Left Termination of the query pattern perm_in_2(g, a) w.r.t. the given Prolog program could successfully be proven:



Prolog
  ↳ PrologToPiTRSProof

Clauses:

app1(.(X, Xs), Ys, .(X, Zs)) :- app1(Xs, Ys, Zs).
app1([], Ys, Ys).
app2(.(X, Xs), Ys, .(X, Zs)) :- app2(Xs, Ys, Zs).
app2([], Ys, Ys).
perm(Xs, .(X, Ys)) :- ','(app2(X1s, .(X, X2s), Xs), ','(app1(X1s, X2s, Zs), perm(Zs, Ys))).
perm([], []).

Queries:

perm(g,a).

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

perm_in([], []) → perm_out([], [])
perm_in(Xs, .(X, Ys)) → U3(Xs, X, Ys, app2_in(X1s, .(X, X2s), Xs))
app2_in([], Ys, Ys) → app2_out([], Ys, Ys)
app2_in(.(X, Xs), Ys, .(X, Zs)) → U2(X, Xs, Ys, Zs, app2_in(Xs, Ys, Zs))
U2(X, Xs, Ys, Zs, app2_out(Xs, Ys, Zs)) → app2_out(.(X, Xs), Ys, .(X, Zs))
U3(Xs, X, Ys, app2_out(X1s, .(X, X2s), Xs)) → U4(Xs, X, Ys, X1s, X2s, app1_in(X1s, X2s, Zs))
app1_in([], Ys, Ys) → app1_out([], Ys, Ys)
app1_in(.(X, Xs), Ys, .(X, Zs)) → U1(X, Xs, Ys, Zs, app1_in(Xs, Ys, Zs))
U1(X, Xs, Ys, Zs, app1_out(Xs, Ys, Zs)) → app1_out(.(X, Xs), Ys, .(X, Zs))
U4(Xs, X, Ys, X1s, X2s, app1_out(X1s, X2s, Zs)) → U5(Xs, X, Ys, perm_in(Zs, Ys))
U5(Xs, X, Ys, perm_out(Zs, Ys)) → perm_out(Xs, .(X, Ys))

The argument filtering Pi contains the following mapping:
perm_in(x1, x2)  =  perm_in(x1)
[]  =  []
perm_out(x1, x2)  =  perm_out(x2)
.(x1, x2)  =  .(x2)
U3(x1, x2, x3, x4)  =  U3(x4)
app2_in(x1, x2, x3)  =  app2_in(x3)
app2_out(x1, x2, x3)  =  app2_out(x1, x2)
U2(x1, x2, x3, x4, x5)  =  U2(x5)
U4(x1, x2, x3, x4, x5, x6)  =  U4(x6)
app1_in(x1, x2, x3)  =  app1_in(x1, x2)
app1_out(x1, x2, x3)  =  app1_out(x3)
U1(x1, x2, x3, x4, x5)  =  U1(x5)
U5(x1, x2, x3, x4)  =  U5(x4)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog



↳ Prolog
  ↳ PrologToPiTRSProof
PiTRS
      ↳ DependencyPairsProof

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

perm_in([], []) → perm_out([], [])
perm_in(Xs, .(X, Ys)) → U3(Xs, X, Ys, app2_in(X1s, .(X, X2s), Xs))
app2_in([], Ys, Ys) → app2_out([], Ys, Ys)
app2_in(.(X, Xs), Ys, .(X, Zs)) → U2(X, Xs, Ys, Zs, app2_in(Xs, Ys, Zs))
U2(X, Xs, Ys, Zs, app2_out(Xs, Ys, Zs)) → app2_out(.(X, Xs), Ys, .(X, Zs))
U3(Xs, X, Ys, app2_out(X1s, .(X, X2s), Xs)) → U4(Xs, X, Ys, X1s, X2s, app1_in(X1s, X2s, Zs))
app1_in([], Ys, Ys) → app1_out([], Ys, Ys)
app1_in(.(X, Xs), Ys, .(X, Zs)) → U1(X, Xs, Ys, Zs, app1_in(Xs, Ys, Zs))
U1(X, Xs, Ys, Zs, app1_out(Xs, Ys, Zs)) → app1_out(.(X, Xs), Ys, .(X, Zs))
U4(Xs, X, Ys, X1s, X2s, app1_out(X1s, X2s, Zs)) → U5(Xs, X, Ys, perm_in(Zs, Ys))
U5(Xs, X, Ys, perm_out(Zs, Ys)) → perm_out(Xs, .(X, Ys))

The argument filtering Pi contains the following mapping:
perm_in(x1, x2)  =  perm_in(x1)
[]  =  []
perm_out(x1, x2)  =  perm_out(x2)
.(x1, x2)  =  .(x2)
U3(x1, x2, x3, x4)  =  U3(x4)
app2_in(x1, x2, x3)  =  app2_in(x3)
app2_out(x1, x2, x3)  =  app2_out(x1, x2)
U2(x1, x2, x3, x4, x5)  =  U2(x5)
U4(x1, x2, x3, x4, x5, x6)  =  U4(x6)
app1_in(x1, x2, x3)  =  app1_in(x1, x2)
app1_out(x1, x2, x3)  =  app1_out(x3)
U1(x1, x2, x3, x4, x5)  =  U1(x5)
U5(x1, x2, x3, x4)  =  U5(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(Xs, .(X, Ys)) → U31(Xs, X, Ys, app2_in(X1s, .(X, X2s), Xs))
PERM_IN(Xs, .(X, Ys)) → APP2_IN(X1s, .(X, X2s), Xs)
APP2_IN(.(X, Xs), Ys, .(X, Zs)) → U21(X, Xs, Ys, Zs, app2_in(Xs, Ys, Zs))
APP2_IN(.(X, Xs), Ys, .(X, Zs)) → APP2_IN(Xs, Ys, Zs)
U31(Xs, X, Ys, app2_out(X1s, .(X, X2s), Xs)) → U41(Xs, X, Ys, X1s, X2s, app1_in(X1s, X2s, Zs))
U31(Xs, X, Ys, app2_out(X1s, .(X, X2s), Xs)) → APP1_IN(X1s, X2s, Zs)
APP1_IN(.(X, Xs), Ys, .(X, Zs)) → U11(X, Xs, Ys, Zs, app1_in(Xs, Ys, Zs))
APP1_IN(.(X, Xs), Ys, .(X, Zs)) → APP1_IN(Xs, Ys, Zs)
U41(Xs, X, Ys, X1s, X2s, app1_out(X1s, X2s, Zs)) → U51(Xs, X, Ys, perm_in(Zs, Ys))
U41(Xs, X, Ys, X1s, X2s, app1_out(X1s, X2s, Zs)) → PERM_IN(Zs, Ys)

The TRS R consists of the following rules:

perm_in([], []) → perm_out([], [])
perm_in(Xs, .(X, Ys)) → U3(Xs, X, Ys, app2_in(X1s, .(X, X2s), Xs))
app2_in([], Ys, Ys) → app2_out([], Ys, Ys)
app2_in(.(X, Xs), Ys, .(X, Zs)) → U2(X, Xs, Ys, Zs, app2_in(Xs, Ys, Zs))
U2(X, Xs, Ys, Zs, app2_out(Xs, Ys, Zs)) → app2_out(.(X, Xs), Ys, .(X, Zs))
U3(Xs, X, Ys, app2_out(X1s, .(X, X2s), Xs)) → U4(Xs, X, Ys, X1s, X2s, app1_in(X1s, X2s, Zs))
app1_in([], Ys, Ys) → app1_out([], Ys, Ys)
app1_in(.(X, Xs), Ys, .(X, Zs)) → U1(X, Xs, Ys, Zs, app1_in(Xs, Ys, Zs))
U1(X, Xs, Ys, Zs, app1_out(Xs, Ys, Zs)) → app1_out(.(X, Xs), Ys, .(X, Zs))
U4(Xs, X, Ys, X1s, X2s, app1_out(X1s, X2s, Zs)) → U5(Xs, X, Ys, perm_in(Zs, Ys))
U5(Xs, X, Ys, perm_out(Zs, Ys)) → perm_out(Xs, .(X, Ys))

