Left Termination of the query pattern qs_in_2(a, g) w.r.t. the given Prolog program could not be shown:



↳ Prolog
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof

Clauses:

qs([], []).
qs(.(X, Xs), Ys) :- ','(part(X, Xs, Littles, Bigs), ','(qs(Littles, Ls), ','(qs(Bigs, Bs), app(Ls, .(X, Bs), Ys)))).
part(X, .(Y, Xs), .(Y, Ls), Bs) :- ','(less(X, Y), part(X, Xs, Ls, Bs)).
part(X, .(Y, Xs), Ls, .(Y, Bs)) :- part(X, Xs, Ls, Bs).
part(X, [], [], []).
app([], X, X).
app(.(X, Xs), Ys, .(X, Zs)) :- app(Xs, Ys, Zs).
less(0, s(X)).
less(s(X), s(Y)) :- less(X, Y).

Queries:

qs(a,g).

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(.(X, Xs), Ys) → U1(X, Xs, Ys, part_in(X, Xs, Littles, Bigs))
part_in(X, [], [], []) → part_out(X, [], [], [])
part_in(X, .(Y, Xs), Ls, .(Y, Bs)) → U7(X, Y, Xs, Ls, Bs, part_in(X, Xs, Ls, Bs))
part_in(X, .(Y, Xs), .(Y, Ls), Bs) → U5(X, Y, Xs, Ls, Bs, less_in(X, Y))
less_in(s(X), s(Y)) → U9(X, Y, less_in(X, Y))
less_in(0, s(X)) → less_out(0, s(X))
U9(X, Y, less_out(X, Y)) → less_out(s(X), s(Y))
U5(X, Y, Xs, Ls, Bs, less_out(X, Y)) → U6(X, Y, Xs, Ls, Bs, part_in(X, Xs, Ls, Bs))
U6(X, Y, Xs, Ls, Bs, part_out(X, Xs, Ls, Bs)) → part_out(X, .(Y, Xs), .(Y, Ls), Bs)
U7(X, Y, Xs, Ls, Bs, part_out(X, Xs, Ls, Bs)) → part_out(X, .(Y, Xs), Ls, .(Y, Bs))
U1(X, Xs, Ys, part_out(X, Xs, Littles, Bigs)) → U2(X, Xs, Ys, Bigs, qs_in(Littles, Ls))
qs_in([], []) → qs_out([], [])
U2(X, Xs, Ys, Bigs, qs_out(Littles, Ls)) → U3(X, Xs, Ys, Ls, qs_in(Bigs, Bs))
U3(X, Xs, Ys, Ls, qs_out(Bigs, Bs)) → U4(X, Xs, Ys, app_in(Ls, .(X, Bs), Ys))
app_in(.(X, Xs), Ys, .(X, Zs)) → U8(X, Xs, Ys, Zs, app_in(Xs, Ys, Zs))
app_in([], X, X) → app_out([], X, X)
U8(X, Xs, Ys, Zs, app_out(Xs, Ys, Zs)) → app_out(.(X, Xs), Ys, .(X, Zs))
U4(X, Xs, Ys, app_out(Ls, .(X, Bs), Ys)) → qs_out(.(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
qs_in(x1, x2)  =  qs_in
.(x1, x2)  =  .(x1, x2)
U1(x1, x2, x3, x4)  =  U1(x4)
part_in(x1, x2, x3, x4)  =  part_in
[]  =  []
part_out(x1, x2, x3, x4)  =  part_out
U7(x1, x2, x3, x4, x5, x6)  =  U7(x6)
U5(x1, x2, x3, x4, x5, x6)  =  U5(x6)
less_in(x1, x2)  =  less_in
s(x1)  =  s(x1)
U9(x1, x2, x3)  =  U9(x3)
0  =  0
less_out(x1, x2)  =  less_out(x1)
U6(x1, x2, x3, x4, x5, x6)  =  U6(x6)
U2(x1, x2, x3, x4, x5)  =  U2(x5)
qs_out(x1, x2)  =  qs_out
U3(x1, x2, x3, x4, x5)  =  U3(x5)
U4(x1, x2, x3, x4)  =  U4(x4)
app_in(x1, x2, x3)  =  app_in
U8(x1, x2, x3, x4, x5)  =  U8(x5)
app_out(x1, x2, x3)  =  app_out

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog



↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
  ↳ PrologToPiTRSProof

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

qs_in(.(X, Xs), Ys) → U1(X, Xs, Ys, part_in(X, Xs, Littles, Bigs))
part_in(X, [], [], []) → part_out(X, [], [], [])
part_in(X, .(Y, Xs), Ls, .(Y, Bs)) → U7(X, Y, Xs, Ls, Bs, part_in(X, Xs, Ls, Bs))
part_in(X, .(Y, Xs), .(Y, Ls), Bs) → U5(X, Y, Xs, Ls, Bs, less_in(X, Y))
less_in(s(X), s(Y)) → U9(X, Y, less_in(X, Y))
less_in(0, s(X)) → less_out(0, s(X))
U9(X, Y, less_out(X, Y)) → less_out(s(X), s(Y))
U5(X, Y, Xs, Ls, Bs, less_out(X, Y)) → U6(X, Y, Xs, Ls, Bs, part_in(X, Xs, Ls, Bs))
U6(X, Y, Xs, Ls, Bs, part_out(X, Xs, Ls, Bs)) → part_out(X, .(Y, Xs), .(Y, Ls), Bs)
U7(X, Y, Xs, Ls, Bs, part_out(X, Xs, Ls, Bs)) → part_out(X, .(Y, Xs), Ls, .(Y, Bs))
U1(X, Xs, Ys, part_out(X, Xs, Littles, Bigs)) → U2(X, Xs, Ys, Bigs, qs_in(Littles, Ls))
qs_in([], []) → qs_out([], [])
U2(X, Xs, Ys, Bigs, qs_out(Littles, Ls)) → U3(X, Xs, Ys, Ls, qs_in(Bigs, Bs))
U3(X, Xs, Ys, Ls, qs_out(Bigs, Bs)) → U4(X, Xs, Ys, app_in(Ls, .(X, Bs), Ys))
app_in(.(X, Xs), Ys, .(X, Zs)) → U8(X, Xs, Ys, Zs, app_in(Xs, Ys, Zs))
app_in([], X, X) → app_out([], X, X)
U8(X, Xs, Ys, Zs, app_out(Xs, Ys, Zs)) → app_out(.(X, Xs), Ys, .(X, Zs))
U4(X, Xs, Ys, app_out(Ls, .(X, Bs), Ys)) → qs_out(.(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
qs_in(x1, x2)  =  qs_in
.(x1, x2)  =  .(x1, x2)
U1(x1, x2, x3, x4)  =  U1(x4)
part_in(x1, x2, x3, x4)  =  part_in
[]  =  []
part_out(x1, x2, x3, x4)  =  part_out
U7(x1, x2, x3, x4, x5, x6)  =  U7(x6)
U5(x1, x2, x3, x4, x5, x6)  =  U5(x6)
less_in(x1, x2)  =  less_in
s(x1)  =  s(x1)
U9(x1, x2, x3)  =  U9(x3)
0  =  0
less_out(x1, x2)  =  less_out(x1)
U6(x1, x2, x3, x4, x5, x6)  =  U6(x6)
U2(x1, x2, x3, x4, x5)  =  U2(x5)
qs_out(x1, x2)  =  qs_out
U3(x1, x2, x3, x4, x5)  =  U3(x5)
U4(x1, x2, x3, x4)  =  U4(x4)
app_in(x1, x2, x3)  =  app_in
U8(x1, x2, x3, x4, x5)  =  U8(x5)
app_out(x1, x2, x3)  =  app_out


Using Dependency Pairs [1,30] we result in the following initial DP problem:
Pi DP problem:
The TRS P consists of the following rules:

QS_IN(.(X, Xs), Ys) → U11(X, Xs, Ys, part_in(X, Xs, Littles, Bigs))
QS_IN(.(X, Xs), Ys) → PART_IN(X, Xs, Littles, Bigs)
PART_IN(X, .(Y, Xs), Ls, .(Y, Bs)) → U71(X, Y, Xs, Ls, Bs, part_in(X, Xs, Ls, Bs))
PART_IN(X, .(Y, Xs), Ls, .(Y, Bs)) → PART_IN(X, Xs, Ls, Bs)
PART_IN(X, .(Y, Xs), .(Y, Ls), Bs) → U51(X, Y, Xs, Ls, Bs, less_in(X, Y))
PART_IN(X, .(Y, Xs), .(Y, Ls), Bs) → LESS_IN(X, Y)
LESS_IN(s(X), s(Y)) → U91(X, Y, less_in(X, Y))
LESS_IN(s(X), s(Y)) → LESS_IN(X, Y)
U51(X, Y, Xs, Ls, Bs, less_out(X, Y)) → U61(X, Y, Xs, Ls, Bs, part_in(X, Xs, Ls, Bs))
U51(X, Y, Xs, Ls, Bs, less_out(X, Y)) → PART_IN(X, Xs, Ls, Bs)
U11(X, Xs, Ys, part_out(X, Xs, Littles, Bigs)) → U21(X, Xs, Ys, Bigs, qs_in(Littles, Ls))
U11(X, Xs, Ys, part_out(X, Xs, Littles, Bigs)) → QS_IN(Littles, Ls)
U21(X, Xs, Ys, Bigs, qs_out(Littles, Ls)) → U31(X, Xs, Ys, Ls, qs_in(Bigs, Bs))
U21(X, Xs, Ys, Bigs, qs_out(Littles, Ls)) → QS_IN(Bigs, Bs)
U31(X, Xs, Ys, Ls, qs_out(Bigs, Bs)) → U41(X, Xs, Ys, app_in(Ls, .(X, Bs), Ys))
U31(X, Xs, Ys, Ls, qs_out(Bigs, Bs)) → APP_IN(Ls, .(X, Bs), Ys)
APP_IN(.(X, Xs), Ys, .(X, Zs)) → U81(X, Xs, Ys, Zs, app_in(Xs, Ys, Zs))
APP_IN(.(X, Xs), Ys, .(X, Zs)) → APP_IN(Xs, Ys, Zs)

