Left Termination proof of /tmp/tmpG-mxbk/avg-bfb.pl

Left Termination of the query pattern avg_in_3(g, a, g) w.r.t. the given Prolog program could successfully be proven:

↳ Prolog

  ↳ PrologToPiTRSProof

Clauses:

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

Queries:

avg(g,a,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:

avg_in(0, s(s(0)), s(0)) → avg_out(0, s(s(0)), s(0))
avg_in(0, s(0), 0) → avg_out(0, s(0), 0)
avg_in(0, 0, 0) → avg_out(0, 0, 0)
avg_in(X, s(s(s(Y))), s(Z)) → U2(X, Y, Z, avg_in(s(X), Y, Z))
avg_in(s(X), Y, Z) → U1(X, Y, Z, avg_in(X, s(Y), Z))
U1(X, Y, Z, avg_out(X, s(Y), Z)) → avg_out(s(X), Y, Z)
U2(X, Y, Z, avg_out(s(X), Y, Z)) → avg_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:

avg_in(0, s(s(0)), s(0)) → avg_out(0, s(s(0)), s(0))
avg_in(0, s(0), 0) → avg_out(0, s(0), 0)
avg_in(0, 0, 0) → avg_out(0, 0, 0)
avg_in(X, s(s(s(Y))), s(Z)) → U2(X, Y, Z, avg_in(s(X), Y, Z))
avg_in(s(X), Y, Z) → U1(X, Y, Z, avg_in(X, s(Y), Z))
U1(X, Y, Z, avg_out(X, s(Y), Z)) → avg_out(s(X), Y, Z)
U2(X, Y, Z, avg_out(s(X), Y, Z)) → avg_out(X, s(s(s(Y))), s(Z))

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

AVG_IN(X, s(s(s(Y))), s(Z)) → U2¹(X, Y, Z, avg_in(s(X), Y, Z))
AVG_IN(X, s(s(s(Y))), s(Z)) → AVG_IN(s(X), Y, Z)
AVG_IN(s(X), Y, Z) → U1¹(X, Y, Z, avg_in(X, s(Y), Z))
AVG_IN(s(X), Y, Z) → AVG_IN(X, s(Y), Z)

The TRS R consists of the following rules:

avg_in(0, s(s(0)), s(0)) → avg_out(0, s(s(0)), s(0))
avg_in(0, s(0), 0) → avg_out(0, s(0), 0)
avg_in(0, 0, 0) → avg_out(0, 0, 0)
avg_in(X, s(s(s(Y))), s(Z)) → U2(X, Y, Z, avg_in(s(X), Y, Z))
avg_in(s(X), Y, Z) → U1(X, Y, Z, avg_in(X, s(Y), Z))
U1(X, Y, Z, avg_out(X, s(Y), Z)) → avg_out(s(X), Y, Z)
U2(X, Y, Z, avg_out(s(X), Y, Z)) → avg_out(X, s(s(s(Y))), s(Z))

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

AVG_IN(X, s(s(s(Y))), s(Z)) → U2¹(X, Y, Z, avg_in(s(X), Y, Z))
AVG_IN(X, s(s(s(Y))), s(Z)) → AVG_IN(s(X), Y, Z)
AVG_IN(s(X), Y, Z) → U1¹(X, Y, Z, avg_in(X, s(Y), Z))
AVG_IN(s(X), Y, Z) → AVG_IN(X, s(Y), Z)

The TRS R consists of the following rules:

avg_in(0, s(s(0)), s(0)) → avg_out(0, s(s(0)), s(0))
avg_in(0, s(0), 0) → avg_out(0, s(0), 0)
avg_in(0, 0, 0) → avg_out(0, 0, 0)
avg_in(X, s(s(s(Y))), s(Z)) → U2(X, Y, Z, avg_in(s(X), Y, Z))
avg_in(s(X), Y, Z) → U1(X, Y, Z, avg_in(X, s(Y), Z))
U1(X, Y, Z, avg_out(X, s(Y), Z)) → avg_out(s(X), Y, Z)
U2(X, Y, Z, avg_out(s(X), Y, Z)) → avg_out(X, s(s(s(Y))), s(Z))

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

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

The TRS R consists of the following rules:

avg_in(0, s(s(0)), s(0)) → avg_out(0, s(s(0)), s(0))
avg_in(0, s(0), 0) → avg_out(0, s(0), 0)
avg_in(0, 0, 0) → avg_out(0, 0, 0)
avg_in(X, s(s(s(Y))), s(Z)) → U2(X, Y, Z, avg_in(s(X), Y, Z))
avg_in(s(X), Y, Z) → U1(X, Y, Z, avg_in(X, s(Y), Z))
U1(X, Y, Z, avg_out(X, s(Y), Z)) → avg_out(s(X), Y, Z)
U2(X, Y, Z, avg_out(s(X), Y, Z)) → avg_out(X, s(s(s(Y))), s(Z))

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

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

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

AVG_IN(s(X), Z) → AVG_IN(X, Z)
AVG_IN(X, s(Z)) → AVG_IN(s(X), Z)

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:

AVG_IN(s(X), Z) → AVG_IN(X, Z)
The graph contains the following edges 1 > 1, 2 >= 2
AVG_IN(X, s(Z)) → AVG_IN(s(X), Z)
The graph contains the following edges 2 > 2