Left Termination proof of /tmp/tmpcz6rC1/permutation-fb.pl

Left Termination of the query pattern perm_in_2(a, g) w.r.t. the given Prolog program could not be shown:

↳ Prolog

  ↳ PrologToPiTRSProof

  ↳ PrologToPiTRSProof

Clauses:

perm([], []).
perm(Xs, .(X, Ys)) :- ','(app(X1s, .(X, X2s), Xs), ','(app(X1s, X2s, Zs), perm(Zs, Ys))).
app([], X, X).
app(.(X, Xs), Ys, .(X, Zs)) :- app(Xs, Ys, Zs).

Queries:

perm(a,g).

We use the technique of [30]. With regard to the inferred argument filtering the predicates were used in the following modes:
perm_in: (f,b)
app_in: (f,f,f)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

perm_in_ag([], []) → perm_out_ag([], [])
perm_in_ag(Xs, .(X, Ys)) → U1_ag(Xs, X, Ys, app_in_aaa(X1s, .(X, X2s), Xs))
app_in_aaa([], X, X) → app_out_aaa([], X, X)
app_in_aaa(.(X, Xs), Ys, .(X, Zs)) → U4_aaa(X, Xs, Ys, Zs, app_in_aaa(Xs, Ys, Zs))
U4_aaa(X, Xs, Ys, Zs, app_out_aaa(Xs, Ys, Zs)) → app_out_aaa(.(X, Xs), Ys, .(X, Zs))
U1_ag(Xs, X, Ys, app_out_aaa(X1s, .(X, X2s), Xs)) → U2_ag(Xs, X, Ys, app_in_aaa(X1s, X2s, Zs))
U2_ag(Xs, X, Ys, app_out_aaa(X1s, X2s, Zs)) → U3_ag(Xs, X, Ys, perm_in_ag(Zs, Ys))
U3_ag(Xs, X, Ys, perm_out_ag(Zs, Ys)) → perm_out_ag(Xs, .(X, Ys))

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:

perm_in_ag([], []) → perm_out_ag([], [])
perm_in_ag(Xs, .(X, Ys)) → U1_ag(Xs, X, Ys, app_in_aaa(X1s, .(X, X2s), Xs))
app_in_aaa([], X, X) → app_out_aaa([], X, X)
app_in_aaa(.(X, Xs), Ys, .(X, Zs)) → U4_aaa(X, Xs, Ys, Zs, app_in_aaa(Xs, Ys, Zs))
U4_aaa(X, Xs, Ys, Zs, app_out_aaa(Xs, Ys, Zs)) → app_out_aaa(.(X, Xs), Ys, .(X, Zs))
U1_ag(Xs, X, Ys, app_out_aaa(X1s, .(X, X2s), Xs)) → U2_ag(Xs, X, Ys, app_in_aaa(X1s, X2s, Zs))
U2_ag(Xs, X, Ys, app_out_aaa(X1s, X2s, Zs)) → U3_ag(Xs, X, Ys, perm_in_ag(Zs, Ys))
U3_ag(Xs, X, Ys, perm_out_ag(Zs, Ys)) → perm_out_ag(Xs, .(X, Ys))

The argument filtering Pi contains the following mapping:
perm_in_ag(x1, x2) = perm_in_ag(x2)
[] = []
perm_out_ag(x1, x2) = perm_out_ag
.(x1, x2) = .(x1, x2)
U1_ag(x1, x2, x3, x4) = U1_ag(x3, x4)
app_in_aaa(x1, x2, x3) = app_in_aaa
app_out_aaa(x1, x2, x3) = app_out_aaa
U4_aaa(x1, x2, x3, x4, x5) = U4_aaa(x5)
U2_ag(x1, x2, x3, x4) = U2_ag(x3, x4)
U3_ag(x1, x2, x3, x4) = U3_ag(x4)

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

PERM_IN_AG(Xs, .(X, Ys)) → U1_AG(Xs, X, Ys, app_in_aaa(X1s, .(X, X2s), Xs))
PERM_IN_AG(Xs, .(X, Ys)) → APP_IN_AAA(X1s, .(X, X2s), Xs)
APP_IN_AAA(.(X, Xs), Ys, .(X, Zs)) → U4_AAA(X, Xs, Ys, Zs, app_in_aaa(Xs, Ys, Zs))
APP_IN_AAA(.(X, Xs), Ys, .(X, Zs)) → APP_IN_AAA(Xs, Ys, Zs)
U1_AG(Xs, X, Ys, app_out_aaa(X1s, .(X, X2s), Xs)) → U2_AG(Xs, X, Ys, app_in_aaa(X1s, X2s, Zs))
U1_AG(Xs, X, Ys, app_out_aaa(X1s, .(X, X2s), Xs)) → APP_IN_AAA(X1s, X2s, Zs)
U2_AG(Xs, X, Ys, app_out_aaa(X1s, X2s, Zs)) → U3_AG(Xs, X, Ys, perm_in_ag(Zs, Ys))
U2_AG(Xs, X, Ys, app_out_aaa(X1s, X2s, Zs)) → PERM_IN_AG(Zs, Ys)