The TRS R consists of the following rules:

qs_in(.(X, Xs), Ys) → U1(X, Xs, Ys, part_in(X, Xs, Littles, Bigs))
part_in(X, [], [], []) → part_out(X, [], [], [])
part_in(X, .(Y, Xs), Ls, .(Y, Bs)) → U7(X, Y, Xs, Ls, Bs, part_in(X, Xs, Ls, Bs))
part_in(X, .(Y, Xs), .(Y, Ls), Bs) → U5(X, Y, Xs, Ls, Bs, less_in(X, Y))
less_in(s(X), s(Y)) → U9(X, Y, less_in(X, Y))
less_in(0, s(X)) → less_out(0, s(X))
U9(X, Y, less_out(X, Y)) → less_out(s(X), s(Y))
U5(X, Y, Xs, Ls, Bs, less_out(X, Y)) → U6(X, Y, Xs, Ls, Bs, part_in(X, Xs, Ls, Bs))
U6(X, Y, Xs, Ls, Bs, part_out(X, Xs, Ls, Bs)) → part_out(X, .(Y, Xs), .(Y, Ls), Bs)
U7(X, Y, Xs, Ls, Bs, part_out(X, Xs, Ls, Bs)) → part_out(X, .(Y, Xs), Ls, .(Y, Bs))
U1(X, Xs, Ys, part_out(X, Xs, Littles, Bigs)) → U2(X, Xs, Ys, Bigs, qs_in(Littles, Ls))
qs_in([], []) → qs_out([], [])
U2(X, Xs, Ys, Bigs, qs_out(Littles, Ls)) → U3(X, Xs, Ys, Ls, qs_in(Bigs, Bs))
U3(X, Xs, Ys, Ls, qs_out(Bigs, Bs)) → U4(X, Xs, Ys, app_in(Ls, .(X, Bs), Ys))
app_in(.(X, Xs), Ys, .(X, Zs)) → U8(X, Xs, Ys, Zs, app_in(Xs, Ys, Zs))
app_in([], X, X) → app_out([], X, X)
U8(X, Xs, Ys, Zs, app_out(Xs, Ys, Zs)) → app_out(.(X, Xs), Ys, .(X, Zs))
U4(X, Xs, Ys, app_out(Ls, .(X, Bs), Ys)) → qs_out(.(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
qs_in(x1, x2)  =  qs_in
.(x1, x2)  =  .(x1, x2)
U1(x1, x2, x3, x4)  =  U1(x4)
part_in(x1, x2, x3, x4)  =  part_in
[]  =  []
part_out(x1, x2, x3, x4)  =  part_out
U7(x1, x2, x3, x4, x5, x6)  =  U7(x6)
U5(x1, x2, x3, x4, x5, x6)  =  U5(x6)
less_in(x1, x2)  =  less_in
s(x1)  =  s(x1)
U9(x1, x2, x3)  =  U9(x3)
0  =  0
less_out(x1, x2)  =  less_out(x1)
U6(x1, x2, x3, x4, x5, x6)  =  U6(x6)
U2(x1, x2, x3, x4, x5)  =  U2(x5)
qs_out(x1, x2)  =  qs_out
U3(x1, x2, x3, x4, x5)  =  U3(x5)
U4(x1, x2, x3, x4)  =  U4(x4)
app_in(x1, x2, x3)  =  app_in
U8(x1, x2, x3, x4, x5)  =  U8(x5)
app_out(x1, x2, x3)  =  app_out
U51(x1, x2, x3, x4, x5, x6)  =  U51(x6)
U41(x1, x2, x3, x4)  =  U41(x4)
U31(x1, x2, x3, x4, x5)  =  U31(x5)
PART_IN(x1, x2, x3, x4)  =  PART_IN
U91(x1, x2, x3)  =  U91(x3)
QS_IN(x1, x2)  =  QS_IN
U71(x1, x2, x3, x4, x5, x6)  =  U71(x6)
APP_IN(x1, x2, x3)  =  APP_IN
LESS_IN(x1, x2)  =  LESS_IN
U81(x1, x2, x3, x4, x5)  =  U81(x5)
U21(x1, x2, x3, x4, x5)  =  U21(x5)
U11(x1, x2, x3, x4)  =  U11(x4)
U61(x1, x2, x3, x4, x5, x6)  =  U61(x6)

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

↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
  ↳ PrologToPiTRSProof

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

QS_IN(.(X, Xs), Ys) → U11(X, Xs, Ys, part_in(X, Xs, Littles, Bigs))
QS_IN(.(X, Xs), Ys) → PART_IN(X, Xs, Littles, Bigs)
PART_IN(X, .(Y, Xs), Ls, .(Y, Bs)) → U71(X, Y, Xs, Ls, Bs, part_in(X, Xs, Ls, Bs))
PART_IN(X, .(Y, Xs), Ls, .(Y, Bs)) → PART_IN(X, Xs, Ls, Bs)
PART_IN(X, .(Y, Xs), .(Y, Ls), Bs) → U51(X, Y, Xs, Ls, Bs, less_in(X, Y))
PART_IN(X, .(Y, Xs), .(Y, Ls), Bs) → LESS_IN(X, Y)
LESS_IN(s(X), s(Y)) → U91(X, Y, less_in(X, Y))
LESS_IN(s(X), s(Y)) → LESS_IN(X, Y)
U51(X, Y, Xs, Ls, Bs, less_out(X, Y)) → U61(X, Y, Xs, Ls, Bs, part_in(X, Xs, Ls, Bs))
U51(X, Y, Xs, Ls, Bs, less_out(X, Y)) → PART_IN(X, Xs, Ls, Bs)
U11(X, Xs, Ys, part_out(X, Xs, Littles, Bigs)) → U21(X, Xs, Ys, Bigs, qs_in(Littles, Ls))
U11(X, Xs, Ys, part_out(X, Xs, Littles, Bigs)) → QS_IN(Littles, Ls)
U21(X, Xs, Ys, Bigs, qs_out(Littles, Ls)) → U31(X, Xs, Ys, Ls, qs_in(Bigs, Bs))
U21(X, Xs, Ys, Bigs, qs_out(Littles, Ls)) → QS_IN(Bigs, Bs)
U31(X, Xs, Ys, Ls, qs_out(Bigs, Bs)) → U41(X, Xs, Ys, app_in(Ls, .(X, Bs), Ys))
U31(X, Xs, Ys, Ls, qs_out(Bigs, Bs)) → APP_IN(Ls, .(X, Bs), Ys)
APP_IN(.(X, Xs), Ys, .(X, Zs)) → U81(X, Xs, Ys, Zs, app_in(Xs, Ys, Zs))
APP_IN(.(X, Xs), Ys, .(X, Zs)) → APP_IN(Xs, Ys, Zs)

The TRS R consists of the following rules:

qs_in(.(X, Xs), Ys) → U1(X, Xs, Ys, part_in(X, Xs, Littles, Bigs))
part_in(X, [], [], []) → part_out(X, [], [], [])
part_in(X, .(Y, Xs), Ls, .(Y, Bs)) → U7(X, Y, Xs, Ls, Bs, part_in(X, Xs, Ls, Bs))
part_in(X, .(Y, Xs), .(Y, Ls), Bs) → U5(X, Y, Xs, Ls, Bs, less_in(X, Y))
less_in(s(X), s(Y)) → U9(X, Y, less_in(X, Y))
less_in(0, s(X)) → less_out(0, s(X))
U9(X, Y, less_out(X, Y)) → less_out(s(X), s(Y))
U5(X, Y, Xs, Ls, Bs, less_out(X, Y)) → U6(X, Y, Xs, Ls, Bs, part_in(X, Xs, Ls, Bs))
U6(X, Y, Xs, Ls, Bs, part_out(X, Xs, Ls, Bs)) → part_out(X, .(Y, Xs), .(Y, Ls), Bs)
U7(X, Y, Xs, Ls, Bs, part_out(X, Xs, Ls, Bs)) → part_out(X, .(Y, Xs), Ls, .(Y, Bs))
U1(X, Xs, Ys, part_out(X, Xs, Littles, Bigs)) → U2(X, Xs, Ys, Bigs, qs_in(Littles, Ls))
qs_in([], []) → qs_out([], [])
U2(X, Xs, Ys, Bigs, qs_out(Littles, Ls)) → U3(X, Xs, Ys, Ls, qs_in(Bigs, Bs))
U3(X, Xs, Ys, Ls, qs_out(Bigs, Bs)) → U4(X, Xs, Ys, app_in(Ls, .(X, Bs), Ys))
app_in(.(X, Xs), Ys, .(X, Zs)) → U8(X, Xs, Ys, Zs, app_in(Xs, Ys, Zs))
app_in([], X, X) → app_out([], X, X)
U8(X, Xs, Ys, Zs, app_out(Xs, Ys, Zs)) → app_out(.(X, Xs), Ys, .(X, Zs))
U4(X, Xs, Ys, app_out(Ls, .(X, Bs), Ys)) → qs_out(.(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
qs_in(x1, x2)  =  qs_in
.(x1, x2)  =  .(x1, x2)
U1(x1, x2, x3, x4)  =  U1(x4)
part_in(x1, x2, x3, x4)  =  part_in
[]  =  []
part_out(x1, x2, x3, x4)  =  part_out
U7(x1, x2, x3, x4, x5, x6)  =  U7(x6)
U5(x1, x2, x3, x4, x5, x6)  =  U5(x6)
less_in(x1, x2)  =  less_in
s(x1)  =  s(x1)
U9(x1, x2, x3)  =  U9(x3)
0  =  0
less_out(x1, x2)  =  less_out(x1)
U6(x1, x2, x3, x4, x5, x6)  =  U6(x6)
U2(x1, x2, x3, x4, x5)  =  U2(x5)
qs_out(x1, x2)  =  qs_out
U3(x1, x2, x3, x4, x5)  =  U3(x5)
U4(x1, x2, x3, x4)  =  U4(x4)
app_in(x1, x2, x3)  =  app_in
U8(x1, x2, x3, x4, x5)  =  U8(x5)
app_out(x1, x2, x3)  =  app_out
U51(x1, x2, x3, x4, x5, x6)  =  U51(x6)
U41(x1, x2, x3, x4)  =  U41(x4)
U31(x1, x2, x3, x4, x5)  =  U31(x5)
PART_IN(x1, x2, x3, x4)  =  PART_IN
U91(x1, x2, x3)  =  U91(x3)
QS_IN(x1, x2)  =  QS_IN
U71(x1, x2, x3, x4, x5, x6)  =  U71(x6)
APP_IN(x1, x2, x3)  =  APP_IN
LESS_IN(x1, x2)  =  LESS_IN
U81(x1, x2, x3, x4, x5)  =  U81(x5)
U21(x1, x2, x3, x4, x5)  =  U21(x5)
U11(x1, x2, x3, x4)  =  U11(x4)
U61(x1, x2, x3, x4, x5, x6)  =  U61(x6)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph [30] contains 4 SCCs with 9 less nodes.

↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
                ↳ UsableRulesProof
              ↳ PiDP
              ↳ PiDP
              ↳ PiDP
  ↳ PrologToPiTRSProof

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

APP_IN(.(X, Xs), Ys, .(X, Zs)) → APP_IN(Xs, Ys, Zs)

The TRS R consists of the following rules:

qs_in(.(X, Xs), Ys) → U1(X, Xs, Ys, part_in(X, Xs, Littles, Bigs))
part_in(X, [], [], []) → part_out(X, [], [], [])
part_in(X, .(Y, Xs), Ls, .(Y, Bs)) → U7(X, Y, Xs, Ls, Bs, part_in(X, Xs, Ls, Bs))
part_in(X, .(Y, Xs), .(Y, Ls), Bs) → U5(X, Y, Xs, Ls, Bs, less_in(X, Y))
less_in(s(X), s(Y)) → U9(X, Y, less_in(X, Y))
less_in(0, s(X)) → less_out(0, s(X))
U9(X, Y, less_out(X, Y)) → less_out(s(X), s(Y))
U5(X, Y, Xs, Ls, Bs, less_out(X, Y)) → U6(X, Y, Xs, Ls, Bs, part_in(X, Xs, Ls, Bs))
U6(X, Y, Xs, Ls, Bs, part_out(X, Xs, Ls, Bs)) → part_out(X, .(Y, Xs), .(Y, Ls), Bs)
U7(X, Y, Xs, Ls, Bs, part_out(X, Xs, Ls, Bs)) → part_out(X, .(Y, Xs), Ls, .(Y, Bs))
U1(X, Xs, Ys, part_out(X, Xs, Littles, Bigs)) → U2(X, Xs, Ys, Bigs, qs_in(Littles, Ls))
qs_in([], []) → qs_out([], [])
U2(X, Xs, Ys, Bigs, qs_out(Littles, Ls)) → U3(X, Xs, Ys, Ls, qs_in(Bigs, Bs))
U3(X, Xs, Ys, Ls, qs_out(Bigs, Bs)) → U4(X, Xs, Ys, app_in(Ls, .(X, Bs), Ys))
app_in(.(X, Xs), Ys, .(X, Zs)) → U8(X, Xs, Ys, Zs, app_in(Xs, Ys, Zs))
app_in([], X, X) → app_out([], X, X)
U8(X, Xs, Ys, Zs, app_out(Xs, Ys, Zs)) → app_out(.(X, Xs), Ys, .(X, Zs))
U4(X, Xs, Ys, app_out(Ls, .(X, Bs), Ys)) → qs_out(.(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
qs_in(x1, x2)  =  qs_in
.(x1, x2)  =  .(x1, x2)
U1(x1, x2, x3, x4)  =  U1(x4)
part_in(x1, x2, x3, x4)  =  part_in
[]  =  []
part_out(x1, x2, x3, x4)  =  part_out
U7(x1, x2, x3, x4, x5, x6)  =  U7(x6)
U5(x1, x2, x3, x4, x5, x6)  =  U5(x6)
less_in(x1, x2)  =  less_in
s(x1)  =  s(x1)
U9(x1, x2, x3)  =  U9(x3)
0  =  0
less_out(x1, x2)  =  less_out(x1)
U6(x1, x2, x3, x4, x5, x6)  =  U6(x6)
U2(x1, x2, x3, x4, x5)  =  U2(x5)
qs_out(x1, x2)  =  qs_out
U3(x1, x2, x3, x4, x5)  =  U3(x5)
U4(x1, x2, x3, x4)  =  U4(x4)
app_in(x1, x2, x3)  =  app_in
U8(x1, x2, x3, x4, x5)  =  U8(x5)
app_out(x1, x2, x3)  =  app_out
APP_IN(x1, x2, x3)  =  APP_IN

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
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
              ↳ PiDP
              ↳ PiDP
              ↳ PiDP
  ↳ PrologToPiTRSProof

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

APP_IN(.(X, Xs), Ys, .(X, Zs)) → APP_IN(Xs, Ys, Zs)

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

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
                        ↳ NonTerminationProof
              ↳ PiDP
              ↳ PiDP
              ↳ PiDP
  ↳ PrologToPiTRSProof

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

APP_IN → APP_IN

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
We used the non-termination processor [17] to show that the DP problem is infinite.
Found a loop by semiunifying a rule from P directly.

The TRS P consists of the following rules:

APP_IN → APP_IN

The TRS R consists of the following rules:none


s = APP_IN evaluates to t =APP_IN

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:



Rewriting sequence

The DP semiunifies directly so there is only one rewrite step from APP_IN to APP_IN.





↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
              ↳ PiDP
              ↳ PiDP
  ↳ PrologToPiTRSProof

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(.(X, Xs), Ys) → U1(X, Xs, Ys, part_in(X, Xs, Littles, Bigs))
part_in(X, [], [], []) → part_out(X, [], [], [])
part_in(X, .(Y, Xs), Ls, .(Y, Bs)) → U7(X, Y, Xs, Ls, Bs, part_in(X, Xs, Ls, Bs))
part_in(X, .(Y, Xs), .(Y, Ls), Bs) → U5(X, Y, Xs, Ls, Bs, less_in(X, Y))
less_in(s(X), s(Y)) → U9(X, Y, less_in(X, Y))
less_in(0, s(X)) → less_out(0, s(X))
U9(X, Y, less_out(X, Y)) → less_out(s(X), s(Y))
U5(X, Y, Xs, Ls, Bs, less_out(X, Y)) → U6(X, Y, Xs, Ls, Bs, part_in(X, Xs, Ls, Bs))
U6(X, Y, Xs, Ls, Bs, part_out(X, Xs, Ls, Bs)) → part_out(X, .(Y, Xs), .(Y, Ls), Bs)
U7(X, Y, Xs, Ls, Bs, part_out(X, Xs, Ls, Bs)) → part_out(X, .(Y, Xs), Ls, .(Y, Bs))
U1(X, Xs, Ys, part_out(X, Xs, Littles, Bigs)) → U2(X, Xs, Ys, Bigs, qs_in(Littles, Ls))
qs_in([], []) → qs_out([], [])
U2(X, Xs, Ys, Bigs, qs_out(Littles, Ls)) → U3(X, Xs, Ys, Ls, qs_in(Bigs, Bs))
U3(X, Xs, Ys, Ls, qs_out(Bigs, Bs)) → U4(X, Xs, Ys, app_in(Ls, .(X, Bs), Ys))
app_in(.(X, Xs), Ys, .(X, Zs)) → U8(X, Xs, Ys, Zs, app_in(Xs, Ys, Zs))
app_in([], X, X) → app_out([], X, X)
U8(X, Xs, Ys, Zs, app_out(Xs, Ys, Zs)) → app_out(.(X, Xs), Ys, .(X, Zs))
U4(X, Xs, Ys, app_out(Ls, .(X, Bs), Ys)) → qs_out(.(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
qs_in(x1, x2)  =  qs_in
.(x1, x2)  =  .(x1, x2)
U1(x1, x2, x3, x4)  =  U1(x4)
part_in(x1, x2, x3, x4)  =  part_in
[]  =  []
part_out(x1, x2, x3, x4)  =  part_out
U7(x1, x2, x3, x4, x5, x6)  =  U7(x6)
U5(x1, x2, x3, x4, x5, x6)  =  U5(x6)
less_in(x1, x2)  =  less_in
s(x1)  =  s(x1)
U9(x1, x2, x3)  =  U9(x3)
0  =  0
less_out(x1, x2)  =  less_out(x1)
U6(x1, x2, x3, x4, x5, x6)  =  U6(x6)
U2(x1, x2, x3, x4, x5)  =  U2(x5)
qs_out(x1, x2)  =  qs_out
U3(x1, x2, x3, x4, x5)  =  U3(x5)
U4(x1, x2, x3, x4)  =  U4(x4)
app_in(x1, x2, x3)  =  app_in
U8(x1, x2, x3, x4, x5)  =  U8(x5)
app_out(x1, x2, x3)  =  app_out
LESS_IN(x1, x2)  =  LESS_IN

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
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
              ↳ PiDP
              ↳ PiDP
  ↳ PrologToPiTRSProof

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

LESS_IN(s(X), s(Y)) → LESS_IN(X, Y)

R is empty.
The argument filtering Pi contains the following mapping:
s(x1)  =  s(x1)
LESS_IN(x1, x2)  =  LESS_IN

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
                        ↳ NonTerminationProof
              ↳ PiDP
              ↳ PiDP
  ↳ PrologToPiTRSProof

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

LESS_IN → LESS_IN

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
We used the non-termination processor [17] to show that the DP problem is infinite.
Found a loop by semiunifying a rule from P directly.

The TRS P consists of the following rules:

LESS_IN → LESS_IN

The TRS R consists of the following rules:none


s = LESS_IN evaluates to t =LESS_IN

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:



Rewriting sequence

The DP semiunifies directly so there is only one rewrite step from LESS_IN to LESS_IN.





↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
              ↳ PiDP
  ↳ PrologToPiTRSProof

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

PART_IN(X, .(Y, Xs), Ls, .(Y, Bs)) → PART_IN(X, Xs, Ls, Bs)
U51(X, Y, Xs, Ls, Bs, less_out(X, Y)) → PART_IN(X, Xs, Ls, Bs)
PART_IN(X, .(Y, Xs), .(Y, Ls), Bs) → U51(X, Y, Xs, Ls, Bs, less_in(X, Y))

The TRS R consists of the following rules:

qs_in(.(X, Xs), Ys) → U1(X, Xs, Ys, part_in(X, Xs, Littles, Bigs))
part_in(X, [], [], []) → part_out(X, [], [], [])
part_in(X, .(Y, Xs), Ls, .(Y, Bs)) → U7(X, Y, Xs, Ls, Bs, part_in(X, Xs, Ls, Bs))
part_in(X, .(Y, Xs), .(Y, Ls), Bs) → U5(X, Y, Xs, Ls, Bs, less_in(X, Y))
less_in(s(X), s(Y)) → U9(X, Y, less_in(X, Y))
less_in(0, s(X)) → less_out(0, s(X))
U9(X, Y, less_out(X, Y)) → less_out(s(X), s(Y))
U5(X, Y, Xs, Ls, Bs, less_out(X, Y)) → U6(X, Y, Xs, Ls, Bs, part_in(X, Xs, Ls, Bs))
U6(X, Y, Xs, Ls, Bs, part_out(X, Xs, Ls, Bs)) → part_out(X, .(Y, Xs), .(Y, Ls), Bs)
U7(X, Y, Xs, Ls, Bs, part_out(X, Xs, Ls, Bs)) → part_out(X, .(Y, Xs), Ls, .(Y, Bs))
U1(X, Xs, Ys, part_out(X, Xs, Littles, Bigs)) → U2(X, Xs, Ys, Bigs, qs_in(Littles, Ls))
qs_in([], []) → qs_out([], [])
U2(X, Xs, Ys, Bigs, qs_out(Littles, Ls)) → U3(X, Xs, Ys, Ls, qs_in(Bigs, Bs))
U3(X, Xs, Ys, Ls, qs_out(Bigs, Bs)) → U4(X, Xs, Ys, app_in(Ls, .(X, Bs), Ys))
app_in(.(X, Xs), Ys, .(X, Zs)) → U8(X, Xs, Ys, Zs, app_in(Xs, Ys, Zs))
app_in([], X, X) → app_out([], X, X)
U8(X, Xs, Ys, Zs, app_out(Xs, Ys, Zs)) → app_out(.(X, Xs), Ys, .(X, Zs))
U4(X, Xs, Ys, app_out(Ls, .(X, Bs), Ys)) → qs_out(.(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
qs_in(x1, x2)  =  qs_in
.(x1, x2)  =  .(x1, x2)
U1(x1, x2, x3, x4)  =  U1(x4)
part_in(x1, x2, x3, x4)  =  part_in
[]  =  []
part_out(x1, x2, x3, x4)  =  part_out
U7(x1, x2, x3, x4, x5, x6)  =  U7(x6)
U5(x1, x2, x3, x4, x5, x6)  =  U5(x6)
less_in(x1, x2)  =  less_in
s(x1)  =  s(x1)
U9(x1, x2, x3)  =  U9(x3)
0  =  0
less_out(x1, x2)  =  less_out(x1)
U6(x1, x2, x3, x4, x5, x6)  =  U6(x6)
U2(x1, x2, x3, x4, x5)  =  U2(x5)
qs_out(x1, x2)  =  qs_out
U3(x1, x2, x3, x4, x5)  =  U3(x5)
U4(x1, x2, x3, x4)  =  U4(x4)
app_in(x1, x2, x3)  =  app_in
U8(x1, x2, x3, x4, x5)  =  U8(x5)
app_out(x1, x2, x3)  =  app_out
U51(x1, x2, x3, x4, x5, x6)  =  U51(x6)
PART_IN(x1, x2, x3, x4)  =  PART_IN

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
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
              ↳ PiDP
  ↳ PrologToPiTRSProof

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

PART_IN(X, .(Y, Xs), Ls, .(Y, Bs)) → PART_IN(X, Xs, Ls, Bs)
U51(X, Y, Xs, Ls, Bs, less_out(X, Y)) → PART_IN(X, Xs, Ls, Bs)
PART_IN(X, .(Y, Xs), .(Y, Ls), Bs) → U51(X, Y, Xs, Ls, Bs, less_in(X, Y))

The TRS R consists of the following rules:

less_in(s(X), s(Y)) → U9(X, Y, less_in(X, Y))
less_in(0, s(X)) → less_out(0, s(X))
U9(X, Y, less_out(X, Y)) → less_out(s(X), s(Y))

The argument filtering Pi contains the following mapping:
.(x1, x2)  =  .(x1, x2)
less_in(x1, x2)  =  less_in
s(x1)  =  s(x1)
U9(x1, x2, x3)  =  U9(x3)
0  =  0
less_out(x1, x2)  =  less_out(x1)
U51(x1, x2, x3, x4, x5, x6)  =  U51(x6)
PART_IN(x1, x2, x3, x4)  =  PART_IN

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
                        ↳ Narrowing
              ↳ PiDP
  ↳ PrologToPiTRSProof

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

PART_IN → U51(less_in)
U51(less_out(X)) → PART_IN
PART_IN → PART_IN

The TRS R consists of the following rules:

less_in → U9(less_in)
less_in → less_out(0)
U9(less_out(X)) → less_out(s(X))

The set Q consists of the following terms:

less_in
U9(x0)

We have to consider all (P,Q,R)-chains.
By narrowing [15] the rule PART_IN → U51(less_in) at position [0] we obtained the following new rules:

PART_IN → U51(less_out(0))
PART_IN → U51(U9(less_in))



↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
                      ↳ QDP
                        ↳ Narrowing
                          ↳ QDP
                            ↳ NonTerminationProof
              ↳ PiDP
  ↳ PrologToPiTRSProof

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

PART_IN → U51(less_out(0))
PART_IN → U51(U9(less_in))
U51(less_out(X)) → PART_IN
PART_IN → PART_IN

The TRS R consists of the following rules:

less_in → U9(less_in)
less_in → less_out(0)
U9(less_out(X)) → less_out(s(X))

The set Q consists of the following terms:

less_in
U9(x0)

We have to consider all (P,Q,R)-chains.
We used the non-termination processor [17] to show that the DP problem is infinite.
Found a loop by semiunifying a rule from P directly.