The argument filtering Pi contains the following mapping:
perm_in(x1, x2)  =  perm_in(x1)
[]  =  []
perm_out(x1, x2)  =  perm_out(x2)
.(x1, x2)  =  .(x2)
U3(x1, x2, x3, x4)  =  U3(x4)
app2_in(x1, x2, x3)  =  app2_in(x3)
app2_out(x1, x2, x3)  =  app2_out(x1, x2)
U2(x1, x2, x3, x4, x5)  =  U2(x5)
U4(x1, x2, x3, x4, x5, x6)  =  U4(x6)
app1_in(x1, x2, x3)  =  app1_in(x1, x2)
app1_out(x1, x2, x3)  =  app1_out(x3)
U1(x1, x2, x3, x4, x5)  =  U1(x5)
U5(x1, x2, x3, x4)  =  U5(x4)
U51(x1, x2, x3, x4)  =  U51(x4)
U31(x1, x2, x3, x4)  =  U31(x4)
APP2_IN(x1, x2, x3)  =  APP2_IN(x3)
U41(x1, x2, x3, x4, x5, x6)  =  U41(x6)
PERM_IN(x1, x2)  =  PERM_IN(x1)
U21(x1, x2, x3, x4, x5)  =  U21(x5)
U11(x1, x2, x3, x4, x5)  =  U11(x5)
APP1_IN(x1, x2, x3)  =  APP1_IN(x1, x2)

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

↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
PiDP
          ↳ DependencyGraphProof

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

PERM_IN(Xs, .(X, Ys)) → U31(Xs, X, Ys, app2_in(X1s, .(X, X2s), Xs))
PERM_IN(Xs, .(X, Ys)) → APP2_IN(X1s, .(X, X2s), Xs)
APP2_IN(.(X, Xs), Ys, .(X, Zs)) → U21(X, Xs, Ys, Zs, app2_in(Xs, Ys, Zs))
APP2_IN(.(X, Xs), Ys, .(X, Zs)) → APP2_IN(Xs, Ys, Zs)
U31(Xs, X, Ys, app2_out(X1s, .(X, X2s), Xs)) → U41(Xs, X, Ys, X1s, X2s, app1_in(X1s, X2s, Zs))
U31(Xs, X, Ys, app2_out(X1s, .(X, X2s), Xs)) → APP1_IN(X1s, X2s, Zs)
APP1_IN(.(X, Xs), Ys, .(X, Zs)) → U11(X, Xs, Ys, Zs, app1_in(Xs, Ys, Zs))
APP1_IN(.(X, Xs), Ys, .(X, Zs)) → APP1_IN(Xs, Ys, Zs)
U41(Xs, X, Ys, X1s, X2s, app1_out(X1s, X2s, Zs)) → U51(Xs, X, Ys, perm_in(Zs, Ys))
U41(Xs, X, Ys, X1s, X2s, app1_out(X1s, X2s, Zs)) → PERM_IN(Zs, Ys)

The TRS R consists of the following rules:

perm_in([], []) → perm_out([], [])
perm_in(Xs, .(X, Ys)) → U3(Xs, X, Ys, app2_in(X1s, .(X, X2s), Xs))
app2_in([], Ys, Ys) → app2_out([], Ys, Ys)
app2_in(.(X, Xs), Ys, .(X, Zs)) → U2(X, Xs, Ys, Zs, app2_in(Xs, Ys, Zs))
U2(X, Xs, Ys, Zs, app2_out(Xs, Ys, Zs)) → app2_out(.(X, Xs), Ys, .(X, Zs))
U3(Xs, X, Ys, app2_out(X1s, .(X, X2s), Xs)) → U4(Xs, X, Ys, X1s, X2s, app1_in(X1s, X2s, Zs))
app1_in([], Ys, Ys) → app1_out([], Ys, Ys)
app1_in(.(X, Xs), Ys, .(X, Zs)) → U1(X, Xs, Ys, Zs, app1_in(Xs, Ys, Zs))
U1(X, Xs, Ys, Zs, app1_out(Xs, Ys, Zs)) → app1_out(.(X, Xs), Ys, .(X, Zs))
U4(Xs, X, Ys, X1s, X2s, app1_out(X1s, X2s, Zs)) → U5(Xs, X, Ys, perm_in(Zs, Ys))
U5(Xs, X, Ys, perm_out(Zs, Ys)) → perm_out(Xs, .(X, Ys))