The TRS R consists of the following rules:

perm_in_ag([], []) → perm_out_ag([], [])
perm_in_ag(Xs, .(X, Ys)) → U1_ag(Xs, X, Ys, app_in_aaa(X1s, .(X, X2s), Xs))
app_in_aaa([], X, X) → app_out_aaa([], X, X)
app_in_aaa(.(X, Xs), Ys, .(X, Zs)) → U4_aaa(X, Xs, Ys, Zs, app_in_aaa(Xs, Ys, Zs))
U4_aaa(X, Xs, Ys, Zs, app_out_aaa(Xs, Ys, Zs)) → app_out_aaa(.(X, Xs), Ys, .(X, Zs))
U1_ag(Xs, X, Ys, app_out_aaa(X1s, .(X, X2s), Xs)) → U2_ag(Xs, X, Ys, app_in_aaa(X1s, X2s, Zs))
U2_ag(Xs, X, Ys, app_out_aaa(X1s, X2s, Zs)) → U3_ag(Xs, X, Ys, perm_in_ag(Zs, Ys))
U3_ag(Xs, X, Ys, perm_out_ag(Zs, Ys)) → perm_out_ag(Xs, .(X, Ys))

The argument filtering Pi contains the following mapping:
perm_in_ag(x1, x2) = perm_in_ag(x2)
[] = []
perm_out_ag(x1, x2) = perm_out_ag
.(x1, x2) = .(x1, x2)
U1_ag(x1, x2, x3, x4) = U1_ag(x3, x4)
app_in_aaa(x1, x2, x3) = app_in_aaa
app_out_aaa(x1, x2, x3) = app_out_aaa
U4_aaa(x1, x2, x3, x4, x5) = U4_aaa(x5)
U2_ag(x1, x2, x3, x4) = U2_ag(x3, x4)
U3_ag(x1, x2, x3, x4) = U3_ag(x4)
APP_IN_AAA(x1, x2, x3) = APP_IN_AAA
U3_AG(x1, x2, x3, x4) = U3_AG(x4)
PERM_IN_AG(x1, x2) = PERM_IN_AG(x2)
U4_AAA(x1, x2, x3, x4, x5) = U4_AAA(x5)
U2_AG(x1, x2, x3, x4) = U2_AG(x3, x4)
U1_AG(x1, x2, x3, x4) = U1_AG(x3, x4)

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:

PERM_IN_AG(Xs, .(X, Ys)) → U1_AG(Xs, X, Ys, app_in_aaa(X1s, .(X, X2s), Xs))
PERM_IN_AG(Xs, .(X, Ys)) → APP_IN_AAA(X1s, .(X, X2s), Xs)
APP_IN_AAA(.(X, Xs), Ys, .(X, Zs)) → U4_AAA(X, Xs, Ys, Zs, app_in_aaa(Xs, Ys, Zs))
APP_IN_AAA(.(X, Xs), Ys, .(X, Zs)) → APP_IN_AAA(Xs, Ys, Zs)
U1_AG(Xs, X, Ys, app_out_aaa(X1s, .(X, X2s), Xs)) → U2_AG(Xs, X, Ys, app_in_aaa(X1s, X2s, Zs))
U1_AG(Xs, X, Ys, app_out_aaa(X1s, .(X, X2s), Xs)) → APP_IN_AAA(X1s, X2s, Zs)
U2_AG(Xs, X, Ys, app_out_aaa(X1s, X2s, Zs)) → U3_AG(Xs, X, Ys, perm_in_ag(Zs, Ys))
U2_AG(Xs, X, Ys, app_out_aaa(X1s, X2s, Zs)) → PERM_IN_AG(Zs, Ys)

The TRS R consists of the following rules:

perm_in_ag([], []) → perm_out_ag([], [])
perm_in_ag(Xs, .(X, Ys)) → U1_ag(Xs, X, Ys, app_in_aaa(X1s, .(X, X2s), Xs))
app_in_aaa([], X, X) → app_out_aaa([], X, X)
app_in_aaa(.(X, Xs), Ys, .(X, Zs)) → U4_aaa(X, Xs, Ys, Zs, app_in_aaa(Xs, Ys, Zs))
U4_aaa(X, Xs, Ys, Zs, app_out_aaa(Xs, Ys, Zs)) → app_out_aaa(.(X, Xs), Ys, .(X, Zs))
U1_ag(Xs, X, Ys, app_out_aaa(X1s, .(X, X2s), Xs)) → U2_ag(Xs, X, Ys, app_in_aaa(X1s, X2s, Zs))
U2_ag(Xs, X, Ys, app_out_aaa(X1s, X2s, Zs)) → U3_ag(Xs, X, Ys, perm_in_ag(Zs, Ys))
U3_ag(Xs, X, Ys, perm_out_ag(Zs, Ys)) → perm_out_ag(Xs, .(X, Ys))

