Left Termination proof of /tmp/tmpMWjrzm/qsort.pl

Left Termination of the query pattern qs_in_2(g, a) w.r.t. the given Prolog program could successfully be proven:

↳ Prolog

  ↳ PrologToPiTRSProof

Clauses:

qs([], []).
qs(cons(X, L), S) :- ','(split(L, X, L1, L2), ','(qs(L1, S1), ','(qs(L2, S2), append(S1, cons(X, S2), S)))).
append([], L, L).
append(cons(X, L1), L2, cons(X, L3)) :- append(L1, L2, L3).
split([], X, [], []).
split(cons(X, L), Y, cons(X, L1), L2) :- ','(less(X, Y), split(L, Y, L1, L2)).
split(cons(X, L), Y, L1, cons(X, L2)) :- ','(geq(X, Y), split(L, Y, L1, L2)).
less(0, s(X)).
less(s(X), s(Y)) :- less(X, Y).
geq(X, X).
geq(s(X), 0).
geq(s(X), s(Y)) :- geq(X, Y).

Queries:

qs(g,a).

We use the technique of [30].Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

qs_in(cons(X, L), S) → U1(X, L, S, split_in(L, X, L1, L2))
split_in(cons(X, L), Y, L1, cons(X, L2)) → U8(X, L, Y, L1, L2, geq_in(X, Y))
geq_in(s(X), s(Y)) → U11(X, Y, geq_in(X, Y))
geq_in(s(X), 0) → geq_out(s(X), 0)
geq_in(X, X) → geq_out(X, X)
U11(X, Y, geq_out(X, Y)) → geq_out(s(X), s(Y))
U8(X, L, Y, L1, L2, geq_out(X, Y)) → U9(X, L, Y, L1, L2, split_in(L, Y, L1, L2))
split_in(cons(X, L), Y, cons(X, L1), L2) → U6(X, L, Y, L1, L2, less_in(X, Y))
less_in(s(X), s(Y)) → U10(X, Y, less_in(X, Y))
less_in(0, s(X)) → less_out(0, s(X))
U10(X, Y, less_out(X, Y)) → less_out(s(X), s(Y))
U6(X, L, Y, L1, L2, less_out(X, Y)) → U7(X, L, Y, L1, L2, split_in(L, Y, L1, L2))
split_in([], X, [], []) → split_out([], X, [], [])
U7(X, L, Y, L1, L2, split_out(L, Y, L1, L2)) → split_out(cons(X, L), Y, cons(X, L1), L2)
U9(X, L, Y, L1, L2, split_out(L, Y, L1, L2)) → split_out(cons(X, L), Y, L1, cons(X, L2))
U1(X, L, S, split_out(L, X, L1, L2)) → U2(X, L, S, L2, qs_in(L1, S1))
qs_in([], []) → qs_out([], [])
U2(X, L, S, L2, qs_out(L1, S1)) → U3(X, L, S, S1, qs_in(L2, S2))
U3(X, L, S, S1, qs_out(L2, S2)) → U4(X, L, S, append_in(S1, cons(X, S2), S))
append_in(cons(X, L1), L2, cons(X, L3)) → U5(X, L1, L2, L3, append_in(L1, L2, L3))
append_in([], L, L) → append_out([], L, L)
U5(X, L1, L2, L3, append_out(L1, L2, L3)) → append_out(cons(X, L1), L2, cons(X, L3))
U4(X, L, S, append_out(S1, cons(X, S2), S)) → qs_out(cons(X, L), S)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

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

qs_in(cons(X, L), S) → U1(X, L, S, split_in(L, X, L1, L2))
split_in(cons(X, L), Y, L1, cons(X, L2)) → U8(X, L, Y, L1, L2, geq_in(X, Y))
geq_in(s(X), s(Y)) → U11(X, Y, geq_in(X, Y))
geq_in(s(X), 0) → geq_out(s(X), 0)
geq_in(X, X) → geq_out(X, X)
U11(X, Y, geq_out(X, Y)) → geq_out(s(X), s(Y))
U8(X, L, Y, L1, L2, geq_out(X, Y)) → U9(X, L, Y, L1, L2, split_in(L, Y, L1, L2))
split_in(cons(X, L), Y, cons(X, L1), L2) → U6(X, L, Y, L1, L2, less_in(X, Y))
less_in(s(X), s(Y)) → U10(X, Y, less_in(X, Y))
less_in(0, s(X)) → less_out(0, s(X))
U10(X, Y, less_out(X, Y)) → less_out(s(X), s(Y))
U6(X, L, Y, L1, L2, less_out(X, Y)) → U7(X, L, Y, L1, L2, split_in(L, Y, L1, L2))
split_in([], X, [], []) → split_out([], X, [], [])
U7(X, L, Y, L1, L2, split_out(L, Y, L1, L2)) → split_out(cons(X, L), Y, cons(X, L1), L2)
U9(X, L, Y, L1, L2, split_out(L, Y, L1, L2)) → split_out(cons(X, L), Y, L1, cons(X, L2))
U1(X, L, S, split_out(L, X, L1, L2)) → U2(X, L, S, L2, qs_in(L1, S1))
qs_in([], []) → qs_out([], [])
U2(X, L, S, L2, qs_out(L1, S1)) → U3(X, L, S, S1, qs_in(L2, S2))
U3(X, L, S, S1, qs_out(L2, S2)) → U4(X, L, S, append_in(S1, cons(X, S2), S))
append_in(cons(X, L1), L2, cons(X, L3)) → U5(X, L1, L2, L3, append_in(L1, L2, L3))
append_in([], L, L) → append_out([], L, L)
U5(X, L1, L2, L3, append_out(L1, L2, L3)) → append_out(cons(X, L1), L2, cons(X, L3))
U4(X, L, S, append_out(S1, cons(X, S2), S)) → qs_out(cons(X, L), S)