The argument filtering Pi contains the following mapping:
perm_in(x1, x2)  =  perm_in(x1)
[]  =  []
perm_out(x1, x2)  =  perm_out(x2)
.(x1, x2)  =  .(x2)
U3(x1, x2, x3, x4)  =  U3(x4)
app2_in(x1, x2, x3)  =  app2_in(x3)
app2_out(x1, x2, x3)  =  app2_out(x1, x2)
U2(x1, x2, x3, x4, x5)  =  U2(x5)
U4(x1, x2, x3, x4, x5, x6)  =  U4(x6)
app1_in(x1, x2, x3)  =  app1_in(x1, x2)
app1_out(x1, x2, x3)  =  app1_out(x3)
U1(x1, x2, x3, x4, x5)  =  U1(x5)
U5(x1, x2, x3, x4)  =  U5(x4)
U51(x1, x2, x3, x4)  =  U51(x4)
U31(x1, x2, x3, x4)  =  U31(x4)
APP2_IN(x1, x2, x3)  =  APP2_IN(x3)
U41(x1, x2, x3, x4, x5, x6)  =  U41(x6)
PERM_IN(x1, x2)  =  PERM_IN(x1)
U21(x1, x2, x3, x4, x5)  =  U21(x5)
U11(x1, x2, x3, x4, x5)  =  U11(x5)
APP1_IN(x1, x2, x3)  =  APP1_IN(x1, x2)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph [30] contains 3 SCCs with 5 less nodes.

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

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

APP1_IN(.(X, Xs), Ys, .(X, Zs)) → APP1_IN(Xs, Ys, Zs)

The TRS R consists of the following rules:

perm_in([], []) → perm_out([], [])
perm_in(Xs, .(X, Ys)) → U3(Xs, X, Ys, app2_in(X1s, .(X, X2s), Xs))
app2_in([], Ys, Ys) → app2_out([], Ys, Ys)
app2_in(.(X, Xs), Ys, .(X, Zs)) → U2(X, Xs, Ys, Zs, app2_in(Xs, Ys, Zs))
U2(X, Xs, Ys, Zs, app2_out(Xs, Ys, Zs)) → app2_out(.(X, Xs), Ys, .(X, Zs))
U3(Xs, X, Ys, app2_out(X1s, .(X, X2s), Xs)) → U4(Xs, X, Ys, X1s, X2s, app1_in(X1s, X2s, Zs))
app1_in([], Ys, Ys) → app1_out([], Ys, Ys)
app1_in(.(X, Xs), Ys, .(X, Zs)) → U1(X, Xs, Ys, Zs, app1_in(Xs, Ys, Zs))
U1(X, Xs, Ys, Zs, app1_out(Xs, Ys, Zs)) → app1_out(.(X, Xs), Ys, .(X, Zs))
U4(Xs, X, Ys, X1s, X2s, app1_out(X1s, X2s, Zs)) → U5(Xs, X, Ys, perm_in(Zs, Ys))
U5(Xs, X, Ys, perm_out(Zs, Ys)) → perm_out(Xs, .(X, Ys))