The argument filtering Pi contains the following mapping:
perm_in_ag(x1, x2) = perm_in_ag(x2)
[] = []
perm_out_ag(x1, x2) = perm_out_ag
.(x1, x2) = .(x1, x2)
U1_ag(x1, x2, x3, x4) = U1_ag(x3, x4)
app_in_aaa(x1, x2, x3) = app_in_aaa
app_out_aaa(x1, x2, x3) = app_out_aaa
U4_aaa(x1, x2, x3, x4, x5) = U4_aaa(x5)
U2_ag(x1, x2, x3, x4) = U2_ag(x3, x4)
U3_ag(x1, x2, x3, x4) = U3_ag(x4)
APP_IN_AAA(x1, x2, x3) = APP_IN_AAA
U3_AG(x1, x2, x3, x4) = U3_AG(x4)
PERM_IN_AG(x1, x2) = PERM_IN_AG(x2)
U4_AAA(x1, x2, x3, x4, x5) = U4_AAA(x5)
U2_AG(x1, x2, x3, x4) = U2_AG(x3, x4)
U1_AG(x1, x2, x3, x4) = U1_AG(x3, x4)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph [30] contains 2 SCCs with 4 less nodes.

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

                ↳ UsableRulesProof

              ↳ PiDP

  ↳ PrologToPiTRSProof

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

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

The TRS R consists of the following rules:

perm_in_ag([], []) → perm_out_ag([], [])
perm_in_ag(Xs, .(X, Ys)) → U1_ag(Xs, X, Ys, app_in_aaa(X1s, .(X, X2s), Xs))
app_in_aaa([], X, X) → app_out_aaa([], X, X)
app_in_aaa(.(X, Xs), Ys, .(X, Zs)) → U4_aaa(X, Xs, Ys, Zs, app_in_aaa(Xs, Ys, Zs))
U4_aaa(X, Xs, Ys, Zs, app_out_aaa(Xs, Ys, Zs)) → app_out_aaa(.(X, Xs), Ys, .(X, Zs))
U1_ag(Xs, X, Ys, app_out_aaa(X1s, .(X, X2s), Xs)) → U2_ag(Xs, X, Ys, app_in_aaa(X1s, X2s, Zs))
U2_ag(Xs, X, Ys, app_out_aaa(X1s, X2s, Zs)) → U3_ag(Xs, X, Ys, perm_in_ag(Zs, Ys))
U3_ag(Xs, X, Ys, perm_out_ag(Zs, Ys)) → perm_out_ag(Xs, .(X, Ys))

The argument filtering Pi contains the following mapping:
perm_in_ag(x1, x2) = perm_in_ag(x2)
[] = []
perm_out_ag(x1, x2) = perm_out_ag
.(x1, x2) = .(x1, x2)
U1_ag(x1, x2, x3, x4) = U1_ag(x3, x4)
app_in_aaa(x1, x2, x3) = app_in_aaa
app_out_aaa(x1, x2, x3) = app_out_aaa
U4_aaa(x1, x2, x3, x4, x5) = U4_aaa(x5)
U2_ag(x1, x2, x3, x4) = U2_ag(x3, x4)
U3_ag(x1, x2, x3, x4) = U3_ag(x4)
APP_IN_AAA(x1, x2, x3) = APP_IN_AAA

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

  ↳ PrologToPiTRSProof

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

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

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

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

  ↳ PrologToPiTRSProof

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

APP_IN_AAA → APP_IN_AAA

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_AAA → APP_IN_AAA

The TRS R consists of the following rules:none

s = APP_IN_AAA evaluates to t =APP_IN_AAA

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

Semiunifier: [ ]
Matcher: [ ]

Rewriting sequence

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

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

  ↳ PrologToPiTRSProof

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

U1_AG(Xs, X, Ys, app_out_aaa(X1s, .(X, X2s), Xs)) → U2_AG(Xs, X, Ys, app_in_aaa(X1s, X2s, Zs))
U2_AG(Xs, X, Ys, app_out_aaa(X1s, X2s, Zs)) → PERM_IN_AG(Zs, Ys)
PERM_IN_AG(Xs, .(X, Ys)) → U1_AG(Xs, X, Ys, app_in_aaa(X1s, .(X, X2s), Xs))

The TRS R consists of the following rules:

perm_in_ag([], []) → perm_out_ag([], [])
perm_in_ag(Xs, .(X, Ys)) → U1_ag(Xs, X, Ys, app_in_aaa(X1s, .(X, X2s), Xs))
app_in_aaa([], X, X) → app_out_aaa([], X, X)
app_in_aaa(.(X, Xs), Ys, .(X, Zs)) → U4_aaa(X, Xs, Ys, Zs, app_in_aaa(Xs, Ys, Zs))
U4_aaa(X, Xs, Ys, Zs, app_out_aaa(Xs, Ys, Zs)) → app_out_aaa(.(X, Xs), Ys, .(X, Zs))
U1_ag(Xs, X, Ys, app_out_aaa(X1s, .(X, X2s), Xs)) → U2_ag(Xs, X, Ys, app_in_aaa(X1s, X2s, Zs))
U2_ag(Xs, X, Ys, app_out_aaa(X1s, X2s, Zs)) → U3_ag(Xs, X, Ys, perm_in_ag(Zs, Ys))
U3_ag(Xs, X, Ys, perm_out_ag(Zs, Ys)) → perm_out_ag(Xs, .(X, Ys))

The argument filtering Pi contains the following mapping:
perm_in_ag(x1, x2) = perm_in_ag(x2)
[] = []
perm_out_ag(x1, x2) = perm_out_ag
.(x1, x2) = .(x1, x2)
U1_ag(x1, x2, x3, x4) = U1_ag(x3, x4)
app_in_aaa(x1, x2, x3) = app_in_aaa
app_out_aaa(x1, x2, x3) = app_out_aaa
U4_aaa(x1, x2, x3, x4, x5) = U4_aaa(x5)
U2_ag(x1, x2, x3, x4) = U2_ag(x3, x4)
U3_ag(x1, x2, x3, x4) = U3_ag(x4)
PERM_IN_AG(x1, x2) = PERM_IN_AG(x2)
U2_AG(x1, x2, x3, x4) = U2_AG(x3, x4)
U1_AG(x1, x2, x3, x4) = U1_AG(x3, x4)

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

  ↳ PrologToPiTRSProof

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

U1_AG(Xs, X, Ys, app_out_aaa(X1s, .(X, X2s), Xs)) → U2_AG(Xs, X, Ys, app_in_aaa(X1s, X2s, Zs))
U2_AG(Xs, X, Ys, app_out_aaa(X1s, X2s, Zs)) → PERM_IN_AG(Zs, Ys)
PERM_IN_AG(Xs, .(X, Ys)) → U1_AG(Xs, X, Ys, app_in_aaa(X1s, .(X, X2s), Xs))

The TRS R consists of the following rules:

app_in_aaa([], X, X) → app_out_aaa([], X, X)
app_in_aaa(.(X, Xs), Ys, .(X, Zs)) → U4_aaa(X, Xs, Ys, Zs, app_in_aaa(Xs, Ys, Zs))
U4_aaa(X, Xs, Ys, Zs, app_out_aaa(Xs, Ys, Zs)) → app_out_aaa(.(X, Xs), Ys, .(X, Zs))

The argument filtering Pi contains the following mapping:
[] = []
.(x1, x2) = .(x1, x2)
app_in_aaa(x1, x2, x3) = app_in_aaa
app_out_aaa(x1, x2, x3) = app_out_aaa
U4_aaa(x1, x2, x3, x4, x5) = U4_aaa(x5)
PERM_IN_AG(x1, x2) = PERM_IN_AG(x2)
U2_AG(x1, x2, x3, x4) = U2_AG(x3, x4)
U1_AG(x1, x2, x3, x4) = U1_AG(x3, 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

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ QDPSizeChangeProof

  ↳ PrologToPiTRSProof

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

PERM_IN_AG(.(X, Ys)) → U1_AG(Ys, app_in_aaa)
U2_AG(Ys, app_out_aaa) → PERM_IN_AG(Ys)
U1_AG(Ys, app_out_aaa) → U2_AG(Ys, app_in_aaa)

The TRS R consists of the following rules:

app_in_aaa → app_out_aaa
app_in_aaa → U4_aaa(app_in_aaa)
U4_aaa(app_out_aaa) → app_out_aaa

The set Q consists of the following terms:

app_in_aaa
U4_aaa(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:

U2_AG(Ys, app_out_aaa) → PERM_IN_AG(Ys)
The graph contains the following edges 1 >= 1
PERM_IN_AG(.(X, Ys)) → U1_AG(Ys, app_in_aaa)
The graph contains the following edges 1 > 1
U1_AG(Ys, app_out_aaa) → U2_AG(Ys, app_in_aaa)
The graph contains the following edges 1 >= 1

We use the technique of [30]. With regard to the inferred argument filtering the predicates were used in the following modes:
perm_in: (f,b)
app_in: (f,f,f)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

perm_in_ag([], []) → perm_out_ag([], [])
perm_in_ag(Xs, .(X, Ys)) → U1_ag(Xs, X, Ys, app_in_aaa(X1s, .(X, X2s), Xs))
app_in_aaa([], X, X) → app_out_aaa([], X, X)
app_in_aaa(.(X, Xs), Ys, .(X, Zs)) → U4_aaa(X, Xs, Ys, Zs, app_in_aaa(Xs, Ys, Zs))
U4_aaa(X, Xs, Ys, Zs, app_out_aaa(Xs, Ys, Zs)) → app_out_aaa(.(X, Xs), Ys, .(X, Zs))
U1_ag(Xs, X, Ys, app_out_aaa(X1s, .(X, X2s), Xs)) → U2_ag(Xs, X, Ys, app_in_aaa(X1s, X2s, Zs))
U2_ag(Xs, X, Ys, app_out_aaa(X1s, X2s, Zs)) → U3_ag(Xs, X, Ys, perm_in_ag(Zs, Ys))
U3_ag(Xs, X, Ys, perm_out_ag(Zs, Ys)) → perm_out_ag(Xs, .(X, Ys))

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:

perm_in_ag([], []) → perm_out_ag([], [])
perm_in_ag(Xs, .(X, Ys)) → U1_ag(Xs, X, Ys, app_in_aaa(X1s, .(X, X2s), Xs))
app_in_aaa([], X, X) → app_out_aaa([], X, X)
app_in_aaa(.(X, Xs), Ys, .(X, Zs)) → U4_aaa(X, Xs, Ys, Zs, app_in_aaa(Xs, Ys, Zs))
U4_aaa(X, Xs, Ys, Zs, app_out_aaa(Xs, Ys, Zs)) → app_out_aaa(.(X, Xs), Ys, .(X, Zs))
U1_ag(Xs, X, Ys, app_out_aaa(X1s, .(X, X2s), Xs)) → U2_ag(Xs, X, Ys, app_in_aaa(X1s, X2s, Zs))
U2_ag(Xs, X, Ys, app_out_aaa(X1s, X2s, Zs)) → U3_ag(Xs, X, Ys, perm_in_ag(Zs, Ys))
U3_ag(Xs, X, Ys, perm_out_ag(Zs, Ys)) → perm_out_ag(Xs, .(X, Ys))

The argument filtering Pi contains the following mapping:
perm_in_ag(x1, x2) = perm_in_ag(x2)
[] = []
perm_out_ag(x1, x2) = perm_out_ag(x2)
.(x1, x2) = .(x1, x2)
U1_ag(x1, x2, x3, x4) = U1_ag(x2, x3, x4)
app_in_aaa(x1, x2, x3) = app_in_aaa
app_out_aaa(x1, x2, x3) = app_out_aaa
U4_aaa(x1, x2, x3, x4, x5) = U4_aaa(x5)
U2_ag(x1, x2, x3, x4) = U2_ag(x2, x3, x4)
U3_ag(x1, x2, x3, x4) = U3_ag(x2, x3, x4)

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

PERM_IN_AG(Xs, .(X, Ys)) → U1_AG(Xs, X, Ys, app_in_aaa(X1s, .(X, X2s), Xs))
PERM_IN_AG(Xs, .(X, Ys)) → APP_IN_AAA(X1s, .(X, X2s), Xs)
APP_IN_AAA(.(X, Xs), Ys, .(X, Zs)) → U4_AAA(X, Xs, Ys, Zs, app_in_aaa(Xs, Ys, Zs))
APP_IN_AAA(.(X, Xs), Ys, .(X, Zs)) → APP_IN_AAA(Xs, Ys, Zs)
U1_AG(Xs, X, Ys, app_out_aaa(X1s, .(X, X2s), Xs)) → U2_AG(Xs, X, Ys, app_in_aaa(X1s, X2s, Zs))
U1_AG(Xs, X, Ys, app_out_aaa(X1s, .(X, X2s), Xs)) → APP_IN_AAA(X1s, X2s, Zs)
U2_AG(Xs, X, Ys, app_out_aaa(X1s, X2s, Zs)) → U3_AG(Xs, X, Ys, perm_in_ag(Zs, Ys))
U2_AG(Xs, X, Ys, app_out_aaa(X1s, X2s, Zs)) → PERM_IN_AG(Zs, Ys)

