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

0 Prolog
↳1 PrologToDTProblemTransformerProof (⇒, 87 ms)
↳2 TRIPLES
↳3 TriplesToPiDPProof (⇒, 0 ms)
↳4 PiDP
↳5 DependencyGraphProof (⇔, 0 ms)
↳6 AND
    ↳7 PiDP
        ↳8 UsableRulesProof (⇔, 0 ms)
        ↳9 PiDP
        ↳10 PiDPToQDPProof (⇔, 4 ms)
        ↳11 QDP
        ↳12 QDPSizeChangeProof (⇔, 0 ms)
        ↳13 YES
    ↳14 PiDP
        ↳15 PiDPToQDPProof (⇔, 0 ms)
        ↳16 QDP
        ↳17 MRRProof (⇔, 143 ms)
        ↳18 QDP
        ↳19 DependencyGraphProof (⇔, 0 ms)
        ↳20 TRUE

(0) Obligation:

Clauses:

ordered([]).
ordered(.(X1, [])).
ordered(.(X, .(Y, Xs))) :- ','(less(X, s(Y)), ordered(.(Y, Xs))).
less(0, s(X2)).
less(s(X), s(Y)) :- less(X, Y).

Query: ordered(g)

(1) PrologToDTProblemTransformerProof (SOUND transformation)

Built DT problem from termination graph DT10.

(2) Obligation:

Triples:

lessB(s(X1), s(X2)) :- lessB(X1, X2).
orderedA(.(0, .(X1, X2))) :- orderedA(.(X1, X2)).
orderedA(.(s(X1), .(X2, X3))) :- lessB(X1, X2).
orderedA(.(s(X1), .(X2, X3))) :- ','(lesscB(X1, X2), orderedA(.(X2, X3))).

Clauses:

orderedcA([]).
orderedcA(.(X1, [])).
orderedcA(.(0, .(X1, X2))) :- orderedcA(.(X1, X2)).
orderedcA(.(s(X1), .(X2, X3))) :- ','(lesscB(X1, X2), orderedcA(.(X2, X3))).
lesscB(0, s(X1)).
lesscB(s(X1), s(X2)) :- lesscB(X1, X2).

Afs:

orderedA(x1)  =  orderedA(x1)

(3) TriplesToPiDPProof (SOUND transformation)

We use the technique of [DT09]. With regard to the inferred argument filtering the predicates were used in the following modes:
orderedA_in: (b)
lessB_in: (b,b)
lesscB_in: (b,b)
Transforming TRIPLES into the following Term Rewriting System:
Pi DP problem:
The TRS P consists of the following rules:

ORDEREDA_IN_G(.(0, .(X1, X2))) → U2_G(X1, X2, orderedA_in_g(.(X1, X2)))
ORDEREDA_IN_G(.(0, .(X1, X2))) → ORDEREDA_IN_G(.(X1, X2))
ORDEREDA_IN_G(.(s(X1), .(X2, X3))) → U3_G(X1, X2, X3, lessB_in_gg(X1, X2))
ORDEREDA_IN_G(.(s(X1), .(X2, X3))) → LESSB_IN_GG(X1, X2)
LESSB_IN_GG(s(X1), s(X2)) → U1_GG(X1, X2, lessB_in_gg(X1, X2))
LESSB_IN_GG(s(X1), s(X2)) → LESSB_IN_GG(X1, X2)
ORDEREDA_IN_G(.(s(X1), .(X2, X3))) → U4_G(X1, X2, X3, lesscB_in_gg(X1, X2))
U4_G(X1, X2, X3, lesscB_out_gg(X1, X2)) → U5_G(X1, X2, X3, orderedA_in_g(.(X2, X3)))
U4_G(X1, X2, X3, lesscB_out_gg(X1, X2)) → ORDEREDA_IN_G(.(X2, X3))

The TRS R consists of the following rules:

lesscB_in_gg(0, s(X1)) → lesscB_out_gg(0, s(X1))
lesscB_in_gg(s(X1), s(X2)) → U10_gg(X1, X2, lesscB_in_gg(X1, X2))
U10_gg(X1, X2, lesscB_out_gg(X1, X2)) → lesscB_out_gg(s(X1), s(X2))

Pi is empty.
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:

ORDEREDA_IN_G(.(0, .(X1, X2))) → U2_G(X1, X2, orderedA_in_g(.(X1, X2)))
ORDEREDA_IN_G(.(0, .(X1, X2))) → ORDEREDA_IN_G(.(X1, X2))
ORDEREDA_IN_G(.(s(X1), .(X2, X3))) → U3_G(X1, X2, X3, lessB_in_gg(X1, X2))
ORDEREDA_IN_G(.(s(X1), .(X2, X3))) → LESSB_IN_GG(X1, X2)
LESSB_IN_GG(s(X1), s(X2)) → U1_GG(X1, X2, lessB_in_gg(X1, X2))
LESSB_IN_GG(s(X1), s(X2)) → LESSB_IN_GG(X1, X2)
ORDEREDA_IN_G(.(s(X1), .(X2, X3))) → U4_G(X1, X2, X3, lesscB_in_gg(X1, X2))
U4_G(X1, X2, X3, lesscB_out_gg(X1, X2)) → U5_G(X1, X2, X3, orderedA_in_g(.(X2, X3)))
U4_G(X1, X2, X3, lesscB_out_gg(X1, X2)) → ORDEREDA_IN_G(.(X2, X3))

