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

(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) PrologToPiTRSProof (SOUND transformation)

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

ordered_in_g([]) → ordered_out_g([])
ordered_in_g(.(X1, [])) → ordered_out_g(.(X1, []))
ordered_in_g(.(X, .(Y, Xs))) → U1_g(X, Y, Xs, less_in_gg(X, s(Y)))
less_in_gg(0, s(X2)) → less_out_gg(0, s(X2))
less_in_gg(s(X), s(Y)) → U3_gg(X, Y, less_in_gg(X, Y))
U3_gg(X, Y, less_out_gg(X, Y)) → less_out_gg(s(X), s(Y))
U1_g(X, Y, Xs, less_out_gg(X, s(Y))) → U2_g(X, Y, Xs, ordered_in_g(.(Y, Xs)))
U2_g(X, Y, Xs, ordered_out_g(.(Y, Xs))) → ordered_out_g(.(X, .(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:

ordered_in_g([]) → ordered_out_g([])
ordered_in_g(.(X1, [])) → ordered_out_g(.(X1, []))
ordered_in_g(.(X, .(Y, Xs))) → U1_g(X, Y, Xs, less_in_gg(X, s(Y)))
less_in_gg(0, s(X2)) → less_out_gg(0, s(X2))
less_in_gg(s(X), s(Y)) → U3_gg(X, Y, less_in_gg(X, Y))
U3_gg(X, Y, less_out_gg(X, Y)) → less_out_gg(s(X), s(Y))
U1_g(X, Y, Xs, less_out_gg(X, s(Y))) → U2_g(X, Y, Xs, ordered_in_g(.(Y, Xs)))
U2_g(X, Y, Xs, ordered_out_g(.(Y, Xs))) → ordered_out_g(.(X, .(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:

ORDERED_IN_G(.(X, .(Y, Xs))) → U1_G(X, Y, Xs, less_in_gg(X, s(Y)))
ORDERED_IN_G(.(X, .(Y, Xs))) → LESS_IN_GG(X, s(Y))
LESS_IN_GG(s(X), s(Y)) → U3_GG(X, Y, less_in_gg(X, Y))
LESS_IN_GG(s(X), s(Y)) → LESS_IN_GG(X, Y)
U1_G(X, Y, Xs, less_out_gg(X, s(Y))) → U2_G(X, Y, Xs, ordered_in_g(.(Y, Xs)))
U1_G(X, Y, Xs, less_out_gg(X, s(Y))) → ORDERED_IN_G(.(Y, Xs))

The TRS R consists of the following rules:

ordered_in_g([]) → ordered_out_g([])
ordered_in_g(.(X1, [])) → ordered_out_g(.(X1, []))
ordered_in_g(.(X, .(Y, Xs))) → U1_g(X, Y, Xs, less_in_gg(X, s(Y)))
less_in_gg(0, s(X2)) → less_out_gg(0, s(X2))
less_in_gg(s(X), s(Y)) → U3_gg(X, Y, less_in_gg(X, Y))
U3_gg(X, Y, less_out_gg(X, Y)) → less_out_gg(s(X), s(Y))
U1_g(X, Y, Xs, less_out_gg(X, s(Y))) → U2_g(X, Y, Xs, ordered_in_g(.(Y, Xs)))
U2_g(X, Y, Xs, ordered_out_g(.(Y, Xs))) → ordered_out_g(.(X, .(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:

ORDERED_IN_G(.(X, .(Y, Xs))) → U1_G(X, Y, Xs, less_in_gg(X, s(Y)))
ORDERED_IN_G(.(X, .(Y, Xs))) → LESS_IN_GG(X, s(Y))
LESS_IN_GG(s(X), s(Y)) → U3_GG(X, Y, less_in_gg(X, Y))
LESS_IN_GG(s(X), s(Y)) → LESS_IN_GG(X, Y)
U1_G(X, Y, Xs, less_out_gg(X, s(Y))) → U2_G(X, Y, Xs, ordered_in_g(.(Y, Xs)))
U1_G(X, Y, Xs, less_out_gg(X, s(Y))) → ORDERED_IN_G(.(Y, Xs))

The TRS R consists of the following rules:

ordered_in_g([]) → ordered_out_g([])
ordered_in_g(.(X1, [])) → ordered_out_g(.(X1, []))
ordered_in_g(.(X, .(Y, Xs))) → U1_g(X, Y, Xs, less_in_gg(X, s(Y)))
less_in_gg(0, s(X2)) → less_out_gg(0, s(X2))
less_in_gg(s(X), s(Y)) → U3_gg(X, Y, less_in_gg(X, Y))
U3_gg(X, Y, less_out_gg(X, Y)) → less_out_gg(s(X), s(Y))
U1_g(X, Y, Xs, less_out_gg(X, s(Y))) → U2_g(X, Y, Xs, ordered_in_g(.(Y, Xs)))
U2_g(X, Y, Xs, ordered_out_g(.(Y, Xs))) → ordered_out_g(.(X, .(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 2 SCCs with 3 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

LESS_IN_GG(s(X), s(Y)) → LESS_IN_GG(X, Y)

The TRS R consists of the following rules:

ordered_in_g([]) → ordered_out_g([])
ordered_in_g(.(X1, [])) → ordered_out_g(.(X1, []))
ordered_in_g(.(X, .(Y, Xs))) → U1_g(X, Y, Xs, less_in_gg(X, s(Y)))
less_in_gg(0, s(X2)) → less_out_gg(0, s(X2))
less_in_gg(s(X), s(Y)) → U3_gg(X, Y, less_in_gg(X, Y))
U3_gg(X, Y, less_out_gg(X, Y)) → less_out_gg(s(X), s(Y))
U1_g(X, Y, Xs, less_out_gg(X, s(Y))) → U2_g(X, Y, Xs, ordered_in_g(.(Y, Xs)))
U2_g(X, Y, Xs, ordered_out_g(.(Y, Xs))) → ordered_out_g(.(X, .(Y, Xs)))

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:

LESS_IN_GG(s(X), s(Y)) → LESS_IN_GG(X, Y)

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:

LESS_IN_GG(s(X), s(Y)) → LESS_IN_GG(X, Y)

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:

• LESS_IN_GG(s(X), s(Y)) → LESS_IN_GG(X, Y)
  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:

U1_G(X, Y, Xs, less_out_gg(X, s(Y))) → ORDERED_IN_G(.(Y, Xs))
ORDERED_IN_G(.(X, .(Y, Xs))) → U1_G(X, Y, Xs, less_in_gg(X, s(Y)))

The TRS R consists of the following rules:

ordered_in_g([]) → ordered_out_g([])
ordered_in_g(.(X1, [])) → ordered_out_g(.(X1, []))
ordered_in_g(.(X, .(Y, Xs))) → U1_g(X, Y, Xs, less_in_gg(X, s(Y)))
less_in_gg(0, s(X2)) → less_out_gg(0, s(X2))
less_in_gg(s(X), s(Y)) → U3_gg(X, Y, less_in_gg(X, Y))
U3_gg(X, Y, less_out_gg(X, Y)) → less_out_gg(s(X), s(Y))
U1_g(X, Y, Xs, less_out_gg(X, s(Y))) → U2_g(X, Y, Xs, ordered_in_g(.(Y, Xs)))
U2_g(X, Y, Xs, ordered_out_g(.(Y, Xs))) → ordered_out_g(.(X, .(Y, Xs)))

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

(15) UsableRulesProof (EQUIVALENT transformation)

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

(16) Obligation:

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

U1_G(X, Y, Xs, less_out_gg(X, s(Y))) → ORDERED_IN_G(.(Y, Xs))
ORDERED_IN_G(.(X, .(Y, Xs))) → U1_G(X, Y, Xs, less_in_gg(X, s(Y)))

The TRS R consists of the following rules:

less_in_gg(0, s(X2)) → less_out_gg(0, s(X2))
less_in_gg(s(X), s(Y)) → U3_gg(X, Y, less_in_gg(X, Y))
U3_gg(X, Y, less_out_gg(X, Y)) → less_out_gg(s(X), s(Y))

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

(17) PiDPToQDPProof (EQUIVALENT transformation)

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

(18) Obligation:

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

U1_G(X, Y, Xs, less_out_gg(X, s(Y))) → ORDERED_IN_G(.(Y, Xs))
ORDERED_IN_G(.(X, .(Y, Xs))) → U1_G(X, Y, Xs, less_in_gg(X, s(Y)))

The TRS R consists of the following rules:

less_in_gg(0, s(X2)) → less_out_gg(0, s(X2))
less_in_gg(s(X), s(Y)) → U3_gg(X, Y, less_in_gg(X, Y))
U3_gg(X, Y, less_out_gg(X, Y)) → less_out_gg(s(X), s(Y))

The set Q consists of the following terms:

less_in_gg(x0, x1)
U3_gg(x0, x1, x2)

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

(19) 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:

ORDERED_IN_G(.(X, .(Y, Xs))) → U1_G(X, Y, Xs, less_in_gg(X, s(Y)))

Used ordering: Polynomial interpretation [POLO]:

POL(.(x₁, x₂)) = 1 + 2·x₁ + x₂   
POL(0) = 0   
POL(ORDERED_IN_G(x₁)) = 2·x₁   
POL(U1_G(x₁, x₂, x₃, x₄)) = 2 + x₁ + 2·x₂ + 2·x₃ + x₄   
POL(U3_gg(x₁, x₂, x₃)) = 2·x₁ + x₂ + x₃   
POL(less_in_gg(x₁, x₂)) = 2·x₁ + x₂   
POL(less_out_gg(x₁, x₂)) = x₁ + x₂   
POL(s(x₁)) = 2·x₁   

(20) Obligation:

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

U1_G(X, Y, Xs, less_out_gg(X, s(Y))) → ORDERED_IN_G(.(Y, Xs))

The TRS R consists of the following rules:

less_in_gg(0, s(X2)) → less_out_gg(0, s(X2))
less_in_gg(s(X), s(Y)) → U3_gg(X, Y, less_in_gg(X, Y))
U3_gg(X, Y, less_out_gg(X, Y)) → less_out_gg(s(X), s(Y))

The set Q consists of the following terms:

less_in_gg(x0, x1)
U3_gg(x0, x1, x2)

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

(21) DependencyGraphProof (EQUIVALENT transformation)

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

(22) TRUE