QS_IN(cons(X, L), S) → U1¹(X, L, S, split_in(L, X, L1, L2))
QS_IN(cons(X, L), S) → SPLIT_IN(L, X, L1, L2)
SPLIT_IN(cons(X, L), Y, L1, cons(X, L2)) → U8¹(X, L, Y, L1, L2, geq_in(X, Y))
SPLIT_IN(cons(X, L), Y, L1, cons(X, L2)) → GEQ_IN(X, Y)
GEQ_IN(s(X), s(Y)) → U11¹(X, Y, geq_in(X, Y))
GEQ_IN(s(X), s(Y)) → GEQ_IN(X, Y)
U8¹(X, L, Y, L1, L2, geq_out(X, Y)) → U9¹(X, L, Y, L1, L2, split_in(L, Y, L1, L2))
U8¹(X, L, Y, L1, L2, geq_out(X, Y)) → SPLIT_IN(L, Y, L1, L2)
SPLIT_IN(cons(X, L), Y, cons(X, L1), L2) → U6¹(X, L, Y, L1, L2, less_in(X, Y))
SPLIT_IN(cons(X, L), Y, cons(X, L1), L2) → LESS_IN(X, Y)
LESS_IN(s(X), s(Y)) → U10¹(X, Y, less_in(X, Y))
LESS_IN(s(X), s(Y)) → LESS_IN(X, Y)
U6¹(X, L, Y, L1, L2, less_out(X, Y)) → U7¹(X, L, Y, L1, L2, split_in(L, Y, L1, L2))
U6¹(X, L, Y, L1, L2, less_out(X, Y)) → SPLIT_IN(L, Y, L1, L2)
U1¹(X, L, S, split_out(L, X, L1, L2)) → U2¹(X, L, S, L2, qs_in(L1, S1))
U1¹(X, L, S, split_out(L, X, L1, L2)) → QS_IN(L1, S1)
U2¹(X, L, S, L2, qs_out(L1, S1)) → U3¹(X, L, S, S1, qs_in(L2, S2))
U2¹(X, L, S, L2, qs_out(L1, S1)) → QS_IN(L2, S2)
U3¹(X, L, S, S1, qs_out(L2, S2)) → U4¹(X, L, S, append_in(S1, cons(X, S2), S))
U3¹(X, L, S, S1, qs_out(L2, S2)) → APPEND_IN(S1, cons(X, S2), S)
APPEND_IN(cons(X, L1), L2, cons(X, L3)) → U5¹(X, L1, L2, L3, append_in(L1, L2, L3))
APPEND_IN(cons(X, L1), L2, cons(X, L3)) → APPEND_IN(L1, L2, L3)

The TRS R consists of the following rules:

qs_in(cons(X, L), S) → U1(X, L, S, split_in(L, X, L1, L2))
split_in(cons(X, L), Y, L1, cons(X, L2)) → U8(X, L, Y, L1, L2, geq_in(X, Y))
geq_in(s(X), s(Y)) → U11(X, Y, geq_in(X, Y))
geq_in(s(X), 0) → geq_out(s(X), 0)
geq_in(X, X) → geq_out(X, X)
U11(X, Y, geq_out(X, Y)) → geq_out(s(X), s(Y))
U8(X, L, Y, L1, L2, geq_out(X, Y)) → U9(X, L, Y, L1, L2, split_in(L, Y, L1, L2))
split_in(cons(X, L), Y, cons(X, L1), L2) → U6(X, L, Y, L1, L2, less_in(X, Y))
less_in(s(X), s(Y)) → U10(X, Y, less_in(X, Y))
less_in(0, s(X)) → less_out(0, s(X))
U10(X, Y, less_out(X, Y)) → less_out(s(X), s(Y))
U6(X, L, Y, L1, L2, less_out(X, Y)) → U7(X, L, Y, L1, L2, split_in(L, Y, L1, L2))
split_in([], X, [], []) → split_out([], X, [], [])
U7(X, L, Y, L1, L2, split_out(L, Y, L1, L2)) → split_out(cons(X, L), Y, cons(X, L1), L2)
U9(X, L, Y, L1, L2, split_out(L, Y, L1, L2)) → split_out(cons(X, L), Y, L1, cons(X, L2))
U1(X, L, S, split_out(L, X, L1, L2)) → U2(X, L, S, L2, qs_in(L1, S1))
qs_in([], []) → qs_out([], [])
U2(X, L, S, L2, qs_out(L1, S1)) → U3(X, L, S, S1, qs_in(L2, S2))
U3(X, L, S, S1, qs_out(L2, S2)) → U4(X, L, S, append_in(S1, cons(X, S2), S))
append_in(cons(X, L1), L2, cons(X, L3)) → U5(X, L1, L2, L3, append_in(L1, L2, L3))
append_in([], L, L) → append_out([], L, L)
U5(X, L1, L2, L3, append_out(L1, L2, L3)) → append_out(cons(X, L1), L2, cons(X, L3))
U4(X, L, S, append_out(S1, cons(X, S2), S)) → qs_out(cons(X, L), S)

The argument filtering Pi contains the following mapping:
qs_in(x1, x2) = qs_in(x1)
cons(x1, x2) = cons(x1, x2)
U1(x1, x2, x3, x4) = U1(x1, x4)
split_in(x1, x2, x3, x4) = split_in(x1, x2)
U8(x1, x2, x3, x4, x5, x6) = U8(x1, x2, x3, x6)
geq_in(x1, x2) = geq_in(x1, x2)
s(x1) = s(x1)
U11(x1, x2, x3) = U11(x3)
0 = 0
geq_out(x1, x2) = geq_out
U9(x1, x2, x3, x4, x5, x6) = U9(x1, x6)
U6(x1, x2, x3, x4, x5, x6) = U6(x1, x2, x3, x6)
less_in(x1, x2) = less_in(x1, x2)
U10(x1, x2, x3) = U10(x3)
less_out(x1, x2) = less_out
U7(x1, x2, x3, x4, x5, x6) = U7(x1, x6)
[] = []
split_out(x1, x2, x3, x4) = split_out(x3, x4)
U2(x1, x2, x3, x4, x5) = U2(x1, x4, x5)
qs_out(x1, x2) = qs_out(x2)
U3(x1, x2, x3, x4, x5) = U3(x1, x4, x5)
U4(x1, x2, x3, x4) = U4(x4)
append_in(x1, x2, x3) = append_in(x1, x2)
U5(x1, x2, x3, x4, x5) = U5(x1, x5)
append_out(x1, x2, x3) = append_out(x3)
U11¹(x1, x2, x3) = U11¹(x3)
U4¹(x1, x2, x3, x4) = U4¹(x4)
APPEND_IN(x1, x2, x3) = APPEND_IN(x1, x2)
U3¹(x1, x2, x3, x4, x5) = U3¹(x1, x4, x5)
U8¹(x1, x2, x3, x4, x5, x6) = U8¹(x1, x2, x3, x6)
QS_IN(x1, x2) = QS_IN(x1)
U7¹(x1, x2, x3, x4, x5, x6) = U7¹(x1, x6)
LESS_IN(x1, x2) = LESS_IN(x1, x2)
SPLIT_IN(x1, x2, x3, x4) = SPLIT_IN(x1, x2)
U5¹(x1, x2, x3, x4, x5) = U5¹(x1, x5)
U10¹(x1, x2, x3) = U10¹(x3)
GEQ_IN(x1, x2) = GEQ_IN(x1, x2)
U9¹(x1, x2, x3, x4, x5, x6) = U9¹(x1, x6)
U2¹(x1, x2, x3, x4, x5) = U2¹(x1, x4, x5)
U1¹(x1, x2, x3, x4) = U1¹(x1, x4)
U6¹(x1, x2, x3, x4, x5, x6) = U6¹(x1, x2, x3, x6)

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

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

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

QS_IN(cons(X, L), S) → U1¹(X, L, S, split_in(L, X, L1, L2))
QS_IN(cons(X, L), S) → SPLIT_IN(L, X, L1, L2)
SPLIT_IN(cons(X, L), Y, L1, cons(X, L2)) → U8¹(X, L, Y, L1, L2, geq_in(X, Y))
SPLIT_IN(cons(X, L), Y, L1, cons(X, L2)) → GEQ_IN(X, Y)
GEQ_IN(s(X), s(Y)) → U11¹(X, Y, geq_in(X, Y))
GEQ_IN(s(X), s(Y)) → GEQ_IN(X, Y)
U8¹(X, L, Y, L1, L2, geq_out(X, Y)) → U9¹(X, L, Y, L1, L2, split_in(L, Y, L1, L2))
U8¹(X, L, Y, L1, L2, geq_out(X, Y)) → SPLIT_IN(L, Y, L1, L2)
SPLIT_IN(cons(X, L), Y, cons(X, L1), L2) → U6¹(X, L, Y, L1, L2, less_in(X, Y))
SPLIT_IN(cons(X, L), Y, cons(X, L1), L2) → LESS_IN(X, Y)
LESS_IN(s(X), s(Y)) → U10¹(X, Y, less_in(X, Y))
LESS_IN(s(X), s(Y)) → LESS_IN(X, Y)
U6¹(X, L, Y, L1, L2, less_out(X, Y)) → U7¹(X, L, Y, L1, L2, split_in(L, Y, L1, L2))
U6¹(X, L, Y, L1, L2, less_out(X, Y)) → SPLIT_IN(L, Y, L1, L2)
U1¹(X, L, S, split_out(L, X, L1, L2)) → U2¹(X, L, S, L2, qs_in(L1, S1))
U1¹(X, L, S, split_out(L, X, L1, L2)) → QS_IN(L1, S1)
U2¹(X, L, S, L2, qs_out(L1, S1)) → U3¹(X, L, S, S1, qs_in(L2, S2))
U2¹(X, L, S, L2, qs_out(L1, S1)) → QS_IN(L2, S2)
U3¹(X, L, S, S1, qs_out(L2, S2)) → U4¹(X, L, S, append_in(S1, cons(X, S2), S))
U3¹(X, L, S, S1, qs_out(L2, S2)) → APPEND_IN(S1, cons(X, S2), S)
APPEND_IN(cons(X, L1), L2, cons(X, L3)) → U5¹(X, L1, L2, L3, append_in(L1, L2, L3))
APPEND_IN(cons(X, L1), L2, cons(X, L3)) → APPEND_IN(L1, L2, L3)

The TRS R consists of the following rules:

qs_in(cons(X, L), S) → U1(X, L, S, split_in(L, X, L1, L2))
split_in(cons(X, L), Y, L1, cons(X, L2)) → U8(X, L, Y, L1, L2, geq_in(X, Y))
geq_in(s(X), s(Y)) → U11(X, Y, geq_in(X, Y))
geq_in(s(X), 0) → geq_out(s(X), 0)
geq_in(X, X) → geq_out(X, X)
U11(X, Y, geq_out(X, Y)) → geq_out(s(X), s(Y))
U8(X, L, Y, L1, L2, geq_out(X, Y)) → U9(X, L, Y, L1, L2, split_in(L, Y, L1, L2))
split_in(cons(X, L), Y, cons(X, L1), L2) → U6(X, L, Y, L1, L2, less_in(X, Y))
less_in(s(X), s(Y)) → U10(X, Y, less_in(X, Y))
less_in(0, s(X)) → less_out(0, s(X))
U10(X, Y, less_out(X, Y)) → less_out(s(X), s(Y))
U6(X, L, Y, L1, L2, less_out(X, Y)) → U7(X, L, Y, L1, L2, split_in(L, Y, L1, L2))
split_in([], X, [], []) → split_out([], X, [], [])
U7(X, L, Y, L1, L2, split_out(L, Y, L1, L2)) → split_out(cons(X, L), Y, cons(X, L1), L2)
U9(X, L, Y, L1, L2, split_out(L, Y, L1, L2)) → split_out(cons(X, L), Y, L1, cons(X, L2))
U1(X, L, S, split_out(L, X, L1, L2)) → U2(X, L, S, L2, qs_in(L1, S1))
qs_in([], []) → qs_out([], [])
U2(X, L, S, L2, qs_out(L1, S1)) → U3(X, L, S, S1, qs_in(L2, S2))
U3(X, L, S, S1, qs_out(L2, S2)) → U4(X, L, S, append_in(S1, cons(X, S2), S))
append_in(cons(X, L1), L2, cons(X, L3)) → U5(X, L1, L2, L3, append_in(L1, L2, L3))
append_in([], L, L) → append_out([], L, L)
U5(X, L1, L2, L3, append_out(L1, L2, L3)) → append_out(cons(X, L1), L2, cons(X, L3))
U4(X, L, S, append_out(S1, cons(X, S2), S)) → qs_out(cons(X, L), S)

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

                ↳ UsableRulesProof

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

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

APPEND_IN(cons(X, L1), L2, cons(X, L3)) → APPEND_IN(L1, L2, L3)

The TRS R consists of the following rules:

qs_in(cons(X, L), S) → U1(X, L, S, split_in(L, X, L1, L2))
split_in(cons(X, L), Y, L1, cons(X, L2)) → U8(X, L, Y, L1, L2, geq_in(X, Y))
geq_in(s(X), s(Y)) → U11(X, Y, geq_in(X, Y))
geq_in(s(X), 0) → geq_out(s(X), 0)
geq_in(X, X) → geq_out(X, X)
U11(X, Y, geq_out(X, Y)) → geq_out(s(X), s(Y))
U8(X, L, Y, L1, L2, geq_out(X, Y)) → U9(X, L, Y, L1, L2, split_in(L, Y, L1, L2))
split_in(cons(X, L), Y, cons(X, L1), L2) → U6(X, L, Y, L1, L2, less_in(X, Y))
less_in(s(X), s(Y)) → U10(X, Y, less_in(X, Y))
less_in(0, s(X)) → less_out(0, s(X))
U10(X, Y, less_out(X, Y)) → less_out(s(X), s(Y))
U6(X, L, Y, L1, L2, less_out(X, Y)) → U7(X, L, Y, L1, L2, split_in(L, Y, L1, L2))
split_in([], X, [], []) → split_out([], X, [], [])
U7(X, L, Y, L1, L2, split_out(L, Y, L1, L2)) → split_out(cons(X, L), Y, cons(X, L1), L2)
U9(X, L, Y, L1, L2, split_out(L, Y, L1, L2)) → split_out(cons(X, L), Y, L1, cons(X, L2))
U1(X, L, S, split_out(L, X, L1, L2)) → U2(X, L, S, L2, qs_in(L1, S1))
qs_in([], []) → qs_out([], [])
U2(X, L, S, L2, qs_out(L1, S1)) → U3(X, L, S, S1, qs_in(L2, S2))
U3(X, L, S, S1, qs_out(L2, S2)) → U4(X, L, S, append_in(S1, cons(X, S2), S))
append_in(cons(X, L1), L2, cons(X, L3)) → U5(X, L1, L2, L3, append_in(L1, L2, L3))
append_in([], L, L) → append_out([], L, L)
U5(X, L1, L2, L3, append_out(L1, L2, L3)) → append_out(cons(X, L1), L2, cons(X, L3))
U4(X, L, S, append_out(S1, cons(X, S2), S)) → qs_out(cons(X, L), S)

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

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

APPEND_IN(cons(X, L1), L2, cons(X, L3)) → APPEND_IN(L1, L2, L3)

R is empty.
The argument filtering Pi contains the following mapping:
cons(x1, x2) = cons(x1, x2)
APPEND_IN(x1, x2, x3) = APPEND_IN(x1, x2)

We have to consider all (P,R,Pi)-chains
Transforming (infinitary) constructor rewriting Pi-DP problem [30] into ordinary QDP problem [15] by application of Pi.

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ QDPSizeChangeProof

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

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

APPEND_IN(cons(X, L1), L2) → APPEND_IN(L1, L2)

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] 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:

APPEND_IN(cons(X, L1), L2) → APPEND_IN(L1, L2)
The graph contains the following edges 1 > 1, 2 >= 2

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

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

LESS_IN(s(X), s(Y)) → LESS_IN(X, Y)

The TRS R consists of the following rules:

qs_in(cons(X, L), S) → U1(X, L, S, split_in(L, X, L1, L2))
split_in(cons(X, L), Y, L1, cons(X, L2)) → U8(X, L, Y, L1, L2, geq_in(X, Y))
geq_in(s(X), s(Y)) → U11(X, Y, geq_in(X, Y))
geq_in(s(X), 0) → geq_out(s(X), 0)
geq_in(X, X) → geq_out(X, X)
U11(X, Y, geq_out(X, Y)) → geq_out(s(X), s(Y))
U8(X, L, Y, L1, L2, geq_out(X, Y)) → U9(X, L, Y, L1, L2, split_in(L, Y, L1, L2))
split_in(cons(X, L), Y, cons(X, L1), L2) → U6(X, L, Y, L1, L2, less_in(X, Y))
less_in(s(X), s(Y)) → U10(X, Y, less_in(X, Y))
less_in(0, s(X)) → less_out(0, s(X))
U10(X, Y, less_out(X, Y)) → less_out(s(X), s(Y))
U6(X, L, Y, L1, L2, less_out(X, Y)) → U7(X, L, Y, L1, L2, split_in(L, Y, L1, L2))
split_in([], X, [], []) → split_out([], X, [], [])
U7(X, L, Y, L1, L2, split_out(L, Y, L1, L2)) → split_out(cons(X, L), Y, cons(X, L1), L2)
U9(X, L, Y, L1, L2, split_out(L, Y, L1, L2)) → split_out(cons(X, L), Y, L1, cons(X, L2))
U1(X, L, S, split_out(L, X, L1, L2)) → U2(X, L, S, L2, qs_in(L1, S1))
qs_in([], []) → qs_out([], [])
U2(X, L, S, L2, qs_out(L1, S1)) → U3(X, L, S, S1, qs_in(L2, S2))
U3(X, L, S, S1, qs_out(L2, S2)) → U4(X, L, S, append_in(S1, cons(X, S2), S))
append_in(cons(X, L1), L2, cons(X, L3)) → U5(X, L1, L2, L3, append_in(L1, L2, L3))
append_in([], L, L) → append_out([], L, L)
U5(X, L1, L2, L3, append_out(L1, L2, L3)) → append_out(cons(X, L1), L2, cons(X, L3))
U4(X, L, S, append_out(S1, cons(X, S2), S)) → qs_out(cons(X, L), S)

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

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

LESS_IN(s(X), s(Y)) → LESS_IN(X, Y)

R is empty.
Pi is empty.
We have to consider all (P,R,Pi)-chains
Transforming (infinitary) constructor rewriting Pi-DP problem [30] into ordinary QDP problem [15] by application of Pi.