The TRS R consists of the following rules:

perm_in_ag([], []) → perm_out_ag([], [])
perm_in_ag(Xs, .(X, Ys)) → U1_ag(Xs, X, Ys, app_in_aaa(X1s, .(X, X2s), Xs))
app_in_aaa([], X, X) → app_out_aaa([], X, X)
app_in_aaa(.(X, Xs), Ys, .(X, Zs)) → U4_aaa(X, Xs, Ys, Zs, app_in_aaa(Xs, Ys, Zs))
U4_aaa(X, Xs, Ys, Zs, app_out_aaa(Xs, Ys, Zs)) → app_out_aaa(.(X, Xs), Ys, .(X, Zs))
U1_ag(Xs, X, Ys, app_out_aaa(X1s, .(X, X2s), Xs)) → U2_ag(Xs, X, Ys, app_in_aaa(X1s, X2s, Zs))
U2_ag(Xs, X, Ys, app_out_aaa(X1s, X2s, Zs)) → U3_ag(Xs, X, Ys, perm_in_ag(Zs, Ys))
U3_ag(Xs, X, Ys, perm_out_ag(Zs, Ys)) → perm_out_ag(Xs, .(X, Ys))

The argument filtering Pi contains the following mapping:
perm_in_ag(x1, x2) = perm_in_ag(x2)
[] = []
perm_out_ag(x1, x2) = perm_out_ag(x2)
.(x1, x2) = .(x1, x2)
U1_ag(x1, x2, x3, x4) = U1_ag(x2, x3, x4)
app_in_aaa(x1, x2, x3) = app_in_aaa
app_out_aaa(x1, x2, x3) = app_out_aaa
U4_aaa(x1, x2, x3, x4, x5) = U4_aaa(x5)
U2_ag(x1, x2, x3, x4) = U2_ag(x2, x3, x4)
U3_ag(x1, x2, x3, x4) = U3_ag(x2, x3, x4)
APP_IN_AAA(x1, x2, x3) = APP_IN_AAA
U3_AG(x1, x2, x3, x4) = U3_AG(x2, x3, x4)
PERM_IN_AG(x1, x2) = PERM_IN_AG(x2)
U4_AAA(x1, x2, x3, x4, x5) = U4_AAA(x5)
U2_AG(x1, x2, x3, x4) = U2_AG(x2, x3, x4)
U1_AG(x1, x2, x3, x4) = U1_AG(x2, x3, x4)

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:

PERM_IN_AG(Xs, .(X, Ys)) → U1_AG(Xs, X, Ys, app_in_aaa(X1s, .(X, X2s), Xs))
PERM_IN_AG(Xs, .(X, Ys)) → APP_IN_AAA(X1s, .(X, X2s), Xs)
APP_IN_AAA(.(X, Xs), Ys, .(X, Zs)) → U4_AAA(X, Xs, Ys, Zs, app_in_aaa(Xs, Ys, Zs))
APP_IN_AAA(.(X, Xs), Ys, .(X, Zs)) → APP_IN_AAA(Xs, Ys, Zs)
U1_AG(Xs, X, Ys, app_out_aaa(X1s, .(X, X2s), Xs)) → U2_AG(Xs, X, Ys, app_in_aaa(X1s, X2s, Zs))
U1_AG(Xs, X, Ys, app_out_aaa(X1s, .(X, X2s), Xs)) → APP_IN_AAA(X1s, X2s, Zs)
U2_AG(Xs, X, Ys, app_out_aaa(X1s, X2s, Zs)) → U3_AG(Xs, X, Ys, perm_in_ag(Zs, Ys))
U2_AG(Xs, X, Ys, app_out_aaa(X1s, X2s, Zs)) → PERM_IN_AG(Zs, Ys)

The TRS R consists of the following rules:

perm_in_ag([], []) → perm_out_ag([], [])
perm_in_ag(Xs, .(X, Ys)) → U1_ag(Xs, X, Ys, app_in_aaa(X1s, .(X, X2s), Xs))
app_in_aaa([], X, X) → app_out_aaa([], X, X)
app_in_aaa(.(X, Xs), Ys, .(X, Zs)) → U4_aaa(X, Xs, Ys, Zs, app_in_aaa(Xs, Ys, Zs))
U4_aaa(X, Xs, Ys, Zs, app_out_aaa(Xs, Ys, Zs)) → app_out_aaa(.(X, Xs), Ys, .(X, Zs))
U1_ag(Xs, X, Ys, app_out_aaa(X1s, .(X, X2s), Xs)) → U2_ag(Xs, X, Ys, app_in_aaa(X1s, X2s, Zs))
U2_ag(Xs, X, Ys, app_out_aaa(X1s, X2s, Zs)) → U3_ag(Xs, X, Ys, perm_in_ag(Zs, Ys))
U3_ag(Xs, X, Ys, perm_out_ag(Zs, Ys)) → perm_out_ag(Xs, .(X, Ys))