The TRS P consists of the following rules:

PART_IN → U51(less_out(0))
PART_IN → U51(U9(less_in))
U51(less_out(X)) → PART_IN
PART_IN → PART_IN

The TRS R consists of the following rules:

less_in → U9(less_in)
less_in → less_out(0)
U9(less_out(X)) → less_out(s(X))


s = PART_IN evaluates to t =PART_IN

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:



Rewriting sequence

The DP semiunifies directly so there is only one rewrite step from PART_IN to PART_IN.





↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ PiDP
              ↳ PiDP
              ↳ PiDP
                ↳ PiDPToQDPProof
  ↳ PrologToPiTRSProof

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

U11(X, Xs, Ys, part_out(X, Xs, Littles, Bigs)) → U21(X, Xs, Ys, Bigs, qs_in(Littles, Ls))
U11(X, Xs, Ys, part_out(X, Xs, Littles, Bigs)) → QS_IN(Littles, Ls)
QS_IN(.(X, Xs), Ys) → U11(X, Xs, Ys, part_in(X, Xs, Littles, Bigs))
U21(X, Xs, Ys, Bigs, qs_out(Littles, Ls)) → QS_IN(Bigs, Bs)

The TRS R consists of the following rules:

qs_in(.(X, Xs), Ys) → U1(X, Xs, Ys, part_in(X, Xs, Littles, Bigs))
part_in(X, [], [], []) → part_out(X, [], [], [])
part_in(X, .(Y, Xs), Ls, .(Y, Bs)) → U7(X, Y, Xs, Ls, Bs, part_in(X, Xs, Ls, Bs))
part_in(X, .(Y, Xs), .(Y, Ls), Bs) → U5(X, Y, Xs, Ls, Bs, less_in(X, Y))
less_in(s(X), s(Y)) → U9(X, Y, less_in(X, Y))
less_in(0, s(X)) → less_out(0, s(X))
U9(X, Y, less_out(X, Y)) → less_out(s(X), s(Y))
U5(X, Y, Xs, Ls, Bs, less_out(X, Y)) → U6(X, Y, Xs, Ls, Bs, part_in(X, Xs, Ls, Bs))
U6(X, Y, Xs, Ls, Bs, part_out(X, Xs, Ls, Bs)) → part_out(X, .(Y, Xs), .(Y, Ls), Bs)
U7(X, Y, Xs, Ls, Bs, part_out(X, Xs, Ls, Bs)) → part_out(X, .(Y, Xs), Ls, .(Y, Bs))
U1(X, Xs, Ys, part_out(X, Xs, Littles, Bigs)) → U2(X, Xs, Ys, Bigs, qs_in(Littles, Ls))
qs_in([], []) → qs_out([], [])
U2(X, Xs, Ys, Bigs, qs_out(Littles, Ls)) → U3(X, Xs, Ys, Ls, qs_in(Bigs, Bs))
U3(X, Xs, Ys, Ls, qs_out(Bigs, Bs)) → U4(X, Xs, Ys, app_in(Ls, .(X, Bs), Ys))
app_in(.(X, Xs), Ys, .(X, Zs)) → U8(X, Xs, Ys, Zs, app_in(Xs, Ys, Zs))
app_in([], X, X) → app_out([], X, X)
U8(X, Xs, Ys, Zs, app_out(Xs, Ys, Zs)) → app_out(.(X, Xs), Ys, .(X, Zs))
U4(X, Xs, Ys, app_out(Ls, .(X, Bs), Ys)) → qs_out(.(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
qs_in(x1, x2)  =  qs_in
.(x1, x2)  =  .(x1, x2)
U1(x1, x2, x3, x4)  =  U1(x4)
part_in(x1, x2, x3, x4)  =  part_in
[]  =  []
part_out(x1, x2, x3, x4)  =  part_out
U7(x1, x2, x3, x4, x5, x6)  =  U7(x6)
U5(x1, x2, x3, x4, x5, x6)  =  U5(x6)
less_in(x1, x2)  =  less_in
s(x1)  =  s(x1)
U9(x1, x2, x3)  =  U9(x3)
0  =  0
less_out(x1, x2)  =  less_out(x1)
U6(x1, x2, x3, x4, x5, x6)  =  U6(x6)
U2(x1, x2, x3, x4, x5)  =  U2(x5)
qs_out(x1, x2)  =  qs_out
U3(x1, x2, x3, x4, x5)  =  U3(x5)
U4(x1, x2, x3, x4)  =  U4(x4)
app_in(x1, x2, x3)  =  app_in
U8(x1, x2, x3, x4, x5)  =  U8(x5)
app_out(x1, x2, x3)  =  app_out
QS_IN(x1, x2)  =  QS_IN
U21(x1, x2, x3, x4, x5)  =  U21(x5)
U11(x1, x2, x3, x4)  =  U11(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
                ↳ PiDPToQDPProof
                  ↳ QDP
                    ↳ Narrowing
  ↳ PrologToPiTRSProof

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

U21(qs_out) → QS_IN
U11(part_out) → QS_IN
U11(part_out) → U21(qs_in)
QS_IN → U11(part_in)

The TRS R consists of the following rules:

qs_in → U1(part_in)
part_in → part_out
part_in → U7(part_in)
part_in → U5(less_in)
less_in → U9(less_in)
less_in → less_out(0)
U9(less_out(X)) → less_out(s(X))
U5(less_out(X)) → U6(part_in)
U6(part_out) → part_out
U7(part_out) → part_out
U1(part_out) → U2(qs_in)
qs_in → qs_out
U2(qs_out) → U3(qs_in)
U3(qs_out) → U4(app_in)
app_in → U8(app_in)
app_in → app_out
U8(app_out) → app_out
U4(app_out) → qs_out

The set Q consists of the following terms:

qs_in
part_in
less_in
U9(x0)
U5(x0)
U6(x0)
U7(x0)
U1(x0)
U2(x0)
U3(x0)
app_in
U8(x0)
U4(x0)

We have to consider all (P,Q,R)-chains.
By narrowing [15] the rule U11(part_out) → U21(qs_in) at position [0] we obtained the following new rules:

U11(part_out) → U21(U1(part_in))
U11(part_out) → U21(qs_out)



↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ PiDP
              ↳ PiDP
              ↳ PiDP
                ↳ PiDPToQDPProof
                  ↳ QDP
                    ↳ Narrowing
                      ↳ QDP
                        ↳ Narrowing
  ↳ PrologToPiTRSProof

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

U11(part_out) → QS_IN
U21(qs_out) → QS_IN
U11(part_out) → U21(U1(part_in))
QS_IN → U11(part_in)
U11(part_out) → U21(qs_out)

The TRS R consists of the following rules:

qs_in → U1(part_in)
part_in → part_out
part_in → U7(part_in)
part_in → U5(less_in)
less_in → U9(less_in)
less_in → less_out(0)
U9(less_out(X)) → less_out(s(X))
U5(less_out(X)) → U6(part_in)
U6(part_out) → part_out
U7(part_out) → part_out
U1(part_out) → U2(qs_in)
qs_in → qs_out
U2(qs_out) → U3(qs_in)
U3(qs_out) → U4(app_in)
app_in → U8(app_in)
app_in → app_out
U8(app_out) → app_out
U4(app_out) → qs_out

The set Q consists of the following terms:

qs_in
part_in
less_in
U9(x0)
U5(x0)
U6(x0)
U7(x0)
U1(x0)
U2(x0)
U3(x0)
app_in
U8(x0)
U4(x0)

We have to consider all (P,Q,R)-chains.
By narrowing [15] the rule QS_IN → U11(part_in) at position [0] we obtained the following new rules:

QS_IN → U11(U7(part_in))
QS_IN → U11(U5(less_in))
QS_IN → U11(part_out)



↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ PiDP
              ↳ PiDP
              ↳ PiDP
                ↳ PiDPToQDPProof
                  ↳ QDP
                    ↳ Narrowing
                      ↳ QDP
                        ↳ Narrowing
                          ↳ QDP
                            ↳ NonTerminationProof
  ↳ PrologToPiTRSProof

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

QS_IN → U11(U7(part_in))
U21(qs_out) → QS_IN
U11(part_out) → QS_IN
QS_IN → U11(part_out)
U11(part_out) → U21(U1(part_in))
QS_IN → U11(U5(less_in))
U11(part_out) → U21(qs_out)

The TRS R consists of the following rules:

qs_in → U1(part_in)
part_in → part_out
part_in → U7(part_in)
part_in → U5(less_in)
less_in → U9(less_in)
less_in → less_out(0)
U9(less_out(X)) → less_out(s(X))
U5(less_out(X)) → U6(part_in)
U6(part_out) → part_out
U7(part_out) → part_out
U1(part_out) → U2(qs_in)
qs_in → qs_out
U2(qs_out) → U3(qs_in)
U3(qs_out) → U4(app_in)
app_in → U8(app_in)
app_in → app_out
U8(app_out) → app_out
U4(app_out) → qs_out

The set Q consists of the following terms:

qs_in
part_in
less_in
U9(x0)
U5(x0)
U6(x0)
U7(x0)
U1(x0)
U2(x0)
U3(x0)
app_in
U8(x0)
U4(x0)

We have to consider all (P,Q,R)-chains.
We used the non-termination processor [17] to show that the DP problem is infinite.
Found a loop by narrowing to the left:

The TRS P consists of the following rules:

QS_IN → U11(U7(part_in))
U21(qs_out) → QS_IN
U11(part_out) → QS_IN
QS_IN → U11(part_out)
U11(part_out) → U21(U1(part_in))
QS_IN → U11(U5(less_in))
U11(part_out) → U21(qs_out)

The TRS R consists of the following rules:

qs_in → U1(part_in)
part_in → part_out
part_in → U7(part_in)
part_in → U5(less_in)
less_in → U9(less_in)
less_in → less_out(0)
U9(less_out(X)) → less_out(s(X))
U5(less_out(X)) → U6(part_in)
U6(part_out) → part_out
U7(part_out) → part_out
U1(part_out) → U2(qs_in)
qs_in → qs_out
U2(qs_out) → U3(qs_in)
U3(qs_out) → U4(app_in)
app_in → U8(app_in)
app_in → app_out
U8(app_out) → app_out
U4(app_out) → qs_out


s = QS_IN evaluates to t =QS_IN

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:



Rewriting sequence

QS_IN → U11(part_out)
with rule QS_IN → U11(part_out) at position [] and matcher [ ]

U11(part_out) → QS_IN
with rule U11(part_out) → QS_IN

Now applying the matcher to the start term leads to a term which is equal to the last term in the rewriting sequence


All these steps are and every following step will be a correct step w.r.t to Q.




