Left Termination of the query pattern ordered_in_1(g) w.r.t. the given Prolog program could successfully be proven:

0 Prolog
↳1 PrologToDTProblemTransformerProof (⇐)
↳2 TRIPLES
↳3 TriplesToPiDPProof (⇐)
↳4 PiDP
↳5 DependencyGraphProof (⇔)
↳6 AND
    ↳7 PiDP
        ↳8 UsableRulesProof (⇔)
        ↳9 PiDP
        ↳10 PiDPToQDPProof (⇔)
        ↳11 QDP
        ↳12 QDPSizeChangeProof (⇔)
        ↳13 YES
    ↳14 PiDP
        ↳15 PiDPToQDPProof (⇔)
        ↳16 QDP
        ↳17 MRRProof (⇔)
        ↳18 QDP
        ↳19 MRRProof (⇔)
        ↳20 QDP
        ↳21 DependencyGraphProof (⇔)
        ↳22 TRUE
    ↳23 PiDP
        ↳24 UsableRulesProof (⇔)
        ↳25 PiDP
        ↳26 PiDPToQDPProof (⇔)
        ↳27 QDP
        ↳28 QDPSizeChangeProof (⇔)
        ↳29 YES

(0) Obligation:

Clauses:

ordered([]).
ordered(.(X, [])).
ordered(.(X, .(Y, Xs))) :- ','(le(X, Y), ordered(.(Y, Xs))).
le(s(X), s(Y)) :- le(X, Y).
le(0, s(0)).
le(0, 0).

Queries:

ordered(g).

(1) PrologToDTProblemTransformerProof (SOUND transformation)

Built DT problem from termination graph.

(2) Obligation:

Triples:

le19(s(T33), s(T34)) :- le19(T33, T34).
ordered1(.(s(T19), .(s(T20), T10))) :- le19(T19, T20).
ordered1(.(s(T19), .(s(T20), T10))) :- ','(lec19(T19, T20), ordered1(.(s(T20), T10))).
ordered1(.(0, .(s(0), T10))) :- ordered1(.(s(0), T10)).
ordered1(.(0, .(0, T10))) :- ordered1(.(0, T10)).

Clauses:

orderedc1([]).
orderedc1(.(T3, [])).
orderedc1(.(s(T19), .(s(T20), T10))) :- ','(lec19(T19, T20), orderedc1(.(s(T20), T10))).
orderedc1(.(0, .(s(0), T10))) :- orderedc1(.(s(0), T10)).
orderedc1(.(0, .(0, T10))) :- orderedc1(.(0, T10)).
lec19(s(T33), s(T34)) :- lec19(T33, T34).
lec19(0, s(0)).
lec19(0, 0).

Afs:

ordered1(x1)  =  ordered1(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:
ordered1_in: (b)
le19_in: (b,b)
lec19_in: (b,b)
Transforming TRIPLES into the following Term Rewriting System:
Pi DP problem:
The TRS P consists of the following rules:

ORDERED1_IN_G(.(s(T19), .(s(T20), T10))) → U2_G(T19, T20, T10, le19_in_gg(T19, T20))
ORDERED1_IN_G(.(s(T19), .(s(T20), T10))) → LE19_IN_GG(T19, T20)
LE19_IN_GG(s(T33), s(T34)) → U1_GG(T33, T34, le19_in_gg(T33, T34))
LE19_IN_GG(s(T33), s(T34)) → LE19_IN_GG(T33, T34)
ORDERED1_IN_G(.(s(T19), .(s(T20), T10))) → U3_G(T19, T20, T10, lec19_in_gg(T19, T20))
U3_G(T19, T20, T10, lec19_out_gg(T19, T20)) → U4_G(T19, T20, T10, ordered1_in_g(.(s(T20), T10)))
U3_G(T19, T20, T10, lec19_out_gg(T19, T20)) → ORDERED1_IN_G(.(s(T20), T10))
ORDERED1_IN_G(.(0, .(s(0), T10))) → U5_G(T10, ordered1_in_g(.(s(0), T10)))
ORDERED1_IN_G(.(0, .(s(0), T10))) → ORDERED1_IN_G(.(s(0), T10))
ORDERED1_IN_G(.(0, .(0, T10))) → U6_G(T10, ordered1_in_g(.(0, T10)))
ORDERED1_IN_G(.(0, .(0, T10))) → ORDERED1_IN_G(.(0, T10))

The TRS R consists of the following rules:

lec19_in_gg(s(T33), s(T34)) → U12_gg(T33, T34, lec19_in_gg(T33, T34))
lec19_in_gg(0, s(0)) → lec19_out_gg(0, s(0))
lec19_in_gg(0, 0) → lec19_out_gg(0, 0)
U12_gg(T33, T34, lec19_out_gg(T33, T34)) → lec19_out_gg(s(T33), s(T34))

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:

ORDERED1_IN_G(.(s(T19), .(s(T20), T10))) → U2_G(T19, T20, T10, le19_in_gg(T19, T20))
ORDERED1_IN_G(.(s(T19), .(s(T20), T10))) → LE19_IN_GG(T19, T20)
LE19_IN_GG(s(T33), s(T34)) → U1_GG(T33, T34, le19_in_gg(T33, T34))
LE19_IN_GG(s(T33), s(T34)) → LE19_IN_GG(T33, T34)
ORDERED1_IN_G(.(s(T19), .(s(T20), T10))) → U3_G(T19, T20, T10, lec19_in_gg(T19, T20))
U3_G(T19, T20, T10, lec19_out_gg(T19, T20)) → U4_G(T19, T20, T10, ordered1_in_g(.(s(T20), T10)))
U3_G(T19, T20, T10, lec19_out_gg(T19, T20)) → ORDERED1_IN_G(.(s(T20), T10))
ORDERED1_IN_G(.(0, .(s(0), T10))) → U5_G(T10, ordered1_in_g(.(s(0), T10)))
ORDERED1_IN_G(.(0, .(s(0), T10))) → ORDERED1_IN_G(.(s(0), T10))
ORDERED1_IN_G(.(0, .(0, T10))) → U6_G(T10, ordered1_in_g(.(0, T10)))
ORDERED1_IN_G(.(0, .(0, T10))) → ORDERED1_IN_G(.(0, T10))

The TRS R consists of the following rules:

lec19_in_gg(s(T33), s(T34)) → U12_gg(T33, T34, lec19_in_gg(T33, T34))
lec19_in_gg(0, s(0)) → lec19_out_gg(0, s(0))
lec19_in_gg(0, 0) → lec19_out_gg(0, 0)
U12_gg(T33, T34, lec19_out_gg(T33, T34)) → lec19_out_gg(s(T33), s(T34))

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LOPSTR] contains 3 SCCs with 7 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

LE19_IN_GG(s(T33), s(T34)) → LE19_IN_GG(T33, T34)

The TRS R consists of the following rules:

lec19_in_gg(s(T33), s(T34)) → U12_gg(T33, T34, lec19_in_gg(T33, T34))
lec19_in_gg(0, s(0)) → lec19_out_gg(0, s(0))
lec19_in_gg(0, 0) → lec19_out_gg(0, 0)
U12_gg(T33, T34, lec19_out_gg(T33, T34)) → lec19_out_gg(s(T33), s(T34))

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:

LE19_IN_GG(s(T33), s(T34)) → LE19_IN_GG(T33, T34)

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:

LE19_IN_GG(s(T33), s(T34)) → LE19_IN_GG(T33, T34)

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:

  • LE19_IN_GG(s(T33), s(T34)) → LE19_IN_GG(T33, T34)
    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:

ORDERED1_IN_G(.(s(T19), .(s(T20), T10))) → U3_G(T19, T20, T10, lec19_in_gg(T19, T20))
U3_G(T19, T20, T10, lec19_out_gg(T19, T20)) → ORDERED1_IN_G(.(s(T20), T10))

The TRS R consists of the following rules:

lec19_in_gg(s(T33), s(T34)) → U12_gg(T33, T34, lec19_in_gg(T33, T34))
lec19_in_gg(0, s(0)) → lec19_out_gg(0, s(0))
lec19_in_gg(0, 0) → lec19_out_gg(0, 0)
U12_gg(T33, T34, lec19_out_gg(T33, T34)) → lec19_out_gg(s(T33), s(T34))

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:

ORDERED1_IN_G(.(s(T19), .(s(T20), T10))) → U3_G(T19, T20, T10, lec19_in_gg(T19, T20))
U3_G(T19, T20, T10, lec19_out_gg(T19, T20)) → ORDERED1_IN_G(.(s(T20), T10))