The argument filtering Pi contains the following mapping:
perm_in_ag(x1, x2) = perm_in_ag(x2)
[] = []
perm_out_ag(x1, x2) = perm_out_ag(x2)
.(x1, x2) = .(x1, x2)
U1_ag(x1, x2, x3, x4) = U1_ag(x2, x3, x4)
app_in_aaa(x1, x2, x3) = app_in_aaa
app_out_aaa(x1, x2, x3) = app_out_aaa
U4_aaa(x1, x2, x3, x4, x5) = U4_aaa(x5)
U2_ag(x1, x2, x3, x4) = U2_ag(x2, x3, x4)
U3_ag(x1, x2, x3, x4) = U3_ag(x2, x3, x4)
APP_IN_AAA(x1, x2, x3) = APP_IN_AAA
U3_AG(x1, x2, x3, x4) = U3_AG(x2, x3, x4)
PERM_IN_AG(x1, x2) = PERM_IN_AG(x2)
U4_AAA(x1, x2, x3, x4, x5) = U4_AAA(x5)
U2_AG(x1, x2, x3, x4) = U2_AG(x2, x3, x4)
U1_AG(x1, x2, x3, x4) = U1_AG(x2, x3, x4)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph [30] contains 2 SCCs with 4 less nodes.

↳ Prolog

  ↳ PrologToPiTRSProof

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

                ↳ UsableRulesProof

              ↳ PiDP

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

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

The TRS R consists of the following rules:

perm_in_ag([], []) → perm_out_ag([], [])
perm_in_ag(Xs, .(X, Ys)) → U1_ag(Xs, X, Ys, app_in_aaa(X1s, .(X, X2s), Xs))
app_in_aaa([], X, X) → app_out_aaa([], X, X)
app_in_aaa(.(X, Xs), Ys, .(X, Zs)) → U4_aaa(X, Xs, Ys, Zs, app_in_aaa(Xs, Ys, Zs))
U4_aaa(X, Xs, Ys, Zs, app_out_aaa(Xs, Ys, Zs)) → app_out_aaa(.(X, Xs), Ys, .(X, Zs))
U1_ag(Xs, X, Ys, app_out_aaa(X1s, .(X, X2s), Xs)) → U2_ag(Xs, X, Ys, app_in_aaa(X1s, X2s, Zs))
U2_ag(Xs, X, Ys, app_out_aaa(X1s, X2s, Zs)) → U3_ag(Xs, X, Ys, perm_in_ag(Zs, Ys))
U3_ag(Xs, X, Ys, perm_out_ag(Zs, Ys)) → perm_out_ag(Xs, .(X, Ys))

The argument filtering Pi contains the following mapping:
perm_in_ag(x1, x2) = perm_in_ag(x2)
[] = []
perm_out_ag(x1, x2) = perm_out_ag(x2)
.(x1, x2) = .(x1, x2)
U1_ag(x1, x2, x3, x4) = U1_ag(x2, x3, x4)
app_in_aaa(x1, x2, x3) = app_in_aaa
app_out_aaa(x1, x2, x3) = app_out_aaa
U4_aaa(x1, x2, x3, x4, x5) = U4_aaa(x5)
U2_ag(x1, x2, x3, x4) = U2_ag(x2, x3, x4)
U3_ag(x1, x2, x3, x4) = U3_ag(x2, x3, x4)
APP_IN_AAA(x1, x2, x3) = APP_IN_AAA

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

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

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

↳ Prolog

  ↳ PrologToPiTRSProof

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ NonTerminationProof

              ↳ PiDP

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

APP_IN_AAA → APP_IN_AAA

The TRS R consists of the following rules:none

s = APP_IN_AAA evaluates to t =APP_IN_AAA

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

Semiunifier: [ ]
Matcher: [ ]

Rewriting sequence

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

↳ Prolog

  ↳ PrologToPiTRSProof

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

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

U1_AG(Xs, X, Ys, app_out_aaa(X1s, .(X, X2s), Xs)) → U2_AG(Xs, X, Ys, app_in_aaa(X1s, X2s, Zs))
U2_AG(Xs, X, Ys, app_out_aaa(X1s, X2s, Zs)) → PERM_IN_AG(Zs, Ys)
PERM_IN_AG(Xs, .(X, Ys)) → U1_AG(Xs, X, Ys, app_in_aaa(X1s, .(X, X2s), Xs))