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(.(X, Xs), Ys) → U1(X, Xs, Ys, part_in(X, Xs, Littles, Bigs))
part_in(X, [], [], []) → part_out(X, [], [], [])
part_in(X, .(Y, Xs), Ls, .(Y, Bs)) → U7(X, Y, Xs, Ls, Bs, part_in(X, Xs, Ls, Bs))
part_in(X, .(Y, Xs), .(Y, Ls), Bs) → U5(X, Y, Xs, Ls, Bs, less_in(X, Y))
less_in(s(X), s(Y)) → U9(X, Y, less_in(X, Y))
less_in(0, s(X)) → less_out(0, s(X))
U9(X, Y, less_out(X, Y)) → less_out(s(X), s(Y))
U5(X, Y, Xs, Ls, Bs, less_out(X, Y)) → U6(X, Y, Xs, Ls, Bs, part_in(X, Xs, Ls, Bs))
U6(X, Y, Xs, Ls, Bs, part_out(X, Xs, Ls, Bs)) → part_out(X, .(Y, Xs), .(Y, Ls), Bs)
U7(X, Y, Xs, Ls, Bs, part_out(X, Xs, Ls, Bs)) → part_out(X, .(Y, Xs), Ls, .(Y, Bs))
U1(X, Xs, Ys, part_out(X, Xs, Littles, Bigs)) → U2(X, Xs, Ys, Bigs, qs_in(Littles, Ls))
qs_in([], []) → qs_out([], [])
U2(X, Xs, Ys, Bigs, qs_out(Littles, Ls)) → U3(X, Xs, Ys, Ls, qs_in(Bigs, Bs))
U3(X, Xs, Ys, Ls, qs_out(Bigs, Bs)) → U4(X, Xs, Ys, app_in(Ls, .(X, Bs), Ys))
app_in(.(X, Xs), Ys, .(X, Zs)) → U8(X, Xs, Ys, Zs, app_in(Xs, Ys, Zs))
app_in([], X, X) → app_out([], X, X)
U8(X, Xs, Ys, Zs, app_out(Xs, Ys, Zs)) → app_out(.(X, Xs), Ys, .(X, Zs))
U4(X, Xs, Ys, app_out(Ls, .(X, Bs), Ys)) → qs_out(.(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
qs_in(x1, x2)  =  qs_in
.(x1, x2)  =  .(x1, x2)
U1(x1, x2, x3, x4)  =  U1(x4)
part_in(x1, x2, x3, x4)  =  part_in
[]  =  []
part_out(x1, x2, x3, x4)  =  part_out
U7(x1, x2, x3, x4, x5, x6)  =  U7(x6)
U5(x1, x2, x3, x4, x5, x6)  =  U5(x6)
less_in(x1, x2)  =  less_in
s(x1)  =  s(x1)
U9(x1, x2, x3)  =  U9(x3)
0  =  0
less_out(x1, x2)  =  less_out(x1)
U6(x1, x2, x3, x4, x5, x6)  =  U6(x6)
U2(x1, x2, x3, x4, x5)  =  U2(x5)
qs_out(x1, x2)  =  qs_out
U3(x1, x2, x3, x4, x5)  =  U3(x5)
U4(x1, x2, x3, x4)  =  U4(x4)
app_in(x1, x2, x3)  =  app_in
U8(x1, x2, x3, x4, x5)  =  U8(x5)
app_out(x1, x2, x3)  =  app_out

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog



↳ Prolog
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof

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

qs_in(.(X, Xs), Ys) → U1(X, Xs, Ys, part_in(X, Xs, Littles, Bigs))
part_in(X, [], [], []) → part_out(X, [], [], [])
part_in(X, .(Y, Xs), Ls, .(Y, Bs)) → U7(X, Y, Xs, Ls, Bs, part_in(X, Xs, Ls, Bs))
part_in(X, .(Y, Xs), .(Y, Ls), Bs) → U5(X, Y, Xs, Ls, Bs, less_in(X, Y))
less_in(s(X), s(Y)) → U9(X, Y, less_in(X, Y))
less_in(0, s(X)) → less_out(0, s(X))
U9(X, Y, less_out(X, Y)) → less_out(s(X), s(Y))
U5(X, Y, Xs, Ls, Bs, less_out(X, Y)) → U6(X, Y, Xs, Ls, Bs, part_in(X, Xs, Ls, Bs))
U6(X, Y, Xs, Ls, Bs, part_out(X, Xs, Ls, Bs)) → part_out(X, .(Y, Xs), .(Y, Ls), Bs)
U7(X, Y, Xs, Ls, Bs, part_out(X, Xs, Ls, Bs)) → part_out(X, .(Y, Xs), Ls, .(Y, Bs))
U1(X, Xs, Ys, part_out(X, Xs, Littles, Bigs)) → U2(X, Xs, Ys, Bigs, qs_in(Littles, Ls))
qs_in([], []) → qs_out([], [])
U2(X, Xs, Ys, Bigs, qs_out(Littles, Ls)) → U3(X, Xs, Ys, Ls, qs_in(Bigs, Bs))
U3(X, Xs, Ys, Ls, qs_out(Bigs, Bs)) → U4(X, Xs, Ys, app_in(Ls, .(X, Bs), Ys))
app_in(.(X, Xs), Ys, .(X, Zs)) → U8(X, Xs, Ys, Zs, app_in(Xs, Ys, Zs))
app_in([], X, X) → app_out([], X, X)
U8(X, Xs, Ys, Zs, app_out(Xs, Ys, Zs)) → app_out(.(X, Xs), Ys, .(X, Zs))
U4(X, Xs, Ys, app_out(Ls, .(X, Bs), Ys)) → qs_out(.(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
qs_in(x1, x2)  =  qs_in
.(x1, x2)  =  .(x1, x2)
U1(x1, x2, x3, x4)  =  U1(x4)
part_in(x1, x2, x3, x4)  =  part_in
[]  =  []
part_out(x1, x2, x3, x4)  =  part_out
U7(x1, x2, x3, x4, x5, x6)  =  U7(x6)
U5(x1, x2, x3, x4, x5, x6)  =  U5(x6)
less_in(x1, x2)  =  less_in
s(x1)  =  s(x1)
U9(x1, x2, x3)  =  U9(x3)
0  =  0
less_out(x1, x2)  =  less_out(x1)
U6(x1, x2, x3, x4, x5, x6)  =  U6(x6)
U2(x1, x2, x3, x4, x5)  =  U2(x5)
qs_out(x1, x2)  =  qs_out
U3(x1, x2, x3, x4, x5)  =  U3(x5)
U4(x1, x2, x3, x4)  =  U4(x4)
app_in(x1, x2, x3)  =  app_in
U8(x1, x2, x3, x4, x5)  =  U8(x5)
app_out(x1, x2, x3)  =  app_out


Using Dependency Pairs [1,30] we result in the following initial DP problem:
Pi DP problem:
The TRS P consists of the following rules:

QS_IN(.(X, Xs), Ys) → U11(X, Xs, Ys, part_in(X, Xs, Littles, Bigs))
QS_IN(.(X, Xs), Ys) → PART_IN(X, Xs, Littles, Bigs)
PART_IN(X, .(Y, Xs), Ls, .(Y, Bs)) → U71(X, Y, Xs, Ls, Bs, part_in(X, Xs, Ls, Bs))
PART_IN(X, .(Y, Xs), Ls, .(Y, Bs)) → PART_IN(X, Xs, Ls, Bs)
PART_IN(X, .(Y, Xs), .(Y, Ls), Bs) → U51(X, Y, Xs, Ls, Bs, less_in(X, Y))
PART_IN(X, .(Y, Xs), .(Y, Ls), Bs) → LESS_IN(X, Y)
LESS_IN(s(X), s(Y)) → U91(X, Y, less_in(X, Y))
LESS_IN(s(X), s(Y)) → LESS_IN(X, Y)
U51(X, Y, Xs, Ls, Bs, less_out(X, Y)) → U61(X, Y, Xs, Ls, Bs, part_in(X, Xs, Ls, Bs))
U51(X, Y, Xs, Ls, Bs, less_out(X, Y)) → PART_IN(X, Xs, Ls, Bs)
U11(X, Xs, Ys, part_out(X, Xs, Littles, Bigs)) → U21(X, Xs, Ys, Bigs, qs_in(Littles, Ls))
U11(X, Xs, Ys, part_out(X, Xs, Littles, Bigs)) → QS_IN(Littles, Ls)
U21(X, Xs, Ys, Bigs, qs_out(Littles, Ls)) → U31(X, Xs, Ys, Ls, qs_in(Bigs, Bs))
U21(X, Xs, Ys, Bigs, qs_out(Littles, Ls)) → QS_IN(Bigs, Bs)
U31(X, Xs, Ys, Ls, qs_out(Bigs, Bs)) → U41(X, Xs, Ys, app_in(Ls, .(X, Bs), Ys))
U31(X, Xs, Ys, Ls, qs_out(Bigs, Bs)) → APP_IN(Ls, .(X, Bs), Ys)
APP_IN(.(X, Xs), Ys, .(X, Zs)) → U81(X, Xs, Ys, Zs, app_in(Xs, Ys, Zs))
APP_IN(.(X, Xs), Ys, .(X, Zs)) → APP_IN(Xs, Ys, Zs)

The TRS R consists of the following rules:

qs_in(.(X, Xs), Ys) → U1(X, Xs, Ys, part_in(X, Xs, Littles, Bigs))
part_in(X, [], [], []) → part_out(X, [], [], [])
part_in(X, .(Y, Xs), Ls, .(Y, Bs)) → U7(X, Y, Xs, Ls, Bs, part_in(X, Xs, Ls, Bs))
part_in(X, .(Y, Xs), .(Y, Ls), Bs) → U5(X, Y, Xs, Ls, Bs, less_in(X, Y))
less_in(s(X), s(Y)) → U9(X, Y, less_in(X, Y))
less_in(0, s(X)) → less_out(0, s(X))
U9(X, Y, less_out(X, Y)) → less_out(s(X), s(Y))
U5(X, Y, Xs, Ls, Bs, less_out(X, Y)) → U6(X, Y, Xs, Ls, Bs, part_in(X, Xs, Ls, Bs))
U6(X, Y, Xs, Ls, Bs, part_out(X, Xs, Ls, Bs)) → part_out(X, .(Y, Xs), .(Y, Ls), Bs)
U7(X, Y, Xs, Ls, Bs, part_out(X, Xs, Ls, Bs)) → part_out(X, .(Y, Xs), Ls, .(Y, Bs))
U1(X, Xs, Ys, part_out(X, Xs, Littles, Bigs)) → U2(X, Xs, Ys, Bigs, qs_in(Littles, Ls))
qs_in([], []) → qs_out([], [])
U2(X, Xs, Ys, Bigs, qs_out(Littles, Ls)) → U3(X, Xs, Ys, Ls, qs_in(Bigs, Bs))
U3(X, Xs, Ys, Ls, qs_out(Bigs, Bs)) → U4(X, Xs, Ys, app_in(Ls, .(X, Bs), Ys))
app_in(.(X, Xs), Ys, .(X, Zs)) → U8(X, Xs, Ys, Zs, app_in(Xs, Ys, Zs))
app_in([], X, X) → app_out([], X, X)
U8(X, Xs, Ys, Zs, app_out(Xs, Ys, Zs)) → app_out(.(X, Xs), Ys, .(X, Zs))
U4(X, Xs, Ys, app_out(Ls, .(X, Bs), Ys)) → qs_out(.(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
qs_in(x1, x2)  =  qs_in
.(x1, x2)  =  .(x1, x2)
U1(x1, x2, x3, x4)  =  U1(x4)
part_in(x1, x2, x3, x4)  =  part_in
[]  =  []
part_out(x1, x2, x3, x4)  =  part_out
U7(x1, x2, x3, x4, x5, x6)  =  U7(x6)
U5(x1, x2, x3, x4, x5, x6)  =  U5(x6)
less_in(x1, x2)  =  less_in
s(x1)  =  s(x1)
U9(x1, x2, x3)  =  U9(x3)
0  =  0
less_out(x1, x2)  =  less_out(x1)
U6(x1, x2, x3, x4, x5, x6)  =  U6(x6)
U2(x1, x2, x3, x4, x5)  =  U2(x5)
qs_out(x1, x2)  =  qs_out
U3(x1, x2, x3, x4, x5)  =  U3(x5)
U4(x1, x2, x3, x4)  =  U4(x4)
app_in(x1, x2, x3)  =  app_in
U8(x1, x2, x3, x4, x5)  =  U8(x5)
app_out(x1, x2, x3)  =  app_out
U51(x1, x2, x3, x4, x5, x6)  =  U51(x6)
U41(x1, x2, x3, x4)  =  U41(x4)
U31(x1, x2, x3, x4, x5)  =  U31(x5)
PART_IN(x1, x2, x3, x4)  =  PART_IN
U91(x1, x2, x3)  =  U91(x3)
QS_IN(x1, x2)  =  QS_IN
U71(x1, x2, x3, x4, x5, x6)  =  U71(x6)
APP_IN(x1, x2, x3)  =  APP_IN
LESS_IN(x1, x2)  =  LESS_IN
U81(x1, x2, x3, x4, x5)  =  U81(x5)
U21(x1, x2, x3, x4, x5)  =  U21(x5)
U11(x1, x2, x3, x4)  =  U11(x4)
U61(x1, x2, x3, x4, x5, x6)  =  U61(x6)

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

↳ Prolog
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof

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

QS_IN(.(X, Xs), Ys) → U11(X, Xs, Ys, part_in(X, Xs, Littles, Bigs))
QS_IN(.(X, Xs), Ys) → PART_IN(X, Xs, Littles, Bigs)
PART_IN(X, .(Y, Xs), Ls, .(Y, Bs)) → U71(X, Y, Xs, Ls, Bs, part_in(X, Xs, Ls, Bs))
PART_IN(X, .(Y, Xs), Ls, .(Y, Bs)) → PART_IN(X, Xs, Ls, Bs)
PART_IN(X, .(Y, Xs), .(Y, Ls), Bs) → U51(X, Y, Xs, Ls, Bs, less_in(X, Y))
PART_IN(X, .(Y, Xs), .(Y, Ls), Bs) → LESS_IN(X, Y)
LESS_IN(s(X), s(Y)) → U91(X, Y, less_in(X, Y))
LESS_IN(s(X), s(Y)) → LESS_IN(X, Y)
U51(X, Y, Xs, Ls, Bs, less_out(X, Y)) → U61(X, Y, Xs, Ls, Bs, part_in(X, Xs, Ls, Bs))
U51(X, Y, Xs, Ls, Bs, less_out(X, Y)) → PART_IN(X, Xs, Ls, Bs)
U11(X, Xs, Ys, part_out(X, Xs, Littles, Bigs)) → U21(X, Xs, Ys, Bigs, qs_in(Littles, Ls))
U11(X, Xs, Ys, part_out(X, Xs, Littles, Bigs)) → QS_IN(Littles, Ls)
U21(X, Xs, Ys, Bigs, qs_out(Littles, Ls)) → U31(X, Xs, Ys, Ls, qs_in(Bigs, Bs))
U21(X, Xs, Ys, Bigs, qs_out(Littles, Ls)) → QS_IN(Bigs, Bs)
U31(X, Xs, Ys, Ls, qs_out(Bigs, Bs)) → U41(X, Xs, Ys, app_in(Ls, .(X, Bs), Ys))
U31(X, Xs, Ys, Ls, qs_out(Bigs, Bs)) → APP_IN(Ls, .(X, Bs), Ys)
APP_IN(.(X, Xs), Ys, .(X, Zs)) → U81(X, Xs, Ys, Zs, app_in(Xs, Ys, Zs))
APP_IN(.(X, Xs), Ys, .(X, Zs)) → APP_IN(Xs, Ys, Zs)

The TRS R consists of the following rules:

qs_in(.(X, Xs), Ys) → U1(X, Xs, Ys, part_in(X, Xs, Littles, Bigs))
part_in(X, [], [], []) → part_out(X, [], [], [])
part_in(X, .(Y, Xs), Ls, .(Y, Bs)) → U7(X, Y, Xs, Ls, Bs, part_in(X, Xs, Ls, Bs))
part_in(X, .(Y, Xs), .(Y, Ls), Bs) → U5(X, Y, Xs, Ls, Bs, less_in(X, Y))
less_in(s(X), s(Y)) → U9(X, Y, less_in(X, Y))
less_in(0, s(X)) → less_out(0, s(X))
U9(X, Y, less_out(X, Y)) → less_out(s(X), s(Y))
U5(X, Y, Xs, Ls, Bs, less_out(X, Y)) → U6(X, Y, Xs, Ls, Bs, part_in(X, Xs, Ls, Bs))
U6(X, Y, Xs, Ls, Bs, part_out(X, Xs, Ls, Bs)) → part_out(X, .(Y, Xs), .(Y, Ls), Bs)
U7(X, Y, Xs, Ls, Bs, part_out(X, Xs, Ls, Bs)) → part_out(X, .(Y, Xs), Ls, .(Y, Bs))
U1(X, Xs, Ys, part_out(X, Xs, Littles, Bigs)) → U2(X, Xs, Ys, Bigs, qs_in(Littles, Ls))
qs_in([], []) → qs_out([], [])
U2(X, Xs, Ys, Bigs, qs_out(Littles, Ls)) → U3(X, Xs, Ys, Ls, qs_in(Bigs, Bs))
U3(X, Xs, Ys, Ls, qs_out(Bigs, Bs)) → U4(X, Xs, Ys, app_in(Ls, .(X, Bs), Ys))
app_in(.(X, Xs), Ys, .(X, Zs)) → U8(X, Xs, Ys, Zs, app_in(Xs, Ys, Zs))
app_in([], X, X) → app_out([], X, X)
U8(X, Xs, Ys, Zs, app_out(Xs, Ys, Zs)) → app_out(.(X, Xs), Ys, .(X, Zs))
U4(X, Xs, Ys, app_out(Ls, .(X, Bs), Ys)) → qs_out(.(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
qs_in(x1, x2)  =  qs_in
.(x1, x2)  =  .(x1, x2)
U1(x1, x2, x3, x4)  =  U1(x4)
part_in(x1, x2, x3, x4)  =  part_in
[]  =  []
part_out(x1, x2, x3, x4)  =  part_out
U7(x1, x2, x3, x4, x5, x6)  =  U7(x6)
U5(x1, x2, x3, x4, x5, x6)  =  U5(x6)
less_in(x1, x2)  =  less_in
s(x1)  =  s(x1)
U9(x1, x2, x3)  =  U9(x3)
0  =  0
less_out(x1, x2)  =  less_out(x1)
U6(x1, x2, x3, x4, x5, x6)  =  U6(x6)
U2(x1, x2, x3, x4, x5)  =  U2(x5)
qs_out(x1, x2)  =  qs_out
U3(x1, x2, x3, x4, x5)  =  U3(x5)
U4(x1, x2, x3, x4)  =  U4(x4)
app_in(x1, x2, x3)  =  app_in
U8(x1, x2, x3, x4, x5)  =  U8(x5)
app_out(x1, x2, x3)  =  app_out
U51(x1, x2, x3, x4, x5, x6)  =  U51(x6)
U41(x1, x2, x3, x4)  =  U41(x4)
U31(x1, x2, x3, x4, x5)  =  U31(x5)
PART_IN(x1, x2, x3, x4)  =  PART_IN
U91(x1, x2, x3)  =  U91(x3)
QS_IN(x1, x2)  =  QS_IN
U71(x1, x2, x3, x4, x5, x6)  =  U71(x6)
APP_IN(x1, x2, x3)  =  APP_IN
LESS_IN(x1, x2)  =  LESS_IN
U81(x1, x2, x3, x4, x5)  =  U81(x5)
U21(x1, x2, x3, x4, x5)  =  U21(x5)
U11(x1, x2, x3, x4)  =  U11(x4)
U61(x1, x2, x3, x4, x5, x6)  =  U61(x6)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph [30] contains 4 SCCs with 9 less nodes.

↳ Prolog
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
                ↳ UsableRulesProof
              ↳ PiDP
              ↳ PiDP
              ↳ PiDP

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

APP_IN(.(X, Xs), Ys, .(X, Zs)) → APP_IN(Xs, Ys, Zs)

The TRS R consists of the following rules:

qs_in(.(X, Xs), Ys) → U1(X, Xs, Ys, part_in(X, Xs, Littles, Bigs))
part_in(X, [], [], []) → part_out(X, [], [], [])
part_in(X, .(Y, Xs), Ls, .(Y, Bs)) → U7(X, Y, Xs, Ls, Bs, part_in(X, Xs, Ls, Bs))
part_in(X, .(Y, Xs), .(Y, Ls), Bs) → U5(X, Y, Xs, Ls, Bs, less_in(X, Y))
less_in(s(X), s(Y)) → U9(X, Y, less_in(X, Y))
less_in(0, s(X)) → less_out(0, s(X))
U9(X, Y, less_out(X, Y)) → less_out(s(X), s(Y))
U5(X, Y, Xs, Ls, Bs, less_out(X, Y)) → U6(X, Y, Xs, Ls, Bs, part_in(X, Xs, Ls, Bs))
U6(X, Y, Xs, Ls, Bs, part_out(X, Xs, Ls, Bs)) → part_out(X, .(Y, Xs), .(Y, Ls), Bs)
U7(X, Y, Xs, Ls, Bs, part_out(X, Xs, Ls, Bs)) → part_out(X, .(Y, Xs), Ls, .(Y, Bs))
U1(X, Xs, Ys, part_out(X, Xs, Littles, Bigs)) → U2(X, Xs, Ys, Bigs, qs_in(Littles, Ls))
qs_in([], []) → qs_out([], [])
U2(X, Xs, Ys, Bigs, qs_out(Littles, Ls)) → U3(X, Xs, Ys, Ls, qs_in(Bigs, Bs))
U3(X, Xs, Ys, Ls, qs_out(Bigs, Bs)) → U4(X, Xs, Ys, app_in(Ls, .(X, Bs), Ys))
app_in(.(X, Xs), Ys, .(X, Zs)) → U8(X, Xs, Ys, Zs, app_in(Xs, Ys, Zs))
app_in([], X, X) → app_out([], X, X)
U8(X, Xs, Ys, Zs, app_out(Xs, Ys, Zs)) → app_out(.(X, Xs), Ys, .(X, Zs))
U4(X, Xs, Ys, app_out(Ls, .(X, Bs), Ys)) → qs_out(.(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
qs_in(x1, x2)  =  qs_in
.(x1, x2)  =  .(x1, x2)
U1(x1, x2, x3, x4)  =  U1(x4)
part_in(x1, x2, x3, x4)  =  part_in
[]  =  []
part_out(x1, x2, x3, x4)  =  part_out
U7(x1, x2, x3, x4, x5, x6)  =  U7(x6)
U5(x1, x2, x3, x4, x5, x6)  =  U5(x6)
less_in(x1, x2)  =  less_in
s(x1)  =  s(x1)
U9(x1, x2, x3)  =  U9(x3)
0  =  0
less_out(x1, x2)  =  less_out(x1)
U6(x1, x2, x3, x4, x5, x6)  =  U6(x6)
U2(x1, x2, x3, x4, x5)  =  U2(x5)
qs_out(x1, x2)  =  qs_out
U3(x1, x2, x3, x4, x5)  =  U3(x5)
U4(x1, x2, x3, x4)  =  U4(x4)
app_in(x1, x2, x3)  =  app_in
U8(x1, x2, x3, x4, x5)  =  U8(x5)
app_out(x1, x2, x3)  =  app_out
APP_IN(x1, x2, x3)  =  APP_IN

