Left Termination proof of /tmp/tmpX6LSVH/incomplete2.pl

Left Termination of the query pattern f_in_1(g) w.r.t. the given Prolog program could successfully be proven:

↳ Prolog

  ↳ PrologToPiTRSProof

Clauses:

f(X) :- g(s(s(s(X)))).
f(s(X)) :- f(X).
g(s(s(s(s(X))))) :- f(X).

Queries:

f(g).

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

f_in_g(X) → U1_g(X, g_in_g(s(s(s(X)))))
g_in_g(s(s(s(s(X))))) → U3_g(X, f_in_g(X))
f_in_g(s(X)) → U2_g(X, f_in_g(X))
U2_g(X, f_out_g(X)) → f_out_g(s(X))
U3_g(X, f_out_g(X)) → g_out_g(s(s(s(s(X)))))
U1_g(X, g_out_g(s(s(s(X))))) → f_out_g(X)

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_g(X) → U1_g(X, g_in_g(s(s(s(X)))))
g_in_g(s(s(s(s(X))))) → U3_g(X, f_in_g(X))
f_in_g(s(X)) → U2_g(X, f_in_g(X))
U2_g(X, f_out_g(X)) → f_out_g(s(X))
U3_g(X, f_out_g(X)) → g_out_g(s(s(s(s(X)))))
U1_g(X, g_out_g(s(s(s(X))))) → f_out_g(X)

The argument filtering Pi contains the following mapping:
f_in_g(x1) = f_in_g(x1)
U1_g(x1, x2) = U1_g(x2)
g_in_g(x1) = g_in_g(x1)
s(x1) = s(x1)
U3_g(x1, x2) = U3_g(x2)
U2_g(x1, x2) = U2_g(x2)
f_out_g(x1) = f_out_g
g_out_g(x1) = g_out_g

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_G(X) → U1_G(X, g_in_g(s(s(s(X)))))
F_IN_G(X) → G_IN_G(s(s(s(X))))
G_IN_G(s(s(s(s(X))))) → U3_G(X, f_in_g(X))
G_IN_G(s(s(s(s(X))))) → F_IN_G(X)
F_IN_G(s(X)) → U2_G(X, f_in_g(X))
F_IN_G(s(X)) → F_IN_G(X)

The TRS R consists of the following rules:

f_in_g(X) → U1_g(X, g_in_g(s(s(s(X)))))
g_in_g(s(s(s(s(X))))) → U3_g(X, f_in_g(X))
f_in_g(s(X)) → U2_g(X, f_in_g(X))
U2_g(X, f_out_g(X)) → f_out_g(s(X))
U3_g(X, f_out_g(X)) → g_out_g(s(s(s(s(X)))))
U1_g(X, g_out_g(s(s(s(X))))) → f_out_g(X)

The argument filtering Pi contains the following mapping:
f_in_g(x1) = f_in_g(x1)
U1_g(x1, x2) = U1_g(x2)
g_in_g(x1) = g_in_g(x1)
s(x1) = s(x1)
U3_g(x1, x2) = U3_g(x2)
U2_g(x1, x2) = U2_g(x2)
f_out_g(x1) = f_out_g
g_out_g(x1) = g_out_g
U2_G(x1, x2) = U2_G(x2)
F_IN_G(x1) = F_IN_G(x1)
U1_G(x1, x2) = U1_G(x2)
U3_G(x1, x2) = U3_G(x2)
G_IN_G(x1) = G_IN_G(x1)

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_G(X) → U1_G(X, g_in_g(s(s(s(X)))))
F_IN_G(X) → G_IN_G(s(s(s(X))))
G_IN_G(s(s(s(s(X))))) → U3_G(X, f_in_g(X))
G_IN_G(s(s(s(s(X))))) → F_IN_G(X)
F_IN_G(s(X)) → U2_G(X, f_in_g(X))
F_IN_G(s(X)) → F_IN_G(X)

The TRS R consists of the following rules:

f_in_g(X) → U1_g(X, g_in_g(s(s(s(X)))))
g_in_g(s(s(s(s(X))))) → U3_g(X, f_in_g(X))
f_in_g(s(X)) → U2_g(X, f_in_g(X))
U2_g(X, f_out_g(X)) → f_out_g(s(X))
U3_g(X, f_out_g(X)) → g_out_g(s(s(s(s(X)))))
U1_g(X, g_out_g(s(s(s(X))))) → f_out_g(X)