The TRS R consists of the following rules:

perm_in_ag([], []) → perm_out_ag([], [])
perm_in_ag(Xs, .(X, Ys)) → U1_ag(Xs, X, Ys, app_in_aaa(X1s, .(X, X2s), Xs))
app_in_aaa([], X, X) → app_out_aaa([], X, X)
app_in_aaa(.(X, Xs), Ys, .(X, Zs)) → U4_aaa(X, Xs, Ys, Zs, app_in_aaa(Xs, Ys, Zs))
U4_aaa(X, Xs, Ys, Zs, app_out_aaa(Xs, Ys, Zs)) → app_out_aaa(.(X, Xs), Ys, .(X, Zs))
U1_ag(Xs, X, Ys, app_out_aaa(X1s, .(X, X2s), Xs)) → U2_ag(Xs, X, Ys, app_in_aaa(X1s, X2s, Zs))
U2_ag(Xs, X, Ys, app_out_aaa(X1s, X2s, Zs)) → U3_ag(Xs, X, Ys, perm_in_ag(Zs, Ys))
U3_ag(Xs, X, Ys, perm_out_ag(Zs, Ys)) → perm_out_ag(Xs, .(X, Ys))

The argument filtering Pi contains the following mapping:
perm_in_ag(x1, x2) = perm_in_ag(x2)
[] = []
perm_out_ag(x1, x2) = perm_out_ag(x2)
.(x1, x2) = .(x1, x2)
U1_ag(x1, x2, x3, x4) = U1_ag(x2, x3, x4)
app_in_aaa(x1, x2, x3) = app_in_aaa
app_out_aaa(x1, x2, x3) = app_out_aaa
U4_aaa(x1, x2, x3, x4, x5) = U4_aaa(x5)
U2_ag(x1, x2, x3, x4) = U2_ag(x2, x3, x4)
U3_ag(x1, x2, x3, x4) = U3_ag(x2, x3, x4)
PERM_IN_AG(x1, x2) = PERM_IN_AG(x2)
U2_AG(x1, x2, x3, x4) = U2_AG(x2, x3, x4)
U1_AG(x1, x2, x3, x4) = U1_AG(x2, x3, x4)

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

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

U1_AG(Xs, X, Ys, app_out_aaa(X1s, .(X, X2s), Xs)) → U2_AG(Xs, X, Ys, app_in_aaa(X1s, X2s, Zs))
U2_AG(Xs, X, Ys, app_out_aaa(X1s, X2s, Zs)) → PERM_IN_AG(Zs, Ys)
PERM_IN_AG(Xs, .(X, Ys)) → U1_AG(Xs, X, Ys, app_in_aaa(X1s, .(X, X2s), Xs))

The TRS R consists of the following rules:

app_in_aaa([], X, X) → app_out_aaa([], X, X)
app_in_aaa(.(X, Xs), Ys, .(X, Zs)) → U4_aaa(X, Xs, Ys, Zs, app_in_aaa(Xs, Ys, Zs))
U4_aaa(X, Xs, Ys, Zs, app_out_aaa(Xs, Ys, Zs)) → app_out_aaa(.(X, Xs), Ys, .(X, Zs))

The argument filtering Pi contains the following mapping:
[] = []
.(x1, x2) = .(x1, x2)
app_in_aaa(x1, x2, x3) = app_in_aaa
app_out_aaa(x1, x2, x3) = app_out_aaa
U4_aaa(x1, x2, x3, x4, x5) = U4_aaa(x5)
PERM_IN_AG(x1, x2) = PERM_IN_AG(x2)
U2_AG(x1, x2, x3, x4) = U2_AG(x2, x3, x4)
U1_AG(x1, x2, x3, x4) = U1_AG(x2, x3, 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

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ QDPSizeChangeProof

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

U2_AG(X, Ys, app_out_aaa) → PERM_IN_AG(Ys)
U1_AG(X, Ys, app_out_aaa) → U2_AG(X, Ys, app_in_aaa)
PERM_IN_AG(.(X, Ys)) → U1_AG(X, Ys, app_in_aaa)

The TRS R consists of the following rules:

app_in_aaa → app_out_aaa
app_in_aaa → U4_aaa(app_in_aaa)
U4_aaa(app_out_aaa) → app_out_aaa

The set Q consists of the following terms:

app_in_aaa
U4_aaa(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:

U2_AG(X, Ys, app_out_aaa) → PERM_IN_AG(Ys)
The graph contains the following edges 2 >= 1
PERM_IN_AG(.(X, Ys)) → U1_AG(X, Ys, app_in_aaa)
The graph contains the following edges 1 > 1, 1 > 2
U1_AG(X, Ys, app_out_aaa) → U2_AG(X, Ys, app_in_aaa)
The graph contains the following edges 1 >= 1, 2 >= 2