xs)))

Pi is empty.

(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:

P_IN_G(cons(s(s(X)), cons(Y, Xs))) → U1_G(X, Y, Xs, p_in_g(cons(X, cons(Y, Xs))))
P_IN_G(cons(s(s(X)), cons(Y, Xs))) → P_IN_G(cons(X, cons(Y, Xs)))
P_IN_G(cons(0, Xs)) → U3_G(Xs, p_in_g(Xs))
P_IN_G(cons(0, Xs)) → P_IN_G(Xs)
U1_G(X, Y, Xs, p_out_g(cons(X, cons(Y, Xs)))) → U2_G(X, Y, Xs, p_in_g(cons(s(s(s(s(Y)))), Xs)))
U1_G(X, Y, Xs, p_out_g(cons(X, cons(Y, Xs)))) → P_IN_G(cons(s(s(s(s(Y)))), Xs))

The TRS R consists of the following rules:

p_in_g(cons(X, nil)) → p_out_g(cons(X, nil))
p_in_g(cons(s(s(X)), cons(Y, Xs))) → U1_g(X, Y, Xs, p_in_g(cons(X, cons(Y, Xs))))
p_in_g(cons(0, Xs)) → U3_g(Xs, p_in_g(Xs))
U3_g(Xs, p_out_g(Xs)) → p_out_g(cons(0, Xs))
U1_g(X, Y, Xs, p_out_g(cons(X, cons(Y, Xs)))) → U2_g(X, Y, Xs, p_in_g(cons(s(s(s(s(Y)))), Xs)))
U2_g(X, Y, Xs, p_out_g(cons(s(s(s(s(Y)))), Xs))) → p_out_g(cons(s(s(X)), cons(Y, Xs)))

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

(4) Obligation:

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

P_IN_G(cons(s(s(X)), cons(Y, Xs))) → U1_G(X, Y, Xs, p_in_g(cons(X, cons(Y, Xs))))
P_IN_G(cons(s(s(X)), cons(Y, Xs))) → P_IN_G(cons(X, cons(Y, Xs)))
P_IN_G(cons(0, Xs)) → U3_G(Xs, p_in_g(Xs))
P_IN_G(cons(0, Xs)) → P_IN_G(Xs)
U1_G(X, Y, Xs, p_out_g(cons(X, cons(Y, Xs)))) → U2_G(X, Y, Xs, p_in_g(cons(s(s(s(s(Y)))), Xs)))
U1_G(X, Y, Xs, p_out_g(cons(X, cons(Y, Xs)))) → P_IN_G(cons(s(s(s(s(Y)))), Xs))

The TRS R consists of the following rules:

p_in_g(cons(X, nil)) → p_out_g(cons(X, nil))
p_in_g(cons(s(s(X)), cons(Y, Xs))) → U1_g(X, Y, Xs, p_in_g(cons(X, cons(Y, Xs))))
p_in_g(cons(0, Xs)) → U3_g(Xs, p_in_g(Xs))
U3_g(Xs, p_out_g(Xs)) → p_out_g(cons(0, Xs))
U1_g(X, Y, Xs, p_out_g(cons(X, cons(Y, Xs)))) → U2_g(X, Y, Xs, p_in_g(cons(s(s(s(s(Y)))), Xs)))
U2_g(X, Y, Xs, p_out_g(cons(s(s(s(s(Y)))), Xs))) → p_out_g(cons(s(s(X)), cons(Y, Xs)))

Pi is empty.
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:

U1_G(X, Y, Xs, p_out_g(cons(X, cons(Y, Xs)))) → P_IN_G(cons(s(s(s(s(Y)))), Xs))
P_IN_G(cons(s(s(X)), cons(Y, Xs))) → U1_G(X, Y, Xs, p_in_g(cons(X, cons(Y, Xs))))
P_IN_G(cons(s(s(X)), cons(Y, Xs))) → P_IN_G(cons(X, cons(Y, Xs)))
P_IN_G(cons(0, Xs)) → P_IN_G(Xs)

The TRS R consists of the following rules:

p_in_g(cons(X, nil)) → p_out_g(cons(X, nil))
p_in_g(cons(s(s(X)), cons(Y, Xs))) → U1_g(X, Y, Xs, p_in_g(cons(X, cons(Y, Xs))))
p_in_g(cons(0, Xs)) → U3_g(Xs, p_in_g(Xs))
U3_g(Xs, p_out_g(Xs)) → p_out_g(cons(0, Xs))
U1_g(X, Y, Xs, p_out_g(cons(X, cons(Y, Xs)))) → U2_g(X, Y, Xs, p_in_g(cons(s(s(s(s(Y)))), Xs)))
U2_g(X, Y, Xs, p_out_g(cons(s(s(s(s(Y)))), Xs))) → p_out_g(cons(s(s(X)), cons(Y, Xs)))

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

(7) PiDPToQDPProof (EQUIVALENT 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:

U1_G(X, Y, Xs, p_out_g(cons(X, cons(Y, Xs)))) → P_IN_G(cons(s(s(s(s(Y)))), Xs))
P_IN_G(cons(s(s(X)), cons(Y, Xs))) → U1_G(X, Y, Xs, p_in_g(cons(X, cons(Y, Xs))))
P_IN_G(cons(s(s(X)), cons(Y, Xs))) → P_IN_G(cons(X, cons(Y, Xs)))
P_IN_G(cons(0, Xs)) → P_IN_G(Xs)