The TRS R consists of the following rules:

lec19_in_gg(s(T33), s(T34)) → U12_gg(T33, T34, lec19_in_gg(T33, T34))
lec19_in_gg(0, s(0)) → lec19_out_gg(0, s(0))
lec19_in_gg(0, 0) → lec19_out_gg(0, 0)
U12_gg(T33, T34, lec19_out_gg(T33, T34)) → lec19_out_gg(s(T33), s(T34))

The set Q consists of the following terms:

lec19_in_gg(x0, x1)
U12_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 rules of the TRS R:

lec19_in_gg(0, s(0)) → lec19_out_gg(0, s(0))
lec19_in_gg(0, 0) → lec19_out_gg(0, 0)

Used ordering: Polynomial interpretation [POLO]:

POL(.(x1, x2)) = 1 + 2·x1 + 2·x2   
POL(0) = 0   
POL(ORDERED1_IN_G(x1)) = x1   
POL(U12_gg(x1, x2, x3)) = x1 + x2 + x3   
POL(U3_G(x1, x2, x3, x4)) = 1 + x1 + 2·x2 + 2·x3 + 2·x4   
POL(lec19_in_gg(x1, x2)) = 1 + x1 + x2   
POL(lec19_out_gg(x1, x2)) = x1 + x2   
POL(s(x1)) = 2·x1   

(18) Obligation:

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

ORDERED1_IN_G(.(s(T19), .(s(T20), T10))) → U3_G(T19, T20, T10, lec19_in_gg(T19, T20))
U3_G(T19, T20, T10, lec19_out_gg(T19, T20)) → ORDERED1_IN_G(.(s(T20), T10))

The TRS R consists of the following rules:

lec19_in_gg(s(T33), s(T34)) → U12_gg(T33, T34, lec19_in_gg(T33, T34))
U12_gg(T33, T34, lec19_out_gg(T33, T34)) → lec19_out_gg(s(T33), s(T34))

The set Q consists of the following terms:

lec19_in_gg(x0, x1)
U12_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:

U3_G(T19, T20, T10, lec19_out_gg(T19, T20)) → ORDERED1_IN_G(.(s(T20), T10))


Used ordering: Polynomial interpretation [POLO]:

POL(.(x1, x2)) = x1 + x2   
POL(ORDERED1_IN_G(x1)) = 2·x1   
POL(U12_gg(x1, x2, x3)) = x1 + 2·x2 + x3   
POL(U3_G(x1, x2, x3, x4)) = 2·x1 + 2·x2 + 2·x3 + x4   
POL(lec19_in_gg(x1, x2)) = 2·x1 + 2·x2   
POL(lec19_out_gg(x1, x2)) = 2 + x1 + 2·x2   
POL(s(x1)) = 2·x1   

(20) Obligation:

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

ORDERED1_IN_G(.(s(T19), .(s(T20), T10))) → U3_G(T19, T20, T10, lec19_in_gg(T19, T20))

The TRS R consists of the following rules:

lec19_in_gg(s(T33), s(T34)) → U12_gg(T33, T34, lec19_in_gg(T33, T34))
U12_gg(T33, T34, lec19_out_gg(T33, T34)) → lec19_out_gg(s(T33), s(T34))

The set Q consists of the following terms:

lec19_in_gg(x0, x1)
U12_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

(23) Obligation:

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

ORDERED1_IN_G(.(0, .(0, T10))) → ORDERED1_IN_G(.(0, T10))

The TRS R consists of the following rules:

lec19_in_gg(s(T33), s(T34)) → U12_gg(T33, T34, lec19_in_gg(T33, T34))
lec19_in_gg(0, s(0)) → lec19_out_gg(0, s(0))
lec19_in_gg(0, 0) → lec19_out_gg(0, 0)
U12_gg(T33, T34, lec19_out_gg(T33, T34)) → lec19_out_gg(s(T33), s(T34))

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

(24) UsableRulesProof (EQUIVALENT transformation)

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

(25) Obligation:

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

ORDERED1_IN_G(.(0, .(0, T10))) → ORDERED1_IN_G(.(0, T10))

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

(26) PiDPToQDPProof (EQUIVALENT transformation)

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

(27) Obligation:

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

ORDERED1_IN_G(.(0, .(0, T10))) → ORDERED1_IN_G(.(0, T10))

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

(28) 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:

  • ORDERED1_IN_G(.(0, .(0, T10))) → ORDERED1_IN_G(.(0, T10))
    The graph contains the following edges 1 > 1

(29) YES