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

0 Prolog
  ↳1 PrologToPiTRSProof (⇐)
    ↳2 PiTRS
    ↳3 DependencyPairsProof (⇔)
    ↳4 PiDP
    ↳5 DependencyGraphProof (⇔)
    ↳6 AND
      ↳7 PiDP
        ↳8 UsableRulesProof (⇔)
        ↳9 PiDP
        ↳10 PiDPToQDPProof (⇔)
        ↳11 QDP
        ↳12 QDPSizeChangeProof (⇔)
        ↳13 TRUE
      ↳14 PiDP
        ↳15 UsableRulesProof (⇔)
        ↳16 PiDP
        ↳17 PiDPToQDPProof (⇐)
        ↳18 QDP
  ↳19 PrologToPiTRSProof (⇐)
    ↳20 PiTRS
    ↳21 DependencyPairsProof (⇔)
    ↳22 PiDP
    ↳23 DependencyGraphProof (⇔)
    ↳24 AND
      ↳25 PiDP
        ↳26 UsableRulesProof (⇔)
        ↳27 PiDP
        ↳28 PiDPToQDPProof (⇔)
        ↳29 QDP
        ↳30 QDPSizeChangeProof (⇔)
        ↳31 TRUE
      ↳32 PiDP
        ↳33 UsableRulesProof (⇔)
        ↳34 PiDP
        ↳35 PiDPToQDPProof (⇔)
        ↳36 QDP
        ↳37 MRRProof (⇔)
        ↳38 QDP
        ↳39 DependencyGraphProof (⇔)
        ↳40 TRUE

(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) PrologToPiTRSProof (SOUND transformation)

We use the technique of [LOPSTR]. With regard to the inferred argument filtering the predicates were used in the following modes:
ordered_in: (b)
le_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(.(X, [])) → ordered_out_g(.(X, []))
ordered_in_g(.(X, .(Y, Xs))) → U1_g(X, Y, Xs, le_in_gg(X, Y))
le_in_gg(s(X), s(Y)) → U3_gg(X, Y, le_in_gg(X, Y))
le_in_gg(0, s(0)) → le_out_gg(0, s(0))
le_in_gg(0, 0) → le_out_gg(0, 0)
U3_gg(X, Y, le_out_gg(X, Y)) → le_out_gg(s(X), s(Y))
U1_g(X, Y, Xs, le_out_gg(X, 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)))

The argument filtering Pi contains the following mapping:
ordered_in_g(x1)  =  ordered_in_g(x1)
[]  =  []
ordered_out_g(x1)  =  ordered_out_g
.(x1, x2)  =  .(x1, x2)
U1_g(x1, x2, x3, x4)  =  U1_g(x2, x3, x4)
le_in_gg(x1, x2)  =  le_in_gg(x1, x2)
s(x1)  =  s(x1)
U3_gg(x1, x2, x3)  =  U3_gg(x3)
0  =  0
le_out_gg(x1, x2)  =  le_out_gg
U2_g(x1, x2, x3, x4)  =  U2_g(x4)

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(.(X, [])) → ordered_out_g(.(X, []))
ordered_in_g(.(X, .(Y, Xs))) → U1_g(X, Y, Xs, le_in_gg(X, Y))
le_in_gg(s(X), s(Y)) → U3_gg(X, Y, le_in_gg(X, Y))
le_in_gg(0, s(0)) → le_out_gg(0, s(0))
le_in_gg(0, 0) → le_out_gg(0, 0)
U3_gg(X, Y, le_out_gg(X, Y)) → le_out_gg(s(X), s(Y))
U1_g(X, Y, Xs, le_out_gg(X, 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)))

The argument filtering Pi contains the following mapping:
ordered_in_g(x1)  =  ordered_in_g(x1)
[]  =  []
ordered_out_g(x1)  =  ordered_out_g
.(x1, x2)  =  .(x1, x2)
U1_g(x1, x2, x3, x4)  =  U1_g(x2, x3, x4)
le_in_gg(x1, x2)  =  le_in_gg(x1, x2)
s(x1)  =  s(x1)
U3_gg(x1, x2, x3)  =  U3_gg(x3)
0  =  0
le_out_gg(x1, x2)  =  le_out_gg
U2_g(x1, x2, x3, x4)  =  U2_g(x4)

(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, le_in_gg(X, Y))
ORDERED_IN_G(.(X, .(Y, Xs))) → LE_IN_GG(X, Y)
LE_IN_GG(s(X), s(Y)) → U3_GG(X, Y, le_in_gg(X, Y))
LE_IN_GG(s(X), s(Y)) → LE_IN_GG(X, Y)
U1_G(X, Y, Xs, le_out_gg(X, Y)) → U2_G(X, Y, Xs, ordered_in_g(.(Y, Xs)))
U1_G(X, Y, Xs, le_out_gg(X, Y)) → ORDERED_IN_G(.(Y, Xs))

The TRS R consists of the following rules:

ordered_in_g([]) → ordered_out_g([])
ordered_in_g(.(X, [])) → ordered_out_g(.(X, []))
ordered_in_g(.(X, .(Y, Xs))) → U1_g(X, Y, Xs, le_in_gg(X, Y))
le_in_gg(s(X), s(Y)) → U3_gg(X, Y, le_in_gg(X, Y))
le_in_gg(0, s(0)) → le_out_gg(0, s(0))
le_in_gg(0, 0) → le_out_gg(0, 0)
U3_gg(X, Y, le_out_gg(X, Y)) → le_out_gg(s(X), s(Y))
U1_g(X, Y, Xs, le_out_gg(X, 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)))

The argument filtering Pi contains the following mapping:
ordered_in_g(x1)  =  ordered_in_g(x1)
[]  =  []
ordered_out_g(x1)  =  ordered_out_g
.(x1, x2)  =  .(x1, x2)
U1_g(x1, x2, x3, x4)  =  U1_g(x2, x3, x4)
le_in_gg(x1, x2)  =  le_in_gg(x1, x2)
s(x1)  =  s(x1)
U3_gg(x1, x2, x3)  =  U3_gg(x3)
0  =  0
le_out_gg(x1, x2)  =  le_out_gg
U2_g(x1, x2, x3, x4)  =  U2_g(x4)
ORDERED_IN_G(x1)  =  ORDERED_IN_G(x1)
U1_G(x1, x2, x3, x4)  =  U1_G(x2, x3, x4)
LE_IN_GG(x1, x2)  =  LE_IN_GG(x1, x2)
U3_GG(x1, x2, x3)  =  U3_GG(x3)
U2_G(x1, x2, x3, x4)  =  U2_G(x4)

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, le_in_gg(X, Y))
ORDERED_IN_G(.(X, .(Y, Xs))) → LE_IN_GG(X, Y)
LE_IN_GG(s(X), s(Y)) → U3_GG(X, Y, le_in_gg(X, Y))
LE_IN_GG(s(X), s(Y)) → LE_IN_GG(X, Y)
U1_G(X, Y, Xs, le_out_gg(X, Y)) → U2_G(X, Y, Xs, ordered_in_g(.(Y, Xs)))
U1_G(X, Y, Xs, le_out_gg(X, Y)) → ORDERED_IN_G(.(Y, Xs))

The TRS R consists of the following rules:

ordered_in_g([]) → ordered_out_g([])
ordered_in_g(.(X, [])) → ordered_out_g(.(X, []))
ordered_in_g(.(X, .(Y, Xs))) → U1_g(X, Y, Xs, le_in_gg(X, Y))
le_in_gg(s(X), s(Y)) → U3_gg(X, Y, le_in_gg(X, Y))
le_in_gg(0, s(0)) → le_out_gg(0, s(0))
le_in_gg(0, 0) → le_out_gg(0, 0)
U3_gg(X, Y, le_out_gg(X, Y)) → le_out_gg(s(X), s(Y))
U1_g(X, Y, Xs, le_out_gg(X, 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)))

The argument filtering Pi contains the following mapping:
ordered_in_g(x1)  =  ordered_in_g(x1)
[]  =  []
ordered_out_g(x1)  =  ordered_out_g
.(x1, x2)  =  .(x1, x2)
U1_g(x1, x2, x3, x4)  =  U1_g(x2, x3, x4)
le_in_gg(x1, x2)  =  le_in_gg(x1, x2)
s(x1)  =  s(x1)
U3_gg(x1, x2, x3)  =  U3_gg(x3)
0  =  0
le_out_gg(x1, x2)  =  le_out_gg
U2_g(x1, x2, x3, x4)  =  U2_g(x4)
ORDERED_IN_G(x1)  =  ORDERED_IN_G(x1)
U1_G(x1, x2, x3, x4)  =  U1_G(x2, x3, x4)
LE_IN_GG(x1, x2)  =  LE_IN_GG(x1, x2)
U3_GG(x1, x2, x3)  =  U3_GG(x3)
U2_G(x1, x2, x3, x4)  =  U2_G(x4)

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:

LE_IN_GG(s(X), s(Y)) → LE_IN_GG(X, Y)

The TRS R consists of the following rules:

ordered_in_g([]) → ordered_out_g([])
ordered_in_g(.(X, [])) → ordered_out_g(.(X, []))
ordered_in_g(.(X, .(Y, Xs))) → U1_g(X, Y, Xs, le_in_gg(X, Y))
le_in_gg(s(X), s(Y)) → U3_gg(X, Y, le_in_gg(X, Y))
le_in_gg(0, s(0)) → le_out_gg(0, s(0))
le_in_gg(0, 0) → le_out_gg(0, 0)
U3_gg(X, Y, le_out_gg(X, Y)) → le_out_gg(s(X), s(Y))
U1_g(X, Y, Xs, le_out_gg(X, 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)))

The argument filtering Pi contains the following mapping:
ordered_in_g(x1)  =  ordered_in_g(x1)
[]  =  []
ordered_out_g(x1)  =  ordered_out_g
.(x1, x2)  =  .(x1, x2)
U1_g(x1, x2, x3, x4)  =  U1_g(x2, x3, x4)
le_in_gg(x1, x2)  =  le_in_gg(x1, x2)
s(x1)  =  s(x1)
U3_gg(x1, x2, x3)  =  U3_gg(x3)
0  =  0
le_out_gg(x1, x2)  =  le_out_gg
U2_g(x1, x2, x3, x4)  =  U2_g(x4)
LE_IN_GG(x1, x2)  =  LE_IN_GG(x1, x2)

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:

LE_IN_GG(s(X), s(Y)) → LE_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:

LE_IN_GG(s(X), s(Y)) → LE_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:

  • LE_IN_GG(s(X), s(Y)) → LE_IN_GG(X, Y)
    The graph contains the following edges 1 > 1, 2 > 2

(13) TRUE

(14) Obligation:

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

U1_G(X, Y, Xs, le_out_gg(X, Y)) → ORDERED_IN_G(.(Y, Xs))
ORDERED_IN_G(.(X, .(Y, Xs))) → U1_G(X, Y, Xs, le_in_gg(X, Y))

The TRS R consists of the following rules:

ordered_in_g([]) → ordered_out_g([])
ordered_in_g(.(X, [])) → ordered_out_g(.(X, []))
ordered_in_g(.(X, .(Y, Xs))) → U1_g(X, Y, Xs, le_in_gg(X, Y))
le_in_gg(s(X), s(Y)) → U3_gg(X, Y, le_in_gg(X, Y))
le_in_gg(0, s(0)) → le_out_gg(0, s(0))
le_in_gg(0, 0) → le_out_gg(0, 0)
U3_gg(X, Y, le_out_gg(X, Y)) → le_out_gg(s(X), s(Y))
U1_g(X, Y, Xs, le_out_gg(X, 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)))

The argument filtering Pi contains the following mapping:
ordered_in_g(x1)  =  ordered_in_g(x1)
[]  =  []
ordered_out_g(x1)  =  ordered_out_g
.(x1, x2)  =  .(x1, x2)
U1_g(x1, x2, x3, x4)  =  U1_g(x2, x3, x4)
le_in_gg(x1, x2)  =  le_in_gg(x1, x2)
s(x1)  =  s(x1)
U3_gg(x1, x2, x3)  =  U3_gg(x3)
0  =  0
le_out_gg(x1, x2)  =  le_out_gg
U2_g(x1, x2, x3, x4)  =  U2_g(x4)
ORDERED_IN_G(x1)  =  ORDERED_IN_G(x1)
U1_G(x1, x2, x3, x4)  =  U1_G(x2, x3, x4)

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, le_out_gg(X, Y)) → ORDERED_IN_G(.(Y, Xs))
ORDERED_IN_G(.(X, .(Y, Xs))) → U1_G(X, Y, Xs, le_in_gg(X, Y))

The TRS R consists of the following rules:

le_in_gg(s(X), s(Y)) → U3_gg(X, Y, le_in_gg(X, Y))
le_in_gg(0, s(0)) → le_out_gg(0, s(0))
le_in_gg(0, 0) → le_out_gg(0, 0)
U3_gg(X, Y, le_out_gg(X, Y)) → le_out_gg(s(X), s(Y))

The argument filtering Pi contains the following mapping:
.(x1, x2)  =  .(x1, x2)
le_in_gg(x1, x2)  =  le_in_gg(x1, x2)
s(x1)  =  s(x1)
U3_gg(x1, x2, x3)  =  U3_gg(x3)
0  =  0
le_out_gg(x1, x2)  =  le_out_gg
ORDERED_IN_G(x1)  =  ORDERED_IN_G(x1)
U1_G(x1, x2, x3, x4)  =  U1_G(x2, x3, x4)

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

(17) PiDPToQDPProof (SOUND 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(Y, Xs, le_out_gg) → ORDERED_IN_G(.(Y, Xs))
ORDERED_IN_G(.(X, .(Y, Xs))) → U1_G(Y, Xs, le_in_gg(X, Y))

The TRS R consists of the following rules:

le_in_gg(s(X), s(Y)) → U3_gg(le_in_gg(X, Y))
le_in_gg(0, s(0)) → le_out_gg
le_in_gg(0, 0) → le_out_gg
U3_gg(le_out_gg) → le_out_gg

The set Q consists of the following terms:

le_in_gg(x0, x1)
U3_gg(x0)

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

(19) PrologToPiTRSProof (SOUND transformation)

We use the technique of [LOPSTR]. With regard to the inferred argument filtering the predicates were used in the following modes:
ordered_in: (b)
le_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(.(X, [])) → ordered_out_g(.(X, []))
ordered_in_g(.(X, .(Y, Xs))) → U1_g(X, Y, Xs, le_in_gg(X, Y))
le_in_gg(s(X), s(Y)) → U3_gg(X, Y, le_in_gg(X, Y))
le_in_gg(0, s(0)) → le_out_gg(0, s(0))
le_in_gg(0, 0) → le_out_gg(0, 0)
U3_gg(X, Y, le_out_gg(X, Y)) → le_out_gg(s(X), s(Y))
U1_g(X, Y, Xs, le_out_gg(X, 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

(20) Obligation:

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

ordered_in_g([]) → ordered_out_g([])
ordered_in_g(.(X, [])) → ordered_out_g(.(X, []))
ordered_in_g(.(X, .(Y, Xs))) → U1_g(X, Y, Xs, le_in_gg(X, Y))
le_in_gg(s(X), s(Y)) → U3_gg(X, Y, le_in_gg(X, Y))
le_in_gg(0, s(0)) → le_out_gg(0, s(0))
le_in_gg(0, 0) → le_out_gg(0, 0)
U3_gg(X, Y, le_out_gg(X, Y)) → le_out_gg(s(X), s(Y))
U1_g(X, Y, Xs, le_out_gg(X, 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.

