Left Termination proof of /tmp/tmpUTUHv7/p.pl

Left Termination of the query pattern p_in_1(g) w.r.t. the given Prolog program could successfully be proven:

↳ Prolog

  ↳ PrologToPiTRSProof

Clauses:

p(0).
p(s(X)) :- ','(geq(X, Y), p(Y)).
geq(X, X).
geq(s(X), Y) :- geq(X, Y).

Queries:

p(g).

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

p_in_g(0) → p_out_g(0)
p_in_g(s(X)) → U1_g(X, geq_in_ga(X, Y))
geq_in_ga(X, X) → geq_out_ga(X, X)
geq_in_ga(s(X), Y) → U3_ga(X, Y, geq_in_ga(X, Y))
U3_ga(X, Y, geq_out_ga(X, Y)) → geq_out_ga(s(X), Y)
U1_g(X, geq_out_ga(X, Y)) → U2_g(X, p_in_g(Y))
U2_g(X, p_out_g(Y)) → p_out_g(s(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:

p_in_g(0) → p_out_g(0)
p_in_g(s(X)) → U1_g(X, geq_in_ga(X, Y))
geq_in_ga(X, X) → geq_out_ga(X, X)
geq_in_ga(s(X), Y) → U3_ga(X, Y, geq_in_ga(X, Y))
U3_ga(X, Y, geq_out_ga(X, Y)) → geq_out_ga(s(X), Y)
U1_g(X, geq_out_ga(X, Y)) → U2_g(X, p_in_g(Y))
U2_g(X, p_out_g(Y)) → p_out_g(s(X))

The argument filtering Pi contains the following mapping:
p_in_g(x1) = p_in_g(x1)
0 = 0
p_out_g(x1) = p_out_g
s(x1) = s(x1)
U1_g(x1, x2) = U1_g(x2)
geq_in_ga(x1, x2) = geq_in_ga(x1)
geq_out_ga(x1, x2) = geq_out_ga(x2)
U3_ga(x1, x2, x3) = U3_ga(x3)
U2_g(x1, x2) = U2_g(x2)

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

P_IN_G(s(X)) → U1_G(X, geq_in_ga(X, Y))
P_IN_G(s(X)) → GEQ_IN_GA(X, Y)
GEQ_IN_GA(s(X), Y) → U3_GA(X, Y, geq_in_ga(X, Y))
GEQ_IN_GA(s(X), Y) → GEQ_IN_GA(X, Y)
U1_G(X, geq_out_ga(X, Y)) → U2_G(X, p_in_g(Y))
U1_G(X, geq_out_ga(X, Y)) → P_IN_G(Y)

The TRS R consists of the following rules:

p_in_g(0) → p_out_g(0)
p_in_g(s(X)) → U1_g(X, geq_in_ga(X, Y))
geq_in_ga(X, X) → geq_out_ga(X, X)
geq_in_ga(s(X), Y) → U3_ga(X, Y, geq_in_ga(X, Y))
U3_ga(X, Y, geq_out_ga(X, Y)) → geq_out_ga(s(X), Y)
U1_g(X, geq_out_ga(X, Y)) → U2_g(X, p_in_g(Y))
U2_g(X, p_out_g(Y)) → p_out_g(s(X))

The argument filtering Pi contains the following mapping:
p_in_g(x1) = p_in_g(x1)
0 = 0
p_out_g(x1) = p_out_g
s(x1) = s(x1)
U1_g(x1, x2) = U1_g(x2)
geq_in_ga(x1, x2) = geq_in_ga(x1)
geq_out_ga(x1, x2) = geq_out_ga(x2)
U3_ga(x1, x2, x3) = U3_ga(x3)
U2_g(x1, x2) = U2_g(x2)
U2_G(x1, x2) = U2_G(x2)
P_IN_G(x1) = P_IN_G(x1)
GEQ_IN_GA(x1, x2) = GEQ_IN_GA(x1)
U3_GA(x1, x2, x3) = U3_GA(x3)
U1_G(x1, x2) = U1_G(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:

P_IN_G(s(X)) → U1_G(X, geq_in_ga(X, Y))
P_IN_G(s(X)) → GEQ_IN_GA(X, Y)
GEQ_IN_GA(s(X), Y) → U3_GA(X, Y, geq_in_ga(X, Y))
GEQ_IN_GA(s(X), Y) → GEQ_IN_GA(X, Y)
U1_G(X, geq_out_ga(X, Y)) → U2_G(X, p_in_g(Y))
U1_G(X, geq_out_ga(X, Y)) → P_IN_G(Y)

The TRS R consists of the following rules:

p_in_g(0) → p_out_g(0)
p_in_g(s(X)) → U1_g(X, geq_in_ga(X, Y))
geq_in_ga(X, X) → geq_out_ga(X, X)
geq_in_ga(s(X), Y) → U3_ga(X, Y, geq_in_ga(X, Y))
U3_ga(X, Y, geq_out_ga(X, Y)) → geq_out_ga(s(X), Y)
U1_g(X, geq_out_ga(X, Y)) → U2_g(X, p_in_g(Y))
U2_g(X, p_out_g(Y)) → p_out_g(s(X))

The argument filtering Pi contains the following mapping:
p_in_g(x1) = p_in_g(x1)
0 = 0
p_out_g(x1) = p_out_g
s(x1) = s(x1)
U1_g(x1, x2) = U1_g(x2)
geq_in_ga(x1, x2) = geq_in_ga(x1)
geq_out_ga(x1, x2) = geq_out_ga(x2)
U3_ga(x1, x2, x3) = U3_ga(x3)
U2_g(x1, x2) = U2_g(x2)
U2_G(x1, x2) = U2_G(x2)
P_IN_G(x1) = P_IN_G(x1)
GEQ_IN_GA(x1, x2) = GEQ_IN_GA(x1)
U3_GA(x1, x2, x3) = U3_GA(x3)
U1_G(x1, x2) = U1_G(x2)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph [30] contains 2 SCCs with 3 less nodes.

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

                ↳ UsableRulesProof

              ↳ PiDP

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

GEQ_IN_GA(s(X), Y) → GEQ_IN_GA(X, Y)

The TRS R consists of the following rules:

p_in_g(0) → p_out_g(0)
p_in_g(s(X)) → U1_g(X, geq_in_ga(X, Y))
geq_in_ga(X, X) → geq_out_ga(X, X)
geq_in_ga(s(X), Y) → U3_ga(X, Y, geq_in_ga(X, Y))
U3_ga(X, Y, geq_out_ga(X, Y)) → geq_out_ga(s(X), Y)
U1_g(X, geq_out_ga(X, Y)) → U2_g(X, p_in_g(Y))
U2_g(X, p_out_g(Y)) → p_out_g(s(X))

The argument filtering Pi contains the following mapping:
p_in_g(x1) = p_in_g(x1)
0 = 0
p_out_g(x1) = p_out_g
s(x1) = s(x1)
U1_g(x1, x2) = U1_g(x2)
geq_in_ga(x1, x2) = geq_in_ga(x1)
geq_out_ga(x1, x2) = geq_out_ga(x2)
U3_ga(x1, x2, x3) = U3_ga(x3)
U2_g(x1, x2) = U2_g(x2)
GEQ_IN_GA(x1, x2) = GEQ_IN_GA(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

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

              ↳ PiDP

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

GEQ_IN_GA(s(X), Y) → GEQ_IN_GA(X, Y)

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

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ QDPSizeChangeProof

              ↳ PiDP

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

GEQ_IN_GA(s(X)) → GEQ_IN_GA(X)

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:

GEQ_IN_GA(s(X)) → GEQ_IN_GA(X)
The graph contains the following edges 1 > 1

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

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

U1_G(X, geq_out_ga(X, Y)) → P_IN_G(Y)
P_IN_G(s(X)) → U1_G(X, geq_in_ga(X, Y))

