Left Termination proof of /tmp/tmpHibAbr/NJ2.pl

Left Termination of the query pattern f_in_3(g, g, a) w.r.t. the given Prolog program could successfully be proven:

↳ Prolog

  ↳ PrologToPiTRSProof

Clauses:

f([], RES, RES).
f(.(Head, Tail), X, RES) :- g(Tail, X, .(Head, Tail), RES).
g(A, B, C, RES) :- f(A, .(B, C), RES).

Queries:

f(g,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:

f_in(.(Head, Tail), X, RES) → U1(Head, Tail, X, RES, g_in(Tail, X, .(Head, Tail), RES))
g_in(A, B, C, RES) → U2(A, B, C, RES, f_in(A, .(B, C), RES))
f_in([], RES, RES) → f_out([], RES, RES)
U2(A, B, C, RES, f_out(A, .(B, C), RES)) → g_out(A, B, C, RES)
U1(Head, Tail, X, RES, g_out(Tail, X, .(Head, Tail), RES)) → f_out(.(Head, Tail), X, RES)

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:

f_in(.(Head, Tail), X, RES) → U1(Head, Tail, X, RES, g_in(Tail, X, .(Head, Tail), RES))
g_in(A, B, C, RES) → U2(A, B, C, RES, f_in(A, .(B, C), RES))
f_in([], RES, RES) → f_out([], RES, RES)
U2(A, B, C, RES, f_out(A, .(B, C), RES)) → g_out(A, B, C, RES)
U1(Head, Tail, X, RES, g_out(Tail, X, .(Head, Tail), RES)) → f_out(.(Head, Tail), X, RES)

The argument filtering Pi contains the following mapping:
f_in(x1, x2, x3) = f_in(x1, x2)
.(x1, x2) = .(x1, x2)
U1(x1, x2, x3, x4, x5) = U1(x5)
g_in(x1, x2, x3, x4) = g_in(x1, x2, x3)
U2(x1, x2, x3, x4, x5) = U2(x5)
[] = []
f_out(x1, x2, x3) = f_out(x3)
g_out(x1, x2, x3, x4) = g_out(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:

F_IN(.(Head, Tail), X, RES) → U1¹(Head, Tail, X, RES, g_in(Tail, X, .(Head, Tail), RES))
F_IN(.(Head, Tail), X, RES) → G_IN(Tail, X, .(Head, Tail), RES)
G_IN(A, B, C, RES) → U2¹(A, B, C, RES, f_in(A, .(B, C), RES))
G_IN(A, B, C, RES) → F_IN(A, .(B, C), RES)

The TRS R consists of the following rules:

f_in(.(Head, Tail), X, RES) → U1(Head, Tail, X, RES, g_in(Tail, X, .(Head, Tail), RES))
g_in(A, B, C, RES) → U2(A, B, C, RES, f_in(A, .(B, C), RES))
f_in([], RES, RES) → f_out([], RES, RES)
U2(A, B, C, RES, f_out(A, .(B, C), RES)) → g_out(A, B, C, RES)
U1(Head, Tail, X, RES, g_out(Tail, X, .(Head, Tail), RES)) → f_out(.(Head, Tail), X, RES)

The argument filtering Pi contains the following mapping:
f_in(x1, x2, x3) = f_in(x1, x2)
.(x1, x2) = .(x1, x2)
U1(x1, x2, x3, x4, x5) = U1(x5)
g_in(x1, x2, x3, x4) = g_in(x1, x2, x3)
U2(x1, x2, x3, x4, x5) = U2(x5)
[] = []
f_out(x1, x2, x3) = f_out(x3)
g_out(x1, x2, x3, x4) = g_out(x4)
G_IN(x1, x2, x3, x4) = G_IN(x1, x2, x3)
F_IN(x1, x2, x3) = F_IN(x1, x2)
U2¹(x1, x2, x3, x4, x5) = U2¹(x5)
U1¹(x1, x2, x3, x4, x5) = U1¹(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:

F_IN(.(Head, Tail), X, RES) → U1¹(Head, Tail, X, RES, g_in(Tail, X, .(Head, Tail), RES))
F_IN(.(Head, Tail), X, RES) → G_IN(Tail, X, .(Head, Tail), RES)
G_IN(A, B, C, RES) → U2¹(A, B, C, RES, f_in(A, .(B, C), RES))
G_IN(A, B, C, RES) → F_IN(A, .(B, C), RES)