(21) 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, le_in_gg(X, Y))
ORDERED_IN_G(.(X, .(Y, Xs))) → LE_IN_GG(X, Y)
LE_IN_GG(s(X), s(Y)) → U3_GG(X, Y, le_in_gg(X, Y))
LE_IN_GG(s(X), s(Y)) → LE_IN_GG(X, Y)
U1_G(X, Y, Xs, le_out_gg(X, Y)) → U2_G(X, Y, Xs, ordered_in_g(.(Y, Xs)))
U1_G(X, Y, Xs, le_out_gg(X, Y)) → ORDERED_IN_G(.(Y, Xs))

The TRS R consists of the following rules:

ordered_in_g([]) → ordered_out_g([])
ordered_in_g(.(X, [])) → ordered_out_g(.(X, []))
ordered_in_g(.(X, .(Y, Xs))) → U1_g(X, Y, Xs, le_in_gg(X, Y))
le_in_gg(s(X), s(Y)) → U3_gg(X, Y, le_in_gg(X, Y))
le_in_gg(0, s(0)) → le_out_gg(0, s(0))
le_in_gg(0, 0) → le_out_gg(0, 0)
U3_gg(X, Y, le_out_gg(X, Y)) → le_out_gg(s(X), s(Y))
U1_g(X, Y, Xs, le_out_gg(X, 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

(22) Obligation:

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

ORDERED_IN_G(.(X, .(Y, Xs))) → U1_G(X, Y, Xs, le_in_gg(X, Y))
ORDERED_IN_G(.(X, .(Y, Xs))) → LE_IN_GG(X, Y)
LE_IN_GG(s(X), s(Y)) → U3_GG(X, Y, le_in_gg(X, Y))
LE_IN_GG(s(X), s(Y)) → LE_IN_GG(X, Y)
U1_G(X, Y, Xs, le_out_gg(X, Y)) → U2_G(X, Y, Xs, ordered_in_g(.(Y, Xs)))
U1_G(X, Y, Xs, le_out_gg(X, Y)) → ORDERED_IN_G(.(Y, Xs))

The TRS R consists of the following rules:

ordered_in_g([]) → ordered_out_g([])
ordered_in_g(.(X, [])) → ordered_out_g(.(X, []))
ordered_in_g(.(X, .(Y, Xs))) → U1_g(X, Y, Xs, le_in_gg(X, Y))
le_in_gg(s(X), s(Y)) → U3_gg(X, Y, le_in_gg(X, Y))
le_in_gg(0, s(0)) → le_out_gg(0, s(0))
le_in_gg(0, 0) → le_out_gg(0, 0)
U3_gg(X, Y, le_out_gg(X, Y)) → le_out_gg(s(X), s(Y))
U1_g(X, Y, Xs, le_out_gg(X, 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

(23) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LOPSTR] contains 2 SCCs with 3 less nodes.

(24) Complex Obligation (AND)

(25) Obligation:

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

LE_IN_GG(s(X), s(Y)) → LE_IN_GG(X, Y)

The TRS R consists of the following rules:

ordered_in_g([]) → ordered_out_g([])
ordered_in_g(.(X, [])) → ordered_out_g(.(X, []))
ordered_in_g(.(X, .(Y, Xs))) → U1_g(X, Y, Xs, le_in_gg(X, Y))
le_in_gg(s(X), s(Y)) → U3_gg(X, Y, le_in_gg(X, Y))
le_in_gg(0, s(0)) → le_out_gg(0, s(0))
le_in_gg(0, 0) → le_out_gg(0, 0)
U3_gg(X, Y, le_out_gg(X, Y)) → le_out_gg(s(X), s(Y))
U1_g(X, Y, Xs, le_out_gg(X, 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

(26) UsableRulesProof (EQUIVALENT transformation)

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

(27) Obligation:

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

LE_IN_GG(s(X), s(Y)) → LE_IN_GG(X, Y)