The argument filtering Pi contains the following mapping:
perm_in(x1, x2)  =  perm_in(x1)
[]  =  []
perm_out(x1, x2)  =  perm_out(x2)
.(x1, x2)  =  .(x2)
U3(x1, x2, x3, x4)  =  U3(x4)
app2_in(x1, x2, x3)  =  app2_in(x3)
app2_out(x1, x2, x3)  =  app2_out(x1, x2)
U2(x1, x2, x3, x4, x5)  =  U2(x5)
U4(x1, x2, x3, x4, x5, x6)  =  U4(x6)
app1_in(x1, x2, x3)  =  app1_in(x1, x2)
app1_out(x1, x2, x3)  =  app1_out(x3)
U1(x1, x2, x3, x4, x5)  =  U1(x5)
U5(x1, x2, x3, x4)  =  U5(x4)
APP1_IN(x1, x2, x3)  =  APP1_IN(x1, x2)

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

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

APP1_IN(.(X, Xs), Ys, .(X, Zs)) → APP1_IN(Xs, Ys, Zs)

R is empty.
The argument filtering Pi contains the following mapping:
.(x1, x2)  =  .(x2)
APP1_IN(x1, x2, x3)  =  APP1_IN(x1, x2)

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

↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
QDP
                        ↳ QDPSizeChangeProof
              ↳ PiDP
              ↳ PiDP

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

APP1_IN(.(Xs), Ys) → APP1_IN(Xs, Ys)

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] we have proven that there are no infinite chains for this DP problem.

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



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

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

APP2_IN(.(X, Xs), Ys, .(X, Zs)) → APP2_IN(Xs, Ys, Zs)

The TRS R consists of the following rules:

perm_in([], []) → perm_out([], [])
perm_in(Xs, .(X, Ys)) → U3(Xs, X, Ys, app2_in(X1s, .(X, X2s), Xs))
app2_in([], Ys, Ys) → app2_out([], Ys, Ys)
app2_in(.(X, Xs), Ys, .(X, Zs)) → U2(X, Xs, Ys, Zs, app2_in(Xs, Ys, Zs))
U2(X, Xs, Ys, Zs, app2_out(Xs, Ys, Zs)) → app2_out(.(X, Xs), Ys, .(X, Zs))
U3(Xs, X, Ys, app2_out(X1s, .(X, X2s), Xs)) → U4(Xs, X, Ys, X1s, X2s, app1_in(X1s, X2s, Zs))
app1_in([], Ys, Ys) → app1_out([], Ys, Ys)
app1_in(.(X, Xs), Ys, .(X, Zs)) → U1(X, Xs, Ys, Zs, app1_in(Xs, Ys, Zs))
U1(X, Xs, Ys, Zs, app1_out(Xs, Ys, Zs)) → app1_out(.(X, Xs), Ys, .(X, Zs))
U4(Xs, X, Ys, X1s, X2s, app1_out(X1s, X2s, Zs)) → U5(Xs, X, Ys, perm_in(Zs, Ys))
U5(Xs, X, Ys, perm_out(Zs, Ys)) → perm_out(Xs, .(X, Ys))

The argument filtering Pi contains the following mapping:
perm_in(x1, x2)  =  perm_in(x1)
[]  =  []
perm_out(x1, x2)  =  perm_out(x2)
.(x1, x2)  =  .(x2)
U3(x1, x2, x3, x4)  =  U3(x4)
app2_in(x1, x2, x3)  =  app2_in(x3)
app2_out(x1, x2, x3)  =  app2_out(x1, x2)
U2(x1, x2, x3, x4, x5)  =  U2(x5)
U4(x1, x2, x3, x4, x5, x6)  =  U4(x6)
app1_in(x1, x2, x3)  =  app1_in(x1, x2)
app1_out(x1, x2, x3)  =  app1_out(x3)
U1(x1, x2, x3, x4, x5)  =  U1(x5)
U5(x1, x2, x3, x4)  =  U5(x4)
APP2_IN(x1, x2, x3)  =  APP2_IN(x3)

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

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

APP2_IN(.(X, Xs), Ys, .(X, Zs)) → APP2_IN(Xs, Ys, Zs)

R is empty.
The argument filtering Pi contains the following mapping:
.(x1, x2)  =  .(x2)
APP2_IN(x1, x2, x3)  =  APP2_IN(x3)

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

↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
QDP
                        ↳ QDPSizeChangeProof
              ↳ PiDP

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

APP2_IN(.(Zs)) → APP2_IN(Zs)

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] we have proven that there are no infinite chains for this DP problem.