The TRS R consists of the following rules:

lesscB_in_gg(0, s(X1)) → lesscB_out_gg(0, s(X1))
lesscB_in_gg(s(X1), s(X2)) → U10_gg(X1, X2, lesscB_in_gg(X1, X2))
U10_gg(X1, X2, lesscB_out_gg(X1, X2)) → lesscB_out_gg(s(X1), s(X2))

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LOPSTR] contains 2 SCCs with 5 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

LESSB_IN_GG(s(X1), s(X2)) → LESSB_IN_GG(X1, X2)

The TRS R consists of the following rules:

lesscB_in_gg(0, s(X1)) → lesscB_out_gg(0, s(X1))
lesscB_in_gg(s(X1), s(X2)) → U10_gg(X1, X2, lesscB_in_gg(X1, X2))
U10_gg(X1, X2, lesscB_out_gg(X1, X2)) → lesscB_out_gg(s(X1), s(X2))

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

(8) UsableRulesProof (EQUIVALENT transformation)

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

(9) Obligation:

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

LESSB_IN_GG(s(X1), s(X2)) → LESSB_IN_GG(X1, X2)

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

(10) PiDPToQDPProof (EQUIVALENT transformation)

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

(11) Obligation:

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

LESSB_IN_GG(s(X1), s(X2)) → LESSB_IN_GG(X1, X2)

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

(12) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

  • LESSB_IN_GG(s(X1), s(X2)) → LESSB_IN_GG(X1, X2)
    The graph contains the following edges 1 > 1, 2 > 2

(13) YES

(14) Obligation:

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

ORDEREDA_IN_G(.(s(X1), .(X2, X3))) → U4_G(X1, X2, X3, lesscB_in_gg(X1, X2))
U4_G(X1, X2, X3, lesscB_out_gg(X1, X2)) → ORDEREDA_IN_G(.(X2, X3))
ORDEREDA_IN_G(.(0, .(X1, X2))) → ORDEREDA_IN_G(.(X1, X2))

The TRS R consists of the following rules:

lesscB_in_gg(0, s(X1)) → lesscB_out_gg(0, s(X1))
lesscB_in_gg(s(X1), s(X2)) → U10_gg(X1, X2, lesscB_in_gg(X1, X2))
U10_gg(X1, X2, lesscB_out_gg(X1, X2)) → lesscB_out_gg(s(X1), s(X2))

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

(15) PiDPToQDPProof (EQUIVALENT transformation)

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

(16) Obligation:

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

ORDEREDA_IN_G(.(s(X1), .(X2, X3))) → U4_G(X1, X2, X3, lesscB_in_gg(X1, X2))
U4_G(X1, X2, X3, lesscB_out_gg(X1, X2)) → ORDEREDA_IN_G(.(X2, X3))
ORDEREDA_IN_G(.(0, .(X1, X2))) → ORDEREDA_IN_G(.(X1, X2))

The TRS R consists of the following rules:

lesscB_in_gg(0, s(X1)) → lesscB_out_gg(0, s(X1))
lesscB_in_gg(s(X1), s(X2)) → U10_gg(X1, X2, lesscB_in_gg(X1, X2))
U10_gg(X1, X2, lesscB_out_gg(X1, X2)) → lesscB_out_gg(s(X1), s(X2))

The set Q consists of the following terms:

lesscB_in_gg(x0, x1)
U10_gg(x0, x1, x2)

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

(17) MRRProof (EQUIVALENT transformation)

By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

ORDEREDA_IN_G(.(s(X1), .(X2, X3))) → U4_G(X1, X2, X3, lesscB_in_gg(X1, X2))
ORDEREDA_IN_G(.(0, .(X1, X2))) → ORDEREDA_IN_G(.(X1, X2))


Used ordering: Polynomial interpretation [POLO]:

POL(.(x1, x2)) = 1 + 2·x1 + 2·x2   
POL(0) = 0   
POL(ORDEREDA_IN_G(x1)) = x1   
POL(U10_gg(x1, x2, x3)) = x1 + x2 + x3   
POL(U4_G(x1, x2, x3, x4)) = 1 + 2·x1 + x2 + 2·x3 + 2·x4   
POL(lesscB_in_gg(x1, x2)) = x1 + x2   
POL(lesscB_out_gg(x1, x2)) = x1 + x2   
POL(s(x1)) = 2·x1   

(18) Obligation:

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

U4_G(X1, X2, X3, lesscB_out_gg(X1, X2)) → ORDEREDA_IN_G(.(X2, X3))

The TRS R consists of the following rules:

lesscB_in_gg(0, s(X1)) → lesscB_out_gg(0, s(X1))
lesscB_in_gg(s(X1), s(X2)) → U10_gg(X1, X2, lesscB_in_gg(X1, X2))
U10_gg(X1, X2, lesscB_out_gg(X1, X2)) → lesscB_out_gg(s(X1), s(X2))

The set Q consists of the following terms:

lesscB_in_gg(x0, x1)
U10_gg(x0, x1, x2)

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

(19) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.

(20) TRUE