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
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
              ↳ PiDP
              ↳ PiDP
              ↳ PiDP

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

APP_IN(.(X, Xs), Ys, .(X, Zs)) → APP_IN(Xs, Ys, Zs)

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

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
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
                      ↳ QDP
                        ↳ NonTerminationProof
              ↳ PiDP
              ↳ PiDP
              ↳ PiDP

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

APP_IN → APP_IN

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
We used the non-termination processor [17] to show that the DP problem is infinite.
Found a loop by semiunifying a rule from P directly.

The TRS P consists of the following rules:

APP_IN → APP_IN

The TRS R consists of the following rules:none


s = APP_IN evaluates to t =APP_IN

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:



Rewriting sequence

The DP semiunifies directly so there is only one rewrite step from APP_IN to APP_IN.





↳ Prolog
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
              ↳ 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(.(X, Xs), Ys) → U1(X, Xs, Ys, part_in(X, Xs, Littles, Bigs))
part_in(X, [], [], []) → part_out(X, [], [], [])
part_in(X, .(Y, Xs), Ls, .(Y, Bs)) → U7(X, Y, Xs, Ls, Bs, part_in(X, Xs, Ls, Bs))
part_in(X, .(Y, Xs), .(Y, Ls), Bs) → U5(X, Y, Xs, Ls, Bs, less_in(X, Y))
less_in(s(X), s(Y)) → U9(X, Y, less_in(X, Y))
less_in(0, s(X)) → less_out(0, s(X))
U9(X, Y, less_out(X, Y)) → less_out(s(X), s(Y))
U5(X, Y, Xs, Ls, Bs, less_out(X, Y)) → U6(X, Y, Xs, Ls, Bs, part_in(X, Xs, Ls, Bs))
U6(X, Y, Xs, Ls, Bs, part_out(X, Xs, Ls, Bs)) → part_out(X, .(Y, Xs), .(Y, Ls), Bs)
U7(X, Y, Xs, Ls, Bs, part_out(X, Xs, Ls, Bs)) → part_out(X, .(Y, Xs), Ls, .(Y, Bs))
U1(X, Xs, Ys, part_out(X, Xs, Littles, Bigs)) → U2(X, Xs, Ys, Bigs, qs_in(Littles, Ls))
qs_in([], []) → qs_out([], [])
U2(X, Xs, Ys, Bigs, qs_out(Littles, Ls)) → U3(X, Xs, Ys, Ls, qs_in(Bigs, Bs))
U3(X, Xs, Ys, Ls, qs_out(Bigs, Bs)) → U4(X, Xs, Ys, app_in(Ls, .(X, Bs), Ys))
app_in(.(X, Xs), Ys, .(X, Zs)) → U8(X, Xs, Ys, Zs, app_in(Xs, Ys, Zs))
app_in([], X, X) → app_out([], X, X)
U8(X, Xs, Ys, Zs, app_out(Xs, Ys, Zs)) → app_out(.(X, Xs), Ys, .(X, Zs))
U4(X, Xs, Ys, app_out(Ls, .(X, Bs), Ys)) → qs_out(.(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
qs_in(x1, x2)  =  qs_in
.(x1, x2)  =  .(x1, x2)
U1(x1, x2, x3, x4)  =  U1(x4)
part_in(x1, x2, x3, x4)  =  part_in
[]  =  []
part_out(x1, x2, x3, x4)  =  part_out
U7(x1, x2, x3, x4, x5, x6)  =  U7(x6)
U5(x1, x2, x3, x4, x5, x6)  =  U5(x6)
less_in(x1, x2)  =  less_in
s(x1)  =  s(x1)
U9(x1, x2, x3)  =  U9(x3)
0  =  0
less_out(x1, x2)  =  less_out(x1)
U6(x1, x2, x3, x4, x5, x6)  =  U6(x6)
U2(x1, x2, x3, x4, x5)  =  U2(x5)
qs_out(x1, x2)  =  qs_out
U3(x1, x2, x3, x4, x5)  =  U3(x5)
U4(x1, x2, x3, x4)  =  U4(x4)
app_in(x1, x2, x3)  =  app_in
U8(x1, x2, x3, x4, x5)  =  U8(x5)
app_out(x1, x2, x3)  =  app_out
LESS_IN(x1, x2)  =  LESS_IN

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
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
              ↳ 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.
The argument filtering Pi contains the following mapping:
s(x1)  =  s(x1)
LESS_IN(x1, x2)  =  LESS_IN

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
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
                      ↳ QDP
                        ↳ NonTerminationProof
              ↳ PiDP
              ↳ PiDP

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

LESS_IN → LESS_IN

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
We used the non-termination processor [17] to show that the DP problem is infinite.
Found a loop by semiunifying a rule from P directly.

The TRS P consists of the following rules:

LESS_IN → LESS_IN

The TRS R consists of the following rules:none


s = LESS_IN evaluates to t =LESS_IN

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:



Rewriting sequence

The DP semiunifies directly so there is only one rewrite step from LESS_IN to LESS_IN.





↳ Prolog
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
              ↳ PiDP

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

PART_IN(X, .(Y, Xs), Ls, .(Y, Bs)) → PART_IN(X, Xs, Ls, Bs)
U51(X, Y, Xs, Ls, Bs, less_out(X, Y)) → PART_IN(X, Xs, Ls, Bs)
PART_IN(X, .(Y, Xs), .(Y, Ls), Bs) → U51(X, Y, Xs, Ls, Bs, less_in(X, Y))

The TRS R consists of the following rules:

qs_in(.(X, Xs), Ys) → U1(X, Xs, Ys, part_in(X, Xs, Littles, Bigs))
part_in(X, [], [], []) → part_out(X, [], [], [])
part_in(X, .(Y, Xs), Ls, .(Y, Bs)) → U7(X, Y, Xs, Ls, Bs, part_in(X, Xs, Ls, Bs))
part_in(X, .(Y, Xs), .(Y, Ls), Bs) → U5(X, Y, Xs, Ls, Bs, less_in(X, Y))
less_in(s(X), s(Y)) → U9(X, Y, less_in(X, Y))
less_in(0, s(X)) → less_out(0, s(X))
U9(X, Y, less_out(X, Y)) → less_out(s(X), s(Y))
U5(X, Y, Xs, Ls, Bs, less_out(X, Y)) → U6(X, Y, Xs, Ls, Bs, part_in(X, Xs, Ls, Bs))
U6(X, Y, Xs, Ls, Bs, part_out(X, Xs, Ls, Bs)) → part_out(X, .(Y, Xs), .(Y, Ls), Bs)
U7(X, Y, Xs, Ls, Bs, part_out(X, Xs, Ls, Bs)) → part_out(X, .(Y, Xs), Ls, .(Y, Bs))
U1(X, Xs, Ys, part_out(X, Xs, Littles, Bigs)) → U2(X, Xs, Ys, Bigs, qs_in(Littles, Ls))
qs_in([], []) → qs_out([], [])
U2(X, Xs, Ys, Bigs, qs_out(Littles, Ls)) → U3(X, Xs, Ys, Ls, qs_in(Bigs, Bs))
U3(X, Xs, Ys, Ls, qs_out(Bigs, Bs)) → U4(X, Xs, Ys, app_in(Ls, .(X, Bs), Ys))
app_in(.(X, Xs), Ys, .(X, Zs)) → U8(X, Xs, Ys, Zs, app_in(Xs, Ys, Zs))
app_in([], X, X) → app_out([], X, X)
U8(X, Xs, Ys, Zs, app_out(Xs, Ys, Zs)) → app_out(.(X, Xs), Ys, .(X, Zs))
U4(X, Xs, Ys, app_out(Ls, .(X, Bs), Ys)) → qs_out(.(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
qs_in(x1, x2)  =  qs_in
.(x1, x2)  =  .(x1, x2)
U1(x1, x2, x3, x4)  =  U1(x4)
part_in(x1, x2, x3, x4)  =  part_in
[]  =  []
part_out(x1, x2, x3, x4)  =  part_out
U7(x1, x2, x3, x4, x5, x6)  =  U7(x6)
U5(x1, x2, x3, x4, x5, x6)  =  U5(x6)
less_in(x1, x2)  =  less_in
s(x1)  =  s(x1)
U9(x1, x2, x3)  =  U9(x3)
0  =  0
less_out(x1, x2)  =  less_out(x1)
U6(x1, x2, x3, x4, x5, x6)  =  U6(x6)
U2(x1, x2, x3, x4, x5)  =  U2(x5)
qs_out(x1, x2)  =  qs_out
U3(x1, x2, x3, x4, x5)  =  U3(x5)
U4(x1, x2, x3, x4)  =  U4(x4)
app_in(x1, x2, x3)  =  app_in
U8(x1, x2, x3, x4, x5)  =  U8(x5)
app_out(x1, x2, x3)  =  app_out
U51(x1, x2, x3, x4, x5, x6)  =  U51(x6)
PART_IN(x1, x2, x3, x4)  =  PART_IN

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
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
              ↳ PiDP

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

PART_IN(X, .(Y, Xs), Ls, .(Y, Bs)) → PART_IN(X, Xs, Ls, Bs)
U51(X, Y, Xs, Ls, Bs, less_out(X, Y)) → PART_IN(X, Xs, Ls, Bs)
PART_IN(X, .(Y, Xs), .(Y, Ls), Bs) → U51(X, Y, Xs, Ls, Bs, less_in(X, Y))

The TRS R consists of the following rules:

less_in(s(X), s(Y)) → U9(X, Y, less_in(X, Y))
less_in(0, s(X)) → less_out(0, s(X))
U9(X, Y, less_out(X, Y)) → less_out(s(X), s(Y))

The argument filtering Pi contains the following mapping:
.(x1, x2)  =  .(x1, x2)
less_in(x1, x2)  =  less_in
s(x1)  =  s(x1)
U9(x1, x2, x3)  =  U9(x3)
0  =  0
less_out(x1, x2)  =  less_out(x1)
U51(x1, x2, x3, x4, x5, x6)  =  U51(x6)
PART_IN(x1, x2, x3, x4)  =  PART_IN

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
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
                      ↳ QDP
                        ↳ Narrowing
              ↳ PiDP

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

PART_IN → U51(less_in)
U51(less_out(X)) → PART_IN
PART_IN → PART_IN

The TRS R consists of the following rules:

less_in → U9(less_in)
less_in → less_out(0)
U9(less_out(X)) → less_out(s(X))

The set Q consists of the following terms:

less_in
U9(x0)

We have to consider all (P,Q,R)-chains.
By narrowing [15] the rule PART_IN → U51(less_in) at position [0] we obtained the following new rules:

PART_IN → U51(less_out(0))
PART_IN → U51(U9(less_in))



↳ Prolog
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
                      ↳ QDP
                        ↳ Narrowing
                          ↳ QDP
                            ↳ NonTerminationProof
              ↳ PiDP

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

PART_IN → U51(less_out(0))
PART_IN → U51(U9(less_in))
U51(less_out(X)) → PART_IN
PART_IN → PART_IN

The TRS R consists of the following rules:

less_in → U9(less_in)
less_in → less_out(0)
U9(less_out(X)) → less_out(s(X))

The set Q consists of the following terms:

less_in
U9(x0)

We have to consider all (P,Q,R)-chains.
We used the non-termination processor [17] to show that the DP problem is infinite.
Found a loop by semiunifying a rule from P directly.