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



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

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

U41(Xs, X, Ys, X1s, X2s, app1_out(X1s, X2s, Zs)) → PERM_IN(Zs, Ys)
U31(Xs, X, Ys, app2_out(X1s, .(X, X2s), Xs)) → U41(Xs, X, Ys, X1s, X2s, app1_in(X1s, X2s, Zs))
PERM_IN(Xs, .(X, Ys)) → U31(Xs, X, Ys, app2_in(X1s, .(X, X2s), Xs))

The TRS R consists of the following rules:

perm_in([], []) → perm_out([], [])
perm_in(Xs, .(X, Ys)) → U3(Xs, X, Ys, app2_in(X1s, .(X, X2s), Xs))
app2_in([], Ys, Ys) → app2_out([], Ys, Ys)
app2_in(.(X, Xs), Ys, .(X, Zs)) → U2(X, Xs, Ys, Zs, app2_in(Xs, Ys, Zs))
U2(X, Xs, Ys, Zs, app2_out(Xs, Ys, Zs)) → app2_out(.(X, Xs), Ys, .(X, Zs))
U3(Xs, X, Ys, app2_out(X1s, .(X, X2s), Xs)) → U4(Xs, X, Ys, X1s, X2s, app1_in(X1s, X2s, Zs))
app1_in([], Ys, Ys) → app1_out([], Ys, Ys)
app1_in(.(X, Xs), Ys, .(X, Zs)) → U1(X, Xs, Ys, Zs, app1_in(Xs, Ys, Zs))
U1(X, Xs, Ys, Zs, app1_out(Xs, Ys, Zs)) → app1_out(.(X, Xs), Ys, .(X, Zs))
U4(Xs, X, Ys, X1s, X2s, app1_out(X1s, X2s, Zs)) → U5(Xs, X, Ys, perm_in(Zs, Ys))
U5(Xs, X, Ys, perm_out(Zs, Ys)) → perm_out(Xs, .(X, Ys))

The argument filtering Pi contains the following mapping:
perm_in(x1, x2)  =  perm_in(x1)
[]  =  []
perm_out(x1, x2)  =  perm_out(x2)
.(x1, x2)  =  .(x2)
U3(x1, x2, x3, x4)  =  U3(x4)
app2_in(x1, x2, x3)  =  app2_in(x3)
app2_out(x1, x2, x3)  =  app2_out(x1, x2)
U2(x1, x2, x3, x4, x5)  =  U2(x5)
U4(x1, x2, x3, x4, x5, x6)  =  U4(x6)
app1_in(x1, x2, x3)  =  app1_in(x1, x2)
app1_out(x1, x2, x3)  =  app1_out(x3)
U1(x1, x2, x3, x4, x5)  =  U1(x5)
U5(x1, x2, x3, x4)  =  U5(x4)
U31(x1, x2, x3, x4)  =  U31(x4)
U41(x1, x2, x3, x4, x5, x6)  =  U41(x6)
PERM_IN(x1, x2)  =  PERM_IN(x1)

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

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

U41(Xs, X, Ys, X1s, X2s, app1_out(X1s, X2s, Zs)) → PERM_IN(Zs, Ys)
U31(Xs, X, Ys, app2_out(X1s, .(X, X2s), Xs)) → U41(Xs, X, Ys, X1s, X2s, app1_in(X1s, X2s, Zs))
PERM_IN(Xs, .(X, Ys)) → U31(Xs, X, Ys, app2_in(X1s, .(X, X2s), Xs))