The argument filtering Pi contains the following mapping:
f_in_g(x1) = f_in_g(x1)
U1_g(x1, x2) = U1_g(x2)
g_in_g(x1) = g_in_g(x1)
s(x1) = s(x1)
U3_g(x1, x2) = U3_g(x2)
U2_g(x1, x2) = U2_g(x2)
f_out_g(x1) = f_out_g
g_out_g(x1) = g_out_g
U2_G(x1, x2) = U2_G(x2)
F_IN_G(x1) = F_IN_G(x1)
U1_G(x1, x2) = U1_G(x2)
U3_G(x1, x2) = U3_G(x2)
G_IN_G(x1) = G_IN_G(x1)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph [30] contains 1 SCC with 3 less nodes.

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ PiDP

              ↳ UsableRulesProof

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

G_IN_G(s(s(s(s(X))))) → F_IN_G(X)
F_IN_G(s(X)) → F_IN_G(X)
F_IN_G(X) → G_IN_G(s(s(s(X))))

The TRS R consists of the following rules:

f_in_g(X) → U1_g(X, g_in_g(s(s(s(X)))))
g_in_g(s(s(s(s(X))))) → U3_g(X, f_in_g(X))
f_in_g(s(X)) → U2_g(X, f_in_g(X))
U2_g(X, f_out_g(X)) → f_out_g(s(X))
U3_g(X, f_out_g(X)) → g_out_g(s(s(s(s(X)))))
U1_g(X, g_out_g(s(s(s(X))))) → f_out_g(X)

The argument filtering Pi contains the following mapping:
f_in_g(x1) = f_in_g(x1)
U1_g(x1, x2) = U1_g(x2)
g_in_g(x1) = g_in_g(x1)
s(x1) = s(x1)
U3_g(x1, x2) = U3_g(x2)
U2_g(x1, x2) = U2_g(x2)
f_out_g(x1) = f_out_g
g_out_g(x1) = g_out_g
F_IN_G(x1) = F_IN_G(x1)
G_IN_G(x1) = G_IN_G(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

            ↳ PiDP

              ↳ UsableRulesProof

                ↳ PiDP

                  ↳ PiDPToQDPProof

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

G_IN_G(s(s(s(s(X))))) → F_IN_G(X)
F_IN_G(s(X)) → F_IN_G(X)
F_IN_G(X) → G_IN_G(s(s(s(X))))

R is empty.
Pi is empty.
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

                      ↳ RFCMatchBoundsDPProof

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

G_IN_G(s(s(s(s(X))))) → F_IN_G(X)
F_IN_G(s(X)) → F_IN_G(X)
F_IN_G(X) → G_IN_G(s(s(s(X))))

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
Termination of the TRS P cup R can be shown by a matchbound [6,7] of 2. This implies finiteness of the given DP problem.
The following rules (P cup R) were used to construct the certificate:

G_IN_G(s(s(s(s(X))))) → F_IN_G(X)
F_IN_G(s(X)) → F_IN_G(X)
F_IN_G(X) → G_IN_G(s(s(s(X))))

The certificate found is represented by the following graph.

The certificate consists of the following enumerated nodes:

60, 61, 62, 64, 63, 65, 67, 66, 70, 69, 68

Node 60 is start node and node 61 is final node.

Those nodes are connect through the following edges:

60 to 61 labelled F_IN_G_1(0), F_IN_G_1(1)
60 to 62 labelled G_IN_G_1(0)
60 to 65 labelled G_IN_G_1(1)
60 to 68 labelled G_IN_G_1(2)
61 to 61 labelled #_1(0)
62 to 63 labelled s_1(0)
64 to 61 labelled s_1(0)
63 to 64 labelled s_1(0)
65 to 66 labelled s_1(1)
67 to 61 labelled s_1(1)
66 to 67 labelled s_1(1)
70 to 61 labelled s_1(2)
69 to 70 labelled s_1(2)
68 to 69 labelled s_1(2)