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ QDPSizeChangeProof

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

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

LESS_IN(s(X), s(Y)) → LESS_IN(X, Y)

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

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

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

              ↳ PiDP

              ↳ PiDP

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

GEQ_IN(s(X), s(Y)) → GEQ_IN(X, Y)

The TRS R consists of the following rules:

qs_in(cons(X, L), S) → U1(X, L, S, split_in(L, X, L1, L2))
split_in(cons(X, L), Y, L1, cons(X, L2)) → U8(X, L, Y, L1, L2, geq_in(X, Y))
geq_in(s(X), s(Y)) → U11(X, Y, geq_in(X, Y))
geq_in(s(X), 0) → geq_out(s(X), 0)
geq_in(X, X) → geq_out(X, X)
U11(X, Y, geq_out(X, Y)) → geq_out(s(X), s(Y))
U8(X, L, Y, L1, L2, geq_out(X, Y)) → U9(X, L, Y, L1, L2, split_in(L, Y, L1, L2))
split_in(cons(X, L), Y, cons(X, L1), L2) → U6(X, L, Y, L1, L2, less_in(X, Y))
less_in(s(X), s(Y)) → U10(X, Y, less_in(X, Y))
less_in(0, s(X)) → less_out(0, s(X))
U10(X, Y, less_out(X, Y)) → less_out(s(X), s(Y))
U6(X, L, Y, L1, L2, less_out(X, Y)) → U7(X, L, Y, L1, L2, split_in(L, Y, L1, L2))
split_in([], X, [], []) → split_out([], X, [], [])
U7(X, L, Y, L1, L2, split_out(L, Y, L1, L2)) → split_out(cons(X, L), Y, cons(X, L1), L2)
U9(X, L, Y, L1, L2, split_out(L, Y, L1, L2)) → split_out(cons(X, L), Y, L1, cons(X, L2))
U1(X, L, S, split_out(L, X, L1, L2)) → U2(X, L, S, L2, qs_in(L1, S1))
qs_in([], []) → qs_out([], [])
U2(X, L, S, L2, qs_out(L1, S1)) → U3(X, L, S, S1, qs_in(L2, S2))
U3(X, L, S, S1, qs_out(L2, S2)) → U4(X, L, S, append_in(S1, cons(X, S2), S))
append_in(cons(X, L1), L2, cons(X, L3)) → U5(X, L1, L2, L3, append_in(L1, L2, L3))
append_in([], L, L) → append_out([], L, L)
U5(X, L1, L2, L3, append_out(L1, L2, L3)) → append_out(cons(X, L1), L2, cons(X, L3))
U4(X, L, S, append_out(S1, cons(X, S2), S)) → qs_out(cons(X, L), S)

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

              ↳ PiDP

              ↳ PiDP

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

GEQ_IN(s(X), s(Y)) → GEQ_IN(X, Y)

R is empty.
Pi is empty.
We have to consider all (P,R,Pi)-chains
Transforming (infinitary) constructor rewriting Pi-DP problem [30] into ordinary QDP problem [15] by application of Pi.

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ QDPSizeChangeProof

              ↳ PiDP

              ↳ PiDP

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

GEQ_IN(s(X), s(Y)) → GEQ_IN(X, Y)

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

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

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

              ↳ PiDP

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

SPLIT_IN(cons(X, L), Y, cons(X, L1), L2) → U6¹(X, L, Y, L1, L2, less_in(X, Y))
SPLIT_IN(cons(X, L), Y, L1, cons(X, L2)) → U8¹(X, L, Y, L1, L2, geq_in(X, Y))
U8¹(X, L, Y, L1, L2, geq_out(X, Y)) → SPLIT_IN(L, Y, L1, L2)
U6¹(X, L, Y, L1, L2, less_out(X, Y)) → SPLIT_IN(L, Y, L1, L2)

The TRS R consists of the following rules:

qs_in(cons(X, L), S) → U1(X, L, S, split_in(L, X, L1, L2))
split_in(cons(X, L), Y, L1, cons(X, L2)) → U8(X, L, Y, L1, L2, geq_in(X, Y))
geq_in(s(X), s(Y)) → U11(X, Y, geq_in(X, Y))
geq_in(s(X), 0) → geq_out(s(X), 0)
geq_in(X, X) → geq_out(X, X)
U11(X, Y, geq_out(X, Y)) → geq_out(s(X), s(Y))
U8(X, L, Y, L1, L2, geq_out(X, Y)) → U9(X, L, Y, L1, L2, split_in(L, Y, L1, L2))
split_in(cons(X, L), Y, cons(X, L1), L2) → U6(X, L, Y, L1, L2, less_in(X, Y))
less_in(s(X), s(Y)) → U10(X, Y, less_in(X, Y))
less_in(0, s(X)) → less_out(0, s(X))
U10(X, Y, less_out(X, Y)) → less_out(s(X), s(Y))
U6(X, L, Y, L1, L2, less_out(X, Y)) → U7(X, L, Y, L1, L2, split_in(L, Y, L1, L2))
split_in([], X, [], []) → split_out([], X, [], [])
U7(X, L, Y, L1, L2, split_out(L, Y, L1, L2)) → split_out(cons(X, L), Y, cons(X, L1), L2)
U9(X, L, Y, L1, L2, split_out(L, Y, L1, L2)) → split_out(cons(X, L), Y, L1, cons(X, L2))
U1(X, L, S, split_out(L, X, L1, L2)) → U2(X, L, S, L2, qs_in(L1, S1))
qs_in([], []) → qs_out([], [])
U2(X, L, S, L2, qs_out(L1, S1)) → U3(X, L, S, S1, qs_in(L2, S2))
U3(X, L, S, S1, qs_out(L2, S2)) → U4(X, L, S, append_in(S1, cons(X, S2), S))
append_in(cons(X, L1), L2, cons(X, L3)) → U5(X, L1, L2, L3, append_in(L1, L2, L3))
append_in([], L, L) → append_out([], L, L)
U5(X, L1, L2, L3, append_out(L1, L2, L3)) → append_out(cons(X, L1), L2, cons(X, L3))
U4(X, L, S, append_out(S1, cons(X, S2), S)) → qs_out(cons(X, L), S)