The TRS R consists of the following rules:

f_in(.(Head, Tail), X, RES) → U1(Head, Tail, X, RES, g_in(Tail, X, .(Head, Tail), RES))
g_in(A, B, C, RES) → U2(A, B, C, RES, f_in(A, .(B, C), RES))
f_in([], RES, RES) → f_out([], RES, RES)
U2(A, B, C, RES, f_out(A, .(B, C), RES)) → g_out(A, B, C, RES)
U1(Head, Tail, X, RES, g_out(Tail, X, .(Head, Tail), RES)) → f_out(.(Head, Tail), X, RES)

The argument filtering Pi contains the following mapping:
f_in(x1, x2, x3) = f_in(x1, x2)
.(x1, x2) = .(x1, x2)
U1(x1, x2, x3, x4, x5) = U1(x5)
g_in(x1, x2, x3, x4) = g_in(x1, x2, x3)
U2(x1, x2, x3, x4, x5) = U2(x5)
[] = []
f_out(x1, x2, x3) = f_out(x3)
g_out(x1, x2, x3, x4) = g_out(x4)
G_IN(x1, x2, x3, x4) = G_IN(x1, x2, x3)
F_IN(x1, x2, x3) = F_IN(x1, x2)
U2¹(x1, x2, x3, x4, x5) = U2¹(x5)
U1¹(x1, x2, x3, x4, x5) = U1¹(x5)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph [30] contains 1 SCC with 2 less nodes.

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ PiDP

              ↳ UsableRulesProof

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

G_IN(A, B, C, RES) → F_IN(A, .(B, C), RES)
F_IN(.(Head, Tail), X, RES) → G_IN(Tail, X, .(Head, Tail), RES)

The TRS R consists of the following rules:

f_in(.(Head, Tail), X, RES) → U1(Head, Tail, X, RES, g_in(Tail, X, .(Head, Tail), RES))
g_in(A, B, C, RES) → U2(A, B, C, RES, f_in(A, .(B, C), RES))
f_in([], RES, RES) → f_out([], RES, RES)
U2(A, B, C, RES, f_out(A, .(B, C), RES)) → g_out(A, B, C, RES)
U1(Head, Tail, X, RES, g_out(Tail, X, .(Head, Tail), RES)) → f_out(.(Head, Tail), X, RES)

The argument filtering Pi contains the following mapping:
f_in(x1, x2, x3) = f_in(x1, x2)
.(x1, x2) = .(x1, x2)
U1(x1, x2, x3, x4, x5) = U1(x5)
g_in(x1, x2, x3, x4) = g_in(x1, x2, x3)
U2(x1, x2, x3, x4, x5) = U2(x5)
[] = []
f_out(x1, x2, x3) = f_out(x3)
g_out(x1, x2, x3, x4) = g_out(x4)
G_IN(x1, x2, x3, x4) = G_IN(x1, x2, x3)
F_IN(x1, x2, x3) = F_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

            ↳ PiDP

              ↳ UsableRulesProof

                ↳ PiDP

                  ↳ PiDPToQDPProof

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

G_IN(A, B, C, RES) → F_IN(A, .(B, C), RES)
F_IN(.(Head, Tail), X, RES) → G_IN(Tail, X, .(Head, Tail), RES)

R is empty.
The argument filtering Pi contains the following mapping:
.(x1, x2) = .(x1, x2)
G_IN(x1, x2, x3, x4) = G_IN(x1, x2, x3)
F_IN(x1, x2, x3) = F_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

            ↳ PiDP

              ↳ UsableRulesProof

                ↳ PiDP

                  ↳ PiDPToQDPProof

                    ↳ QDP

                      ↳ QDPSizeChangeProof

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

F_IN(.(Head, Tail), X) → G_IN(Tail, X, .(Head, Tail))
G_IN(A, B, C) → F_IN(A, .(B, C))

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:

F_IN(.(Head, Tail), X) → G_IN(Tail, X, .(Head, Tail))
The graph contains the following edges 1 > 1, 2 >= 2, 1 >= 3
G_IN(A, B, C) → F_IN(A, .(B, C))
The graph contains the following edges 1 >= 1