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

(28) PiDPToQDPProof (EQUIVALENT transformation)

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

(29) Obligation:

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

LE_IN_GG(s(X), s(Y)) → LE_IN_GG(X, Y)

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

(30) 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:

  • LE_IN_GG(s(X), s(Y)) → LE_IN_GG(X, Y)
    The graph contains the following edges 1 > 1, 2 > 2

(31) TRUE

(32) Obligation:

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

U1_G(X, Y, Xs, le_out_gg(X, Y)) → ORDERED_IN_G(.(Y, Xs))
ORDERED_IN_G(.(X, .(Y, Xs))) → U1_G(X, Y, Xs, le_in_gg(X, Y))

The TRS R consists of the following rules:

ordered_in_g([]) → ordered_out_g([])
ordered_in_g(.(X, [])) → ordered_out_g(.(X, []))
ordered_in_g(.(X, .(Y, Xs))) → U1_g(X, Y, Xs, le_in_gg(X, Y))
le_in_gg(s(X), s(Y)) → U3_gg(X, Y, le_in_gg(X, Y))
le_in_gg(0, s(0)) → le_out_gg(0, s(0))
le_in_gg(0, 0) → le_out_gg(0, 0)
U3_gg(X, Y, le_out_gg(X, Y)) → le_out_gg(s(X), s(Y))
U1_g(X, Y, Xs, le_out_gg(X, 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

(33) UsableRulesProof (EQUIVALENT transformation)

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

(34) Obligation:

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

U1_G(X, Y, Xs, le_out_gg(X, Y)) → ORDERED_IN_G(.(Y, Xs))
ORDERED_IN_G(.(X, .(Y, Xs))) → U1_G(X, Y, Xs, le_in_gg(X, Y))

The TRS R consists of the following rules:

le_in_gg(s(X), s(Y)) → U3_gg(X, Y, le_in_gg(X, Y))
le_in_gg(0, s(0)) → le_out_gg(0, s(0))
le_in_gg(0, 0) → le_out_gg(0, 0)
U3_gg(X, Y, le_out_gg(X, Y)) → le_out_gg(s(X), s(Y))

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

(35) PiDPToQDPProof (EQUIVALENT transformation)

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

(36) Obligation:

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

U1_G(X, Y, Xs, le_out_gg(X, Y)) → ORDERED_IN_G(.(Y, Xs))
ORDERED_IN_G(.(X, .(Y, Xs))) → U1_G(X, Y, Xs, le_in_gg(X, Y))

The TRS R consists of the following rules:

le_in_gg(s(X), s(Y)) → U3_gg(X, Y, le_in_gg(X, Y))
le_in_gg(0, s(0)) → le_out_gg(0, s(0))
le_in_gg(0, 0) → le_out_gg(0, 0)
U3_gg(X, Y, le_out_gg(X, Y)) → le_out_gg(s(X), s(Y))

The set Q consists of the following terms:

le_in_gg(x0, x1)
U3_gg(x0, x1, x2)

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

(37) 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, le_in_gg(X, 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)) = x1 + 2·x2 + x3   
POL(le_in_gg(x1, x2)) = 2·x1 + 2·x2   
POL(le_out_gg(x1, x2)) = x1 + 2·x2   
POL(s(x1)) = 2·x1   

(38) Obligation:

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

U1_G(X, Y, Xs, le_out_gg(X, Y)) → ORDERED_IN_G(.(Y, Xs))

The TRS R consists of the following rules:

le_in_gg(s(X), s(Y)) → U3_gg(X, Y, le_in_gg(X, Y))
le_in_gg(0, s(0)) → le_out_gg(0, s(0))
le_in_gg(0, 0) → le_out_gg(0, 0)
U3_gg(X, Y, le_out_gg(X, Y)) → le_out_gg(s(X), s(Y))

The set Q consists of the following terms:

le_in_gg(x0, x1)
U3_gg(x0, x1, x2)

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

(39) DependencyGraphProof (EQUIVALENT transformation)

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

(40) TRUE