The argument filtering Pi contains the following mapping:
qs_in(x1, x2) = qs_in(x1)
cons(x1, x2) = cons(x1, x2)
U1(x1, x2, x3, x4) = U1(x1, x4)
split_in(x1, x2, x3, x4) = split_in(x1, x2)
U8(x1, x2, x3, x4, x5, x6) = U8(x1, x2, x3, x6)
geq_in(x1, x2) = geq_in(x1, x2)
s(x1) = s(x1)
U11(x1, x2, x3) = U11(x3)
0 = 0
geq_out(x1, x2) = geq_out
U9(x1, x2, x3, x4, x5, x6) = U9(x1, x6)
U6(x1, x2, x3, x4, x5, x6) = U6(x1, x2, x3, x6)
less_in(x1, x2) = less_in(x1, x2)
U10(x1, x2, x3) = U10(x3)
less_out(x1, x2) = less_out
U7(x1, x2, x3, x4, x5, x6) = U7(x1, x6)
[] = []
split_out(x1, x2, x3, x4) = split_out(x3, x4)
U2(x1, x2, x3, x4, x5) = U2(x1, x4, x5)
qs_out(x1, x2) = qs_out(x2)
U3(x1, x2, x3, x4, x5) = U3(x1, x4, x5)
U4(x1, x2, x3, x4) = U4(x4)
append_in(x1, x2, x3) = append_in(x1, x2)
U5(x1, x2, x3, x4, x5) = U5(x1, x5)
append_out(x1, x2, x3) = append_out(x3)
U8¹(x1, x2, x3, x4, x5, x6) = U8¹(x1, x2, x3, x6)
SPLIT_IN(x1, x2, x3, x4) = SPLIT_IN(x1, x2)
U6¹(x1, x2, x3, x4, x5, x6) = U6¹(x1, x2, x3, x6)

We have to consider all (P,R,Pi)-chains
For (infinitary) constructor rewriting [30] we can delete all non-usable rules from R.

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

              ↳ PiDP

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

SPLIT_IN(cons(X, L), Y, cons(X, L1), L2) → U6¹(X, L, Y, L1, L2, less_in(X, Y))
SPLIT_IN(cons(X, L), Y, L1, cons(X, L2)) → U8¹(X, L, Y, L1, L2, geq_in(X, Y))
U8¹(X, L, Y, L1, L2, geq_out(X, Y)) → SPLIT_IN(L, Y, L1, L2)
U6¹(X, L, Y, L1, L2, less_out(X, Y)) → SPLIT_IN(L, Y, L1, L2)

The TRS R consists of the following rules:

less_in(s(X), s(Y)) → U10(X, Y, less_in(X, Y))
less_in(0, s(X)) → less_out(0, s(X))
geq_in(s(X), s(Y)) → U11(X, Y, geq_in(X, Y))
geq_in(s(X), 0) → geq_out(s(X), 0)
geq_in(X, X) → geq_out(X, X)
U10(X, Y, less_out(X, Y)) → less_out(s(X), s(Y))
U11(X, Y, geq_out(X, Y)) → geq_out(s(X), s(Y))

The argument filtering Pi contains the following mapping:
cons(x1, x2) = cons(x1, x2)
geq_in(x1, x2) = geq_in(x1, x2)
s(x1) = s(x1)
U11(x1, x2, x3) = U11(x3)
0 = 0
geq_out(x1, x2) = geq_out
less_in(x1, x2) = less_in(x1, x2)
U10(x1, x2, x3) = U10(x3)
less_out(x1, x2) = less_out
U8¹(x1, x2, x3, x4, x5, x6) = U8¹(x1, x2, x3, x6)
SPLIT_IN(x1, x2, x3, x4) = SPLIT_IN(x1, x2)
U6¹(x1, x2, x3, x4, x5, x6) = U6¹(x1, x2, x3, x6)

We have to consider all (P,R,Pi)-chains
Transforming (infinitary) constructor rewriting Pi-DP problem [30] into ordinary QDP problem [15] by application of Pi.

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ QDPSizeChangeProof

              ↳ PiDP

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

SPLIT_IN(cons(X, L), Y) → U8¹(X, L, Y, geq_in(X, Y))
U6¹(X, L, Y, less_out) → SPLIT_IN(L, Y)
U8¹(X, L, Y, geq_out) → SPLIT_IN(L, Y)
SPLIT_IN(cons(X, L), Y) → U6¹(X, L, Y, less_in(X, Y))

The TRS R consists of the following rules:

less_in(s(X), s(Y)) → U10(less_in(X, Y))
less_in(0, s(X)) → less_out
geq_in(s(X), s(Y)) → U11(geq_in(X, Y))
geq_in(s(X), 0) → geq_out
geq_in(X, X) → geq_out
U10(less_out) → less_out
U11(geq_out) → geq_out

The set Q consists of the following terms:

less_in(x0, x1)
geq_in(x0, x1)
U10(x0)
U11(x0)

We have to consider all (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] 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:

U6¹(X, L, Y, less_out) → SPLIT_IN(L, Y)
The graph contains the following edges 2 >= 1, 3 >= 2
U8¹(X, L, Y, geq_out) → SPLIT_IN(L, Y)
The graph contains the following edges 2 >= 1, 3 >= 2
SPLIT_IN(cons(X, L), Y) → U8¹(X, L, Y, geq_in(X, Y))
The graph contains the following edges 1 > 1, 1 > 2, 2 >= 3
SPLIT_IN(cons(X, L), Y) → U6¹(X, L, Y, less_in(X, Y))
The graph contains the following edges 1 > 1, 1 > 2, 2 >= 3

