Left Termination of the query pattern app2_in_3(a, g, g) 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).

Queries:

app2(a,g,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:

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))

The argument filtering Pi contains the following mapping:
app2_in(x1, x2, x3)  =  app2_in(x2, x3)
[]  =  []
app2_out(x1, x2, x3)  =  app2_out(x1)
.(x1, x2)  =  .(x1, x2)
U2(x1, x2, x3, x4, x5)  =  U2(x1, x5)

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:

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))

The argument filtering Pi contains the following mapping:
app2_in(x1, x2, x3)  =  app2_in(x2, x3)
[]  =  []
app2_out(x1, x2, x3)  =  app2_out(x1)
.(x1, x2)  =  .(x1, x2)
U2(x1, x2, x3, x4, x5)  =  U2(x1, x5)


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

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)

The TRS R consists of the following rules:

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))

The argument filtering Pi contains the following mapping:
app2_in(x1, x2, x3)  =  app2_in(x2, x3)
[]  =  []
app2_out(x1, x2, x3)  =  app2_out(x1)
.(x1, x2)  =  .(x1, x2)
U2(x1, x2, x3, x4, x5)  =  U2(x1, x5)
APP2_IN(x1, x2, x3)  =  APP2_IN(x2, x3)
U21(x1, x2, x3, x4, x5)  =  U21(x1, x5)

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:

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)

The TRS R consists of the following rules:

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))

The argument filtering Pi contains the following mapping:
app2_in(x1, x2, x3)  =  app2_in(x2, x3)
[]  =  []
app2_out(x1, x2, x3)  =  app2_out(x1)
.(x1, x2)  =  .(x1, x2)
U2(x1, x2, x3, x4, x5)  =  U2(x1, x5)
APP2_IN(x1, x2, x3)  =  APP2_IN(x2, x3)
U21(x1, x2, x3, x4, x5)  =  U21(x1, x5)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph [30] contains 1 SCC with 1 less node.

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

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:

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))

The argument filtering Pi contains the following mapping:
app2_in(x1, x2, x3)  =  app2_in(x2, x3)
[]  =  []
app2_out(x1, x2, x3)  =  app2_out(x1)
.(x1, x2)  =  .(x1, x2)
U2(x1, x2, x3, x4, x5)  =  U2(x1, x5)
APP2_IN(x1, x2, x3)  =  APP2_IN(x2, 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
            ↳ PiDP
              ↳ UsableRulesProof
                ↳ PiDP
                  ↳ PiDPToQDPProof

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)  =  .(x1, x2)
APP2_IN(x1, x2, x3)  =  APP2_IN(x2, 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
            ↳ PiDP
              ↳ UsableRulesProof
                ↳ PiDP
                  ↳ PiDPToQDPProof
                    ↳ QDP
                      ↳ QDPSizeChangeProof

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

APP2_IN(Ys, .(X, Zs)) → APP2_IN(Ys, 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: