Left Termination proof of /tmp/tmpQfpc

Left Termination of the query pattern average_in_3(g, g, a) w.r.t. the given Prolog program could successfully be proven:

↳ Prolog

  ↳ PrologToPiTRSProof

Clauses:

average(0, 0, 0).
average(0, s(0), 0).
average(0, s(s(0)), s(0)).
average(s(X), Y, Z) :- average(X, s(Y), Z).
average(X, s(s(s(Y))), s(Z)) :- average(s(X), Y, Z).

Queries:

average(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:

average_in(X, s(s(s(Y))), s(Z)) → U2(X, Y, Z, average_in(s(X), Y, Z))
average_in(s(X), Y, Z) → U1(X, Y, Z, average_in(X, s(Y), Z))
average_in(0, s(s(0)), s(0)) → average_out(0, s(s(0)), s(0))
average_in(0, s(0), 0) → average_out(0, s(0), 0)
average_in(0, 0, 0) → average_out(0, 0, 0)
U1(X, Y, Z, average_out(X, s(Y), Z)) → average_out(s(X), Y, Z)
U2(X, Y, Z, average_out(s(X), Y, Z)) → average_out(X, s(s(s(Y))), s(Z))

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:

average_in(X, s(s(s(Y))), s(Z)) → U2(X, Y, Z, average_in(s(X), Y, Z))
average_in(s(X), Y, Z) → U1(X, Y, Z, average_in(X, s(Y), Z))
average_in(0, s(s(0)), s(0)) → average_out(0, s(s(0)), s(0))
average_in(0, s(0), 0) → average_out(0, s(0), 0)
average_in(0, 0, 0) → average_out(0, 0, 0)
U1(X, Y, Z, average_out(X, s(Y), Z)) → average_out(s(X), Y, Z)
U2(X, Y, Z, average_out(s(X), Y, Z)) → average_out(X, s(s(s(Y))), s(Z))

The argument filtering Pi contains the following mapping:
average_in(x1, x2, x3) = average_in(x1, x2)
s(x1) = s(x1)
U2(x1, x2, x3, x4) = U2(x4)
U1(x1, x2, x3, x4) = U1(x4)
0 = 0
average_out(x1, x2, x3) = average_out(x3)

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

AVERAGE_IN(X, s(s(s(Y))), s(Z)) → U2¹(X, Y, Z, average_in(s(X), Y, Z))
AVERAGE_IN(X, s(s(s(Y))), s(Z)) → AVERAGE_IN(s(X), Y, Z)
AVERAGE_IN(s(X), Y, Z) → U1¹(X, Y, Z, average_in(X, s(Y), Z))
AVERAGE_IN(s(X), Y, Z) → AVERAGE_IN(X, s(Y), Z)

The TRS R consists of the following rules:

average_in(X, s(s(s(Y))), s(Z)) → U2(X, Y, Z, average_in(s(X), Y, Z))
average_in(s(X), Y, Z) → U1(X, Y, Z, average_in(X, s(Y), Z))
average_in(0, s(s(0)), s(0)) → average_out(0, s(s(0)), s(0))
average_in(0, s(0), 0) → average_out(0, s(0), 0)
average_in(0, 0, 0) → average_out(0, 0, 0)
U1(X, Y, Z, average_out(X, s(Y), Z)) → average_out(s(X), Y, Z)
U2(X, Y, Z, average_out(s(X), Y, Z)) → average_out(X, s(s(s(Y))), s(Z))

The argument filtering Pi contains the following mapping:
average_in(x1, x2, x3) = average_in(x1, x2)
s(x1) = s(x1)
U2(x1, x2, x3, x4) = U2(x4)
U1(x1, x2, x3, x4) = U1(x4)
0 = 0
average_out(x1, x2, x3) = average_out(x3)
AVERAGE_IN(x1, x2, x3) = AVERAGE_IN(x1, x2)
U2¹(x1, x2, x3, x4) = U2¹(x4)
U1¹(x1, x2, x3, x4) = U1¹(x4)

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:

AVERAGE_IN(X, s(s(s(Y))), s(Z)) → U2¹(X, Y, Z, average_in(s(X), Y, Z))
AVERAGE_IN(X, s(s(s(Y))), s(Z)) → AVERAGE_IN(s(X), Y, Z)
AVERAGE_IN(s(X), Y, Z) → U1¹(X, Y, Z, average_in(X, s(Y), Z))
AVERAGE_IN(s(X), Y, Z) → AVERAGE_IN(X, s(Y), Z)

The TRS R consists of the following rules:

average_in(X, s(s(s(Y))), s(Z)) → U2(X, Y, Z, average_in(s(X), Y, Z))
average_in(s(X), Y, Z) → U1(X, Y, Z, average_in(X, s(Y), Z))
average_in(0, s(s(0)), s(0)) → average_out(0, s(s(0)), s(0))
average_in(0, s(0), 0) → average_out(0, s(0), 0)
average_in(0, 0, 0) → average_out(0, 0, 0)
U1(X, Y, Z, average_out(X, s(Y), Z)) → average_out(s(X), Y, Z)
U2(X, Y, Z, average_out(s(X), Y, Z)) → average_out(X, s(s(s(Y))), s(Z))

The argument filtering Pi contains the following mapping:
average_in(x1, x2, x3) = average_in(x1, x2)
s(x1) = s(x1)
U2(x1, x2, x3, x4) = U2(x4)
U1(x1, x2, x3, x4) = U1(x4)
0 = 0
average_out(x1, x2, x3) = average_out(x3)
AVERAGE_IN(x1, x2, x3) = AVERAGE_IN(x1, x2)
U2¹(x1, x2, x3, x4) = U2¹(x4)
U1¹(x1, x2, x3, x4) = U1¹(x4)

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:

AVERAGE_IN(X, s(s(s(Y))), s(Z)) → AVERAGE_IN(s(X), Y, Z)
AVERAGE_IN(s(X), Y, Z) → AVERAGE_IN(X, s(Y), Z)

The TRS R consists of the following rules:

average_in(X, s(s(s(Y))), s(Z)) → U2(X, Y, Z, average_in(s(X), Y, Z))
average_in(s(X), Y, Z) → U1(X, Y, Z, average_in(X, s(Y), Z))
average_in(0, s(s(0)), s(0)) → average_out(0, s(s(0)), s(0))
average_in(0, s(0), 0) → average_out(0, s(0), 0)
average_in(0, 0, 0) → average_out(0, 0, 0)
U1(X, Y, Z, average_out(X, s(Y), Z)) → average_out(s(X), Y, Z)
U2(X, Y, Z, average_out(s(X), Y, Z)) → average_out(X, s(s(s(Y))), s(Z))

The argument filtering Pi contains the following mapping:
average_in(x1, x2, x3) = average_in(x1, x2)
s(x1) = s(x1)
U2(x1, x2, x3, x4) = U2(x4)
U1(x1, x2, x3, x4) = U1(x4)
0 = 0
average_out(x1, x2, x3) = average_out(x3)
AVERAGE_IN(x1, x2, x3) = AVERAGE_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:

AVERAGE_IN(X, s(s(s(Y))), s(Z)) → AVERAGE_IN(s(X), Y, Z)
AVERAGE_IN(s(X), Y, Z) → AVERAGE_IN(X, s(Y), Z)

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

                      ↳ RuleRemovalProof

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

AVERAGE_IN(s(X), Y) → AVERAGE_IN(X, s(Y))
AVERAGE_IN(X, s(s(s(Y)))) → AVERAGE_IN(s(X), Y)

R is empty.
Q is empty.
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:

AVERAGE_IN(X, s(s(s(Y)))) → AVERAGE_IN(s(X), Y)

Used ordering: POLO with Polynomial interpretation [25]:

POL(AVERAGE_IN(x₁, x₂)) = 2·x₁ + 2·x₂
POL(s(x₁)) = 1 + x₁

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ PiDP

              ↳ UsableRulesProof

                ↳ PiDP

                  ↳ PiDPToQDPProof

                    ↳ QDP

                      ↳ RuleRemovalProof

                        ↳ QDP

                          ↳ QDPSizeChangeProof

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

AVERAGE_IN(s(X), Y) → AVERAGE_IN(X, s(Y))

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:

AVERAGE_IN(s(X), Y) → AVERAGE_IN(X, s(Y))
The graph contains the following edges 1 > 1