The TRS R consists of the following rules:

p_in_g(cons(X, nil)) → p_out_g(cons(X, nil))
p_in_g(cons(s(s(X)), cons(Y, Xs))) → U1_g(X, Y, Xs, p_in_g(cons(X, cons(Y, Xs))))
p_in_g(cons(0, Xs)) → U3_g(Xs, p_in_g(Xs))
U3_g(Xs, p_out_g(Xs)) → p_out_g(cons(0, Xs))
U1_g(X, Y, Xs, p_out_g(cons(X, cons(Y, Xs)))) → U2_g(X, Y, Xs, p_in_g(cons(s(s(s(s(Y)))), Xs)))
U2_g(X, Y, Xs, p_out_g(cons(s(s(s(s(Y)))), Xs))) → p_out_g(cons(s(s(X)), cons(Y, Xs)))

The set Q consists of the following terms:

p_in_g(x0)
U3_g(x0, x1)
U1_g(x0, x1, x2, x3)
U2_g(x0, x1, x2, x3)

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

(9) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.


P_IN_G(cons(s(s(X)), cons(Y, Xs))) → U1_G(X, Y, Xs, p_in_g(cons(X, cons(Y, Xs))))
P_IN_G(cons(0, Xs)) → P_IN_G(Xs)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(0) = 1   
POL(P_IN_G(x1)) = x1   
POL(U1_G(x1, x2, x3, x4)) = 1 + x2 + x3   
POL(U1_g(x1, x2, x3, x4)) = 0   
POL(U2_g(x1, x2, x3, x4)) = 0   
POL(U3_g(x1, x2)) = 0   
POL(cons(x1, x2)) = 1 + x1 + x2   
POL(nil) = 1   
POL(p_in_g(x1)) = x1   
POL(p_out_g(x1)) = 1   
POL(s(x1)) = x1   

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
none

(10) Obligation:

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

U1_G(X, Y, Xs, p_out_g(cons(X, cons(Y, Xs)))) → P_IN_G(cons(s(s(s(s(Y)))), Xs))
P_IN_G(cons(s(s(X)), cons(Y, Xs))) → P_IN_G(cons(X, cons(Y, Xs)))

The TRS R consists of the following rules:

p_in_g(cons(X, nil)) → p_out_g(cons(X, nil))
p_in_g(cons(s(s(X)), cons(Y, Xs))) → U1_g(X, Y, Xs, p_in_g(cons(X, cons(Y, Xs))))
p_in_g(cons(0, Xs)) → U3_g(Xs, p_in_g(Xs))
U3_g(Xs, p_out_g(Xs)) → p_out_g(cons(0, Xs))
U1_g(X, Y, Xs, p_out_g(cons(X, cons(Y, Xs)))) → U2_g(X, Y, Xs, p_in_g(cons(s(s(s(s(Y)))), Xs)))
U2_g(X, Y, Xs, p_out_g(cons(s(s(s(s(Y)))), Xs))) → p_out_g(cons(s(s(X)), cons(Y, Xs)))

The set Q consists of the following terms:

p_in_g(x0)
U3_g(x0, x1)
U1_g(x0, x1, x2, x3)
U2_g(x0, x1, x2, x3)

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

(11) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.

(12) Obligation:

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

P_IN_G(cons(s(s(X)), cons(Y, Xs))) → P_IN_G(cons(X, cons(Y, Xs)))

The TRS R consists of the following rules:

p_in_g(cons(X, nil)) → p_out_g(cons(X, nil))
p_in_g(cons(s(s(X)), cons(Y, Xs))) → U1_g(X, Y, Xs, p_in_g(cons(X, cons(Y, Xs))))
p_in_g(cons(0, Xs)) → U3_g(Xs, p_in_g(Xs))
U3_g(Xs, p_out_g(Xs)) → p_out_g(cons(0, Xs))
U1_g(X, Y, Xs, p_out_g(cons(X, cons(Y, Xs)))) → U2_g(X, Y, Xs, p_in_g(cons(s(s(s(s(Y)))), Xs)))
U2_g(X, Y, Xs, p_out_g(cons(s(s(s(s(Y)))), Xs))) → p_out_g(cons(s(s(X)), cons(Y, Xs)))

The set Q consists of the following terms:

p_in_g(x0)
U3_g(x0, x1)
U1_g(x0, x1, x2, x3)
U2_g(x0, x1, x2, x3)

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

(13) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(14) Obligation:

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

P_IN_G(cons(s(s(X)), cons(Y, Xs))) → P_IN_G(cons(X, cons(Y, Xs)))

R is empty.
The set Q consists of the following terms:

p_in_g(x0)
U3_g(x0, x1)
U1_g(x0, x1, x2, x3)
U2_g(x0, x1, x2, x3)

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

(15) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

p_in_g(x0)
U3_g(x0, x1)
U1_g(x0, x1, x2, x3)
U2_g(x0, x1, x2, x3)

(16) Obligation:

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

P_IN_G(cons(s(s(X)), cons(Y, Xs))) → P_IN_G(cons(X, cons(Y, Xs)))

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

(17) 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:

P_IN_G(cons(s(s(X)), cons(Y, Xs))) → P_IN_G(cons(X, cons(Y, Xs)))
No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [POLO]:

POL(P_IN_G(x1)) = 2·x1   
POL(cons(x1, x2)) = 2·x1 + x2   
POL(s(x1)) = 2·x1   

(18) Obligation:

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

(19) PisEmptyProof (EQUIVALENT transformation)

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

(20) YES