The TRS R consists of the following rules:

app1_in([], Ys, Ys) → app1_out([], Ys, Ys)
app1_in(.(X, Xs), Ys, .(X, Zs)) → U1(X, Xs, Ys, Zs, app1_in(Xs, Ys, Zs))
app2_in([], Ys, Ys) → app2_out([], Ys, Ys)
app2_in(.(X, Xs), Ys, .(X, Zs)) → U2(X, Xs, Ys, Zs, app2_in(Xs, Ys, Zs))
U1(X, Xs, Ys, Zs, app1_out(Xs, Ys, Zs)) → app1_out(.(X, Xs), Ys, .(X, Zs))
U2(X, Xs, Ys, Zs, app2_out(Xs, Ys, Zs)) → app2_out(.(X, Xs), Ys, .(X, Zs))

The argument filtering Pi contains the following mapping:
[]  =  []
.(x1, x2)  =  .(x2)
app2_in(x1, x2, x3)  =  app2_in(x3)
app2_out(x1, x2, x3)  =  app2_out(x1, x2)
U2(x1, x2, x3, x4, x5)  =  U2(x5)
app1_in(x1, x2, x3)  =  app1_in(x1, x2)
app1_out(x1, x2, x3)  =  app1_out(x3)
U1(x1, x2, x3, x4, x5)  =  U1(x5)
U31(x1, x2, x3, x4)  =  U31(x4)
U41(x1, x2, x3, x4, x5, x6)  =  U41(x6)
PERM_IN(x1, x2)  =  PERM_IN(x1)

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
                        ↳ RuleRemovalProof

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

U31(app2_out(X1s, .(X2s))) → U41(app1_in(X1s, X2s))
PERM_IN(Xs) → U31(app2_in(Xs))
U41(app1_out(Zs)) → PERM_IN(Zs)

The TRS R consists of the following rules:

app1_in([], Ys) → app1_out(Ys)
app1_in(.(Xs), Ys) → U1(app1_in(Xs, Ys))
app2_in(Ys) → app2_out([], Ys)
app2_in(.(Zs)) → U2(app2_in(Zs))
U1(app1_out(Zs)) → app1_out(.(Zs))
U2(app2_out(Xs, Ys)) → app2_out(.(Xs), Ys)

The set Q consists of the following terms:

app1_in(x0, x1)
app2_in(x0)
U1(x0)
U2(x0)

We have to consider all (P,Q,R)-chains.
By using the rule removal processor [15] with the following polynomial ordering [25], at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

U31(app2_out(X1s, .(X2s))) → U41(app1_in(X1s, X2s))

Strictly oriented rules of the TRS R:

app2_in(Ys) → app2_out([], Ys)

Used ordering: POLO with Polynomial interpretation [25]:

POL(.(x1)) = 1 + x1   
POL(PERM_IN(x1)) = 1 + 2·x1   
POL(U1(x1)) = 2 + x1   
POL(U2(x1)) = 2 + x1   
POL(U31(x1)) = x1   
POL(U41(x1)) = 1 + x1   
POL([]) = 0   
POL(app1_in(x1, x2)) = 2·x1 + 2·x2   
POL(app1_out(x1)) = 2·x1   
POL(app2_in(x1)) = 1 + 2·x1   
POL(app2_out(x1, x2)) = 2·x1 + 2·x2   



↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
                      ↳ QDP
                        ↳ RuleRemovalProof
QDP
                            ↳ DependencyGraphProof

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

PERM_IN(Xs) → U31(app2_in(Xs))
U41(app1_out(Zs)) → PERM_IN(Zs)

The TRS R consists of the following rules:

app1_in([], Ys) → app1_out(Ys)
app1_in(.(Xs), Ys) → U1(app1_in(Xs, Ys))
app2_in(.(Zs)) → U2(app2_in(Zs))
U1(app1_out(Zs)) → app1_out(.(Zs))
U2(app2_out(Xs, Ys)) → app2_out(.(Xs), Ys)

The set Q consists of the following terms:

app1_in(x0, x1)
app2_in(x0)
U1(x0)
U2(x0)

We have to consider all (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 2 less nodes.