The TRS R consists of the following rules:

p_in_g(0) → p_out_g(0)
p_in_g(s(X)) → U1_g(X, geq_in_ga(X, Y))
geq_in_ga(X, X) → geq_out_ga(X, X)
geq_in_ga(s(X), Y) → U3_ga(X, Y, geq_in_ga(X, Y))
U3_ga(X, Y, geq_out_ga(X, Y)) → geq_out_ga(s(X), Y)
U1_g(X, geq_out_ga(X, Y)) → U2_g(X, p_in_g(Y))
U2_g(X, p_out_g(Y)) → p_out_g(s(X))

The argument filtering Pi contains the following mapping:
p_in_g(x1) = p_in_g(x1)
0 = 0
p_out_g(x1) = p_out_g
s(x1) = s(x1)
U1_g(x1, x2) = U1_g(x2)
geq_in_ga(x1, x2) = geq_in_ga(x1)
geq_out_ga(x1, x2) = geq_out_ga(x2)
U3_ga(x1, x2, x3) = U3_ga(x3)
U2_g(x1, x2) = U2_g(x2)
P_IN_G(x1) = P_IN_G(x1)
U1_G(x1, x2) = U1_G(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

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

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

U1_G(X, geq_out_ga(X, Y)) → P_IN_G(Y)
P_IN_G(s(X)) → U1_G(X, geq_in_ga(X, Y))

The TRS R consists of the following rules:

geq_in_ga(X, X) → geq_out_ga(X, X)
geq_in_ga(s(X), Y) → U3_ga(X, Y, geq_in_ga(X, Y))
U3_ga(X, Y, geq_out_ga(X, Y)) → geq_out_ga(s(X), Y)

The argument filtering Pi contains the following mapping:
s(x1) = s(x1)
geq_in_ga(x1, x2) = geq_in_ga(x1)
geq_out_ga(x1, x2) = geq_out_ga(x2)
U3_ga(x1, x2, x3) = U3_ga(x3)
P_IN_G(x1) = P_IN_G(x1)
U1_G(x1, x2) = U1_G(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

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ UsableRulesReductionPairsProof

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

P_IN_G(s(X)) → U1_G(geq_in_ga(X))
U1_G(geq_out_ga(Y)) → P_IN_G(Y)

The TRS R consists of the following rules:

geq_in_ga(X) → geq_out_ga(X)
geq_in_ga(s(X)) → U3_ga(geq_in_ga(X))
U3_ga(geq_out_ga(Y)) → geq_out_ga(Y)

The set Q consists of the following terms:

geq_in_ga(x0)
U3_ga(x0)

We have to consider all (P,Q,R)-chains.
By using the usable rules with reduction pair processor [15] with a polynomial ordering [25], all dependency pairs and the corresponding usable rules [17] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.

The following dependency pairs can be deleted:

P_IN_G(s(X)) → U1_G(geq_in_ga(X))

The following rules are removed from R:

geq_in_ga(s(X)) → U3_ga(geq_in_ga(X))
U3_ga(geq_out_ga(Y)) → geq_out_ga(Y)

Used ordering: POLO with Polynomial interpretation [25]:

POL(P_IN_G(x₁)) = x₁
POL(U1_G(x₁)) = 2·x₁
POL(U3_ga(x₁)) = 1 + x₁
POL(geq_in_ga(x₁)) = x₁
POL(geq_out_ga(x₁)) = x₁
POL(s(x₁)) = 1 + 2·x₁

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ UsableRulesReductionPairsProof

                          ↳ QDP

                            ↳ DependencyGraphProof

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

U1_G(geq_out_ga(Y)) → P_IN_G(Y)

The TRS R consists of the following rules:

geq_in_ga(X) → geq_out_ga(X)

The set Q consists of the following terms:

geq_in_ga(x0)
U3_ga(x0)

We have to consider all (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 1 less node.