The TRS P consists of the following rules:

PART_IN → U51(less_out(0))
PART_IN → U51(U9(less_in))
U51(less_out(X)) → PART_IN
PART_IN → PART_IN

The TRS R consists of the following rules:

less_in → U9(less_in)
less_in → less_out(0)
U9(less_out(X)) → less_out(s(X))


s = PART_IN evaluates to t =PART_IN

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:



Rewriting sequence

The DP semiunifies directly so there is only one rewrite step from PART_IN to PART_IN.





↳ Prolog
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ PiDP
              ↳ PiDP
              ↳ PiDP
                ↳ PiDPToQDPProof

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

U11(X, Xs, Ys, part_out(X, Xs, Littles, Bigs)) → U21(X, Xs, Ys, Bigs, qs_in(Littles, Ls))
U11(X, Xs, Ys, part_out(X, Xs, Littles, Bigs)) → QS_IN(Littles, Ls)
QS_IN(.(X, Xs), Ys) → U11(X, Xs, Ys, part_in(X, Xs, Littles, Bigs))
U21(X, Xs, Ys, Bigs, qs_out(Littles, Ls)) → QS_IN(Bigs, Bs)

The TRS R consists of the following rules:

qs_in(.(X, Xs), Ys) → U1(X, Xs, Ys, part_in(X, Xs, Littles, Bigs))
part_in(X, [], [], []) → part_out(X, [], [], [])
part_in(X, .(Y, Xs), Ls, .(Y, Bs)) → U7(X, Y, Xs, Ls, Bs, part_in(X, Xs, Ls, Bs))
part_in(X, .(Y, Xs), .(Y, Ls), Bs) → U5(X, Y, Xs, Ls, Bs, less_in(X, Y))
less_in(s(X), s(Y)) → U9(X, Y, less_in(X, Y))
less_in(0, s(X)) → less_out(0, s(X))
U9(X, Y, less_out(X, Y)) → less_out(s(X), s(Y))
U5(X, Y, Xs, Ls, Bs, less_out(X, Y)) → U6(X, Y, Xs, Ls, Bs, part_in(X, Xs, Ls, Bs))
U6(X, Y, Xs, Ls, Bs, part_out(X, Xs, Ls, Bs)) → part_out(X, .(Y, Xs), .(Y, Ls), Bs)
U7(X, Y, Xs, Ls, Bs, part_out(X, Xs, Ls, Bs)) → part_out(X, .(Y, Xs), Ls, .(Y, Bs))
U1(X, Xs, Ys, part_out(X, Xs, Littles, Bigs)) → U2(X, Xs, Ys, Bigs, qs_in(Littles, Ls))
qs_in([], []) → qs_out([], [])
U2(X, Xs, Ys, Bigs, qs_out(Littles, Ls)) → U3(X, Xs, Ys, Ls, qs_in(Bigs, Bs))
U3(X, Xs, Ys, Ls, qs_out(Bigs, Bs)) → U4(X, Xs, Ys, app_in(Ls, .(X, Bs), Ys))
app_in(.(X, Xs), Ys, .(X, Zs)) → U8(X, Xs, Ys, Zs, app_in(Xs, Ys, Zs))
app_in([], X, X) → app_out([], X, X)
U8(X, Xs, Ys, Zs, app_out(Xs, Ys, Zs)) → app_out(.(X, Xs), Ys, .(X, Zs))
U4(X, Xs, Ys, app_out(Ls, .(X, Bs), Ys)) → qs_out(.(X, Xs), Ys)

The argument filtering Pi contains the following mapping:
qs_in(x1, x2)  =  qs_in
.(x1, x2)  =  .(x1, x2)
U1(x1, x2, x3, x4)  =  U1(x4)
part_in(x1, x2, x3, x4)  =  part_in
[]  =  []
part_out(x1, x2, x3, x4)  =  part_out
U7(x1, x2, x3, x4, x5, x6)  =  U7(x6)
U5(x1, x2, x3, x4, x5, x6)  =  U5(x6)
less_in(x1, x2)  =  less_in
s(x1)  =  s(x1)
U9(x1, x2, x3)  =  U9(x3)
0  =  0
less_out(x1, x2)  =  less_out(x1)
U6(x1, x2, x3, x4, x5, x6)  =  U6(x6)
U2(x1, x2, x3, x4, x5)  =  U2(x5)
qs_out(x1, x2)  =  qs_out
U3(x1, x2, x3, x4, x5)  =  U3(x5)
U4(x1, x2, x3, x4)  =  U4(x4)
app_in(x1, x2, x3)  =  app_in
U8(x1, x2, x3, x4, x5)  =  U8(x5)
app_out(x1, x2, x3)  =  app_out
QS_IN(x1, x2)  =  QS_IN
U21(x1, x2, x3, x4, x5)  =  U21(x5)
U11(x1, x2, x3, x4)  =  U11(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
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ PiDP
              ↳ PiDP
              ↳ PiDP
                ↳ PiDPToQDPProof
                  ↳ QDP
                    ↳ Narrowing

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

U21(qs_out) → QS_IN
U11(part_out) → QS_IN
U11(part_out) → U21(qs_in)
QS_IN → U11(part_in)

The TRS R consists of the following rules:

qs_in → U1(part_in)
part_in → part_out
part_in → U7(part_in)
part_in → U5(less_in)
less_in → U9(less_in)
less_in → less_out(0)
U9(less_out(X)) → less_out(s(X))
U5(less_out(X)) → U6(part_in)
U6(part_out) → part_out
U7(part_out) → part_out
U1(part_out) → U2(qs_in)
qs_in → qs_out
U2(qs_out) → U3(qs_in)
U3(qs_out) → U4(app_in)
app_in → U8(app_in)
app_in → app_out
U8(app_out) → app_out
U4(app_out) → qs_out

The set Q consists of the following terms:

qs_in
part_in
less_in
U9(x0)
U5(x0)
U6(x0)
U7(x0)
U1(x0)
U2(x0)
U3(x0)
app_in
U8(x0)
U4(x0)

We have to consider all (P,Q,R)-chains.
By narrowing [15] the rule U11(part_out) → U21(qs_in) at position [0] we obtained the following new rules:

U11(part_out) → U21(U1(part_in))
U11(part_out) → U21(qs_out)



↳ Prolog
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ PiDP
              ↳ PiDP
              ↳ PiDP
                ↳ PiDPToQDPProof
                  ↳ QDP
                    ↳ Narrowing
                      ↳ QDP
                        ↳ Narrowing

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

U11(part_out) → QS_IN
U21(qs_out) → QS_IN
U11(part_out) → U21(U1(part_in))
QS_IN → U11(part_in)
U11(part_out) → U21(qs_out)

The TRS R consists of the following rules:

qs_in → U1(part_in)
part_in → part_out
part_in → U7(part_in)
part_in → U5(less_in)
less_in → U9(less_in)
less_in → less_out(0)
U9(less_out(X)) → less_out(s(X))
U5(less_out(X)) → U6(part_in)
U6(part_out) → part_out
U7(part_out) → part_out
U1(part_out) → U2(qs_in)
qs_in → qs_out
U2(qs_out) → U3(qs_in)
U3(qs_out) → U4(app_in)
app_in → U8(app_in)
app_in → app_out
U8(app_out) → app_out
U4(app_out) → qs_out

The set Q consists of the following terms:

qs_in
part_in
less_in
U9(x0)
U5(x0)
U6(x0)
U7(x0)
U1(x0)
U2(x0)
U3(x0)
app_in
U8(x0)
U4(x0)

We have to consider all (P,Q,R)-chains.
By narrowing [15] the rule QS_IN → U11(part_in) at position [0] we obtained the following new rules:

QS_IN → U11(U7(part_in))
QS_IN → U11(U5(less_in))
QS_IN → U11(part_out)



↳ Prolog
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ PiDP
              ↳ PiDP
              ↳ PiDP
                ↳ PiDPToQDPProof
                  ↳ QDP
                    ↳ Narrowing
                      ↳ QDP
                        ↳ Narrowing
                          ↳ QDP
                            ↳ NonTerminationProof

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

QS_IN → U11(U7(part_in))
U21(qs_out) → QS_IN
U11(part_out) → QS_IN
QS_IN → U11(part_out)
U11(part_out) → U21(U1(part_in))
QS_IN → U11(U5(less_in))
U11(part_out) → U21(qs_out)

The TRS R consists of the following rules:

qs_in → U1(part_in)
part_in → part_out
part_in → U7(part_in)
part_in → U5(less_in)
less_in → U9(less_in)
less_in → less_out(0)
U9(less_out(X)) → less_out(s(X))
U5(less_out(X)) → U6(part_in)
U6(part_out) → part_out
U7(part_out) → part_out
U1(part_out) → U2(qs_in)
qs_in → qs_out
U2(qs_out) → U3(qs_in)
U3(qs_out) → U4(app_in)
app_in → U8(app_in)
app_in → app_out
U8(app_out) → app_out
U4(app_out) → qs_out

The set Q consists of the following terms:

qs_in
part_in
less_in
U9(x0)
U5(x0)
U6(x0)
U7(x0)
U1(x0)
U2(x0)
U3(x0)
app_in
U8(x0)
U4(x0)

We have to consider all (P,Q,R)-chains.
We used the non-termination processor [17] to show that the DP problem is infinite.
Found a loop by narrowing to the left:

The TRS P consists of the following rules:

QS_IN → U11(U7(part_in))
U21(qs_out) → QS_IN
U11(part_out) → QS_IN
QS_IN → U11(part_out)
U11(part_out) → U21(U1(part_in))
QS_IN → U11(U5(less_in))
U11(part_out) → U21(qs_out)

The TRS R consists of the following rules:

qs_in → U1(part_in)
part_in → part_out
part_in → U7(part_in)
part_in → U5(less_in)
less_in → U9(less_in)
less_in → less_out(0)
U9(less_out(X)) → less_out(s(X))
U5(less_out(X)) → U6(part_in)
U6(part_out) → part_out
U7(part_out) → part_out
U1(part_out) → U2(qs_in)
qs_in → qs_out
U2(qs_out) → U3(qs_in)
U3(qs_out) → U4(app_in)
app_in → U8(app_in)
app_in → app_out
U8(app_out) → app_out
U4(app_out) → qs_out


s = QS_IN evaluates to t =QS_IN

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:



Rewriting sequence

QS_IN → U11(part_out)
with rule QS_IN → U11(part_out) at position [] and matcher [ ]

U11(part_out) → QS_IN
with rule U11(part_out) → QS_IN

Now applying the matcher to the start term leads to a term which is equal to the last term in the rewriting sequence


All these steps are and every following step will be a correct step w.r.t to Q.