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

                ↳ PiDPToQDPProof

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

U1¹(X, L, S, split_out(L, X, L1, L2)) → QS_IN(L1, S1)
U2¹(X, L, S, L2, qs_out(L1, S1)) → QS_IN(L2, S2)
U1¹(X, L, S, split_out(L, X, L1, L2)) → U2¹(X, L, S, L2, qs_in(L1, S1))
QS_IN(cons(X, L), S) → U1¹(X, L, S, split_in(L, X, L1, L2))

The TRS R consists of the following rules:

qs_in(cons(X, L), S) → U1(X, L, S, split_in(L, X, L1, L2))
split_in(cons(X, L), Y, L1, cons(X, L2)) → U8(X, L, Y, L1, L2, geq_in(X, Y))
geq_in(s(X), s(Y)) → U11(X, Y, geq_in(X, Y))
geq_in(s(X), 0) → geq_out(s(X), 0)
geq_in(X, X) → geq_out(X, X)
U11(X, Y, geq_out(X, Y)) → geq_out(s(X), s(Y))
U8(X, L, Y, L1, L2, geq_out(X, Y)) → U9(X, L, Y, L1, L2, split_in(L, Y, L1, L2))
split_in(cons(X, L), Y, cons(X, L1), L2) → U6(X, L, Y, L1, L2, less_in(X, Y))
less_in(s(X), s(Y)) → U10(X, Y, less_in(X, Y))
less_in(0, s(X)) → less_out(0, s(X))
U10(X, Y, less_out(X, Y)) → less_out(s(X), s(Y))
U6(X, L, Y, L1, L2, less_out(X, Y)) → U7(X, L, Y, L1, L2, split_in(L, Y, L1, L2))
split_in([], X, [], []) → split_out([], X, [], [])
U7(X, L, Y, L1, L2, split_out(L, Y, L1, L2)) → split_out(cons(X, L), Y, cons(X, L1), L2)
U9(X, L, Y, L1, L2, split_out(L, Y, L1, L2)) → split_out(cons(X, L), Y, L1, cons(X, L2))
U1(X, L, S, split_out(L, X, L1, L2)) → U2(X, L, S, L2, qs_in(L1, S1))
qs_in([], []) → qs_out([], [])
U2(X, L, S, L2, qs_out(L1, S1)) → U3(X, L, S, S1, qs_in(L2, S2))
U3(X, L, S, S1, qs_out(L2, S2)) → U4(X, L, S, append_in(S1, cons(X, S2), S))
append_in(cons(X, L1), L2, cons(X, L3)) → U5(X, L1, L2, L3, append_in(L1, L2, L3))
append_in([], L, L) → append_out([], L, L)
U5(X, L1, L2, L3, append_out(L1, L2, L3)) → append_out(cons(X, L1), L2, cons(X, L3))
U4(X, L, S, append_out(S1, cons(X, S2), S)) → qs_out(cons(X, L), S)

The argument filtering Pi contains the following mapping:
qs_in(x1, x2) = qs_in(x1)
cons(x1, x2) = cons(x1, x2)
U1(x1, x2, x3, x4) = U1(x1, x4)
split_in(x1, x2, x3, x4) = split_in(x1, x2)
U8(x1, x2, x3, x4, x5, x6) = U8(x1, x2, x3, x6)
geq_in(x1, x2) = geq_in(x1, x2)
s(x1) = s(x1)
U11(x1, x2, x3) = U11(x3)
0 = 0
geq_out(x1, x2) = geq_out
U9(x1, x2, x3, x4, x5, x6) = U9(x1, x6)
U6(x1, x2, x3, x4, x5, x6) = U6(x1, x2, x3, x6)
less_in(x1, x2) = less_in(x1, x2)
U10(x1, x2, x3) = U10(x3)
less_out(x1, x2) = less_out
U7(x1, x2, x3, x4, x5, x6) = U7(x1, x6)
[] = []
split_out(x1, x2, x3, x4) = split_out(x3, x4)
U2(x1, x2, x3, x4, x5) = U2(x1, x4, x5)
qs_out(x1, x2) = qs_out(x2)
U3(x1, x2, x3, x4, x5) = U3(x1, x4, x5)
U4(x1, x2, x3, x4) = U4(x4)
append_in(x1, x2, x3) = append_in(x1, x2)
U5(x1, x2, x3, x4, x5) = U5(x1, x5)
append_out(x1, x2, x3) = append_out(x3)
QS_IN(x1, x2) = QS_IN(x1)
U2¹(x1, x2, x3, x4, x5) = U2¹(x1, x4, x5)
U1¹(x1, x2, x3, x4) = U1¹(x1, x4)

We have to consider all (P,R,Pi)-chains
Transforming (infinitary) constructor rewriting Pi-DP problem [30] into ordinary QDP problem [15] by application of Pi.

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

                ↳ PiDPToQDPProof

                  ↳ QDP

                    ↳ QDPOrderProof

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

U2¹(X, L2, qs_out(S1)) → QS_IN(L2)
QS_IN(cons(X, L)) → U1¹(X, split_in(L, X))
U1¹(X, split_out(L1, L2)) → U2¹(X, L2, qs_in(L1))
U1¹(X, split_out(L1, L2)) → QS_IN(L1)

The TRS R consists of the following rules:

qs_in(cons(X, L)) → U1(X, split_in(L, X))
split_in(cons(X, L), Y) → U8(X, L, Y, geq_in(X, Y))
geq_in(s(X), s(Y)) → U11(geq_in(X, Y))
geq_in(s(X), 0) → geq_out
geq_in(X, X) → geq_out
U11(geq_out) → geq_out
U8(X, L, Y, geq_out) → U9(X, split_in(L, Y))
split_in(cons(X, L), Y) → U6(X, L, Y, less_in(X, Y))
less_in(s(X), s(Y)) → U10(less_in(X, Y))
less_in(0, s(X)) → less_out
U10(less_out) → less_out
U6(X, L, Y, less_out) → U7(X, split_in(L, Y))
split_in([], X) → split_out([], [])
U7(X, split_out(L1, L2)) → split_out(cons(X, L1), L2)
U9(X, split_out(L1, L2)) → split_out(L1, cons(X, L2))
U1(X, split_out(L1, L2)) → U2(X, L2, qs_in(L1))
qs_in([]) → qs_out([])
U2(X, L2, qs_out(S1)) → U3(X, S1, qs_in(L2))
U3(X, S1, qs_out(S2)) → U4(append_in(S1, cons(X, S2)))
append_in(cons(X, L1), L2) → U5(X, append_in(L1, L2))
append_in([], L) → append_out(L)
U5(X, append_out(L3)) → append_out(cons(X, L3))
U4(append_out(S)) → qs_out(S)

The set Q consists of the following terms:

qs_in(x0)
split_in(x0, x1)
geq_in(x0, x1)
U11(x0)
U8(x0, x1, x2, x3)
less_in(x0, x1)
U10(x0)
U6(x0, x1, x2, x3)
U7(x0, x1)
U9(x0, x1)
U1(x0, x1)
U2(x0, x1, x2)
U3(x0, x1, x2)
append_in(x0, x1)
U5(x0, x1)
U4(x0)

We have to consider all (P,Q,R)-chains.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

QS_IN(cons(X, L)) → U1¹(X, split_in(L, X))

The remaining pairs can at least be oriented weakly.

U2¹(X, L2, qs_out(S1)) → QS_IN(L2)
U1¹(X, split_out(L1, L2)) → U2¹(X, L2, qs_in(L1))
U1¹(X, split_out(L1, L2)) → QS_IN(L1)

Used ordering: Polynomial interpretation [25]:

POL(0) = 0
POL(QS_IN(x₁)) = x₁
POL(U1(x₁, x₂)) = 0
POL(U10(x₁)) = 0
POL(U11(x₁)) = 1
POL(U1¹(x₁, x₂)) = x₂
POL(U2(x₁, x₂, x₃)) = 0
POL(U2¹(x₁, x₂, x₃)) = x₂
POL(U3(x₁, x₂, x₃)) = 0
POL(U4(x₁)) = 0
POL(U5(x₁, x₂)) = 1 + x₂
POL(U6(x₁, x₂, x₃, x₄)) = 1 + x₂
POL(U7(x₁, x₂)) = 1 + x₂
POL(U8(x₁, x₂, x₃, x₄)) = x₂ + x₄
POL(U9(x₁, x₂)) = 1 + x₂
POL([]) = 0
POL(append_in(x₁, x₂)) = 1 + x₁ + x₂
POL(append_out(x₁)) = 1 + x₁
POL(cons(x₁, x₂)) = 1 + x₂
POL(geq_in(x₁, x₂)) = 1
POL(geq_out) = 1
POL(less_in(x₁, x₂)) = 0
POL(less_out) = 0
POL(qs_in(x₁)) = 0
POL(qs_out(x₁)) = 0
POL(s(x₁)) = 0
POL(split_in(x₁, x₂)) = x₁
POL(split_out(x₁, x₂)) = x₁ + x₂

The following usable rules [17] were oriented:

U8(X, L, Y, geq_out) → U9(X, split_in(L, Y))
split_in([], X) → split_out([], [])
geq_in(s(X), s(Y)) → U11(geq_in(X, Y))
split_in(cons(X, L), Y) → U8(X, L, Y, geq_in(X, Y))
U6(X, L, Y, less_out) → U7(X, split_in(L, Y))
split_in(cons(X, L), Y) → U6(X, L, Y, less_in(X, Y))
geq_in(s(X), 0) → geq_out
U9(X, split_out(L1, L2)) → split_out(L1, cons(X, L2))
geq_in(X, X) → geq_out
U7(X, split_out(L1, L2)) → split_out(cons(X, L1), L2)
U11(geq_out) → geq_out

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

                ↳ PiDPToQDPProof

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ DependencyGraphProof

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

U2¹(X, L2, qs_out(S1)) → QS_IN(L2)
U1¹(X, split_out(L1, L2)) → U2¹(X, L2, qs_in(L1))
U1¹(X, split_out(L1, L2)) → QS_IN(L1)

The TRS R consists of the following rules:

qs_in(cons(X, L)) → U1(X, split_in(L, X))
split_in(cons(X, L), Y) → U8(X, L, Y, geq_in(X, Y))
geq_in(s(X), s(Y)) → U11(geq_in(X, Y))
geq_in(s(X), 0) → geq_out
geq_in(X, X) → geq_out
U11(geq_out) → geq_out
U8(X, L, Y, geq_out) → U9(X, split_in(L, Y))
split_in(cons(X, L), Y) → U6(X, L, Y, less_in(X, Y))
less_in(s(X), s(Y)) → U10(less_in(X, Y))
less_in(0, s(X)) → less_out
U10(less_out) → less_out
U6(X, L, Y, less_out) → U7(X, split_in(L, Y))
split_in([], X) → split_out([], [])
U7(X, split_out(L1, L2)) → split_out(cons(X, L1), L2)
U9(X, split_out(L1, L2)) → split_out(L1, cons(X, L2))
U1(X, split_out(L1, L2)) → U2(X, L2, qs_in(L1))
qs_in([]) → qs_out([])
U2(X, L2, qs_out(S1)) → U3(X, S1, qs_in(L2))
U3(X, S1, qs_out(S2)) → U4(append_in(S1, cons(X, S2)))
append_in(cons(X, L1), L2) → U5(X, append_in(L1, L2))
append_in([], L) → append_out(L)
U5(X, append_out(L3)) → append_out(cons(X, L3))
U4(append_out(S)) → qs_out(S)

The set Q consists of the following terms:

qs_in(x0)
split_in(x0, x1)
geq_in(x0, x1)
U11(x0)
U8(x0, x1, x2, x3)
less_in(x0, x1)
U10(x0)
U6(x0, x1, x2, x3)
U7(x0, x1)
U9(x0, x1)
U1(x0, x1)
U2(x0, x1, x2)
U3(x0, x1, x2)
append_in(x0, x1)
U5(x0, x1)
U4(x0)

We have to consider all (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 3 less nodes.