Left Termination proof of /tmp/tmpC4KNtC/avg.pl

Left Termination of the query pattern avg(b,b,f) w.r.t. the given Prolog program could successfully be proven:

↳ PROLOG

  ↳ PrologToPiTRSProof

avg₃(s₁(X), Y, Z) :- avg₃(X, s₁(Y), Z).
avg₃(X, s₁³(Y), s₁(Z)) :- avg₃(s₁(X), Y, Z).
avg₃(0₀, 0₀, 0₀).
avg₃(0₀, s₁(0₀), 0₀).
avg₃(0₀, s₁²(0₀), s₁(0₀)).

With regard to the inferred argument filtering the predicates were used in the following modes:
avg₃: (b,b,f)
Transforming PROLOG into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

avg_3_in_gga₃(s_1₁(X), Y, Z) -> if_avg_3_in_1_gga₄(X, Y, Z, avg_3_in_gga₃(X, s_1₁(Y), Z))
avg_3_in_gga₃(X, s_1₁(s_1₁(s_1₁(Y))), s_1₁(Z)) -> if_avg_3_in_2_gga₄(X, Y, Z, avg_3_in_gga₃(s_1₁(X), Y, Z))
avg_3_in_gga₃(0_0, 0_0, 0_0) -> avg_3_out_gga₃(0_0, 0_0, 0_0)
avg_3_in_gga₃(0_0, s_1₁(0_0), 0_0) -> avg_3_out_gga₃(0_0, s_1₁(0_0), 0_0)
avg_3_in_gga₃(0_0, s_1₁(s_1₁(0_0)), s_1₁(0_0)) -> avg_3_out_gga₃(0_0, s_1₁(s_1₁(0_0)), s_1₁(0_0))
if_avg_3_in_2_gga₄(X, Y, Z, avg_3_out_gga₃(s_1₁(X), Y, Z)) -> avg_3_out_gga₃(X, s_1₁(s_1₁(s_1₁(Y))), s_1₁(Z))
if_avg_3_in_1_gga₄(X, Y, Z, avg_3_out_gga₃(X, s_1₁(Y), Z)) -> avg_3_out_gga₃(s_1₁(X), Y, 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_3_in_gga₃(s_1₁(X), Y, Z) -> if_avg_3_in_1_gga₄(X, Y, Z, avg_3_in_gga₃(X, s_1₁(Y), Z))
avg_3_in_gga₃(X, s_1₁(s_1₁(s_1₁(Y))), s_1₁(Z)) -> if_avg_3_in_2_gga₄(X, Y, Z, avg_3_in_gga₃(s_1₁(X), Y, Z))
avg_3_in_gga₃(0_0, 0_0, 0_0) -> avg_3_out_gga₃(0_0, 0_0, 0_0)
avg_3_in_gga₃(0_0, s_1₁(0_0), 0_0) -> avg_3_out_gga₃(0_0, s_1₁(0_0), 0_0)
avg_3_in_gga₃(0_0, s_1₁(s_1₁(0_0)), s_1₁(0_0)) -> avg_3_out_gga₃(0_0, s_1₁(s_1₁(0_0)), s_1₁(0_0))
if_avg_3_in_2_gga₄(X, Y, Z, avg_3_out_gga₃(s_1₁(X), Y, Z)) -> avg_3_out_gga₃(X, s_1₁(s_1₁(s_1₁(Y))), s_1₁(Z))
if_avg_3_in_1_gga₄(X, Y, Z, avg_3_out_gga₃(X, s_1₁(Y), Z)) -> avg_3_out_gga₃(s_1₁(X), Y, Z)

AVG_3_IN_GGA₃(s_1₁(X), Y, Z) -> IF_AVG_3_IN_1_GGA₄(X, Y, Z, avg_3_in_gga₃(X, s_1₁(Y), Z))
AVG_3_IN_GGA₃(s_1₁(X), Y, Z) -> AVG_3_IN_GGA₃(X, s_1₁(Y), Z)
AVG_3_IN_GGA₃(X, s_1₁(s_1₁(s_1₁(Y))), s_1₁(Z)) -> IF_AVG_3_IN_2_GGA₄(X, Y, Z, avg_3_in_gga₃(s_1₁(X), Y, Z))
AVG_3_IN_GGA₃(X, s_1₁(s_1₁(s_1₁(Y))), s_1₁(Z)) -> AVG_3_IN_GGA₃(s_1₁(X), Y, Z)

The TRS R consists of the following rules:

avg_3_in_gga₃(s_1₁(X), Y, Z) -> if_avg_3_in_1_gga₄(X, Y, Z, avg_3_in_gga₃(X, s_1₁(Y), Z))
avg_3_in_gga₃(X, s_1₁(s_1₁(s_1₁(Y))), s_1₁(Z)) -> if_avg_3_in_2_gga₄(X, Y, Z, avg_3_in_gga₃(s_1₁(X), Y, Z))
avg_3_in_gga₃(0_0, 0_0, 0_0) -> avg_3_out_gga₃(0_0, 0_0, 0_0)
avg_3_in_gga₃(0_0, s_1₁(0_0), 0_0) -> avg_3_out_gga₃(0_0, s_1₁(0_0), 0_0)
avg_3_in_gga₃(0_0, s_1₁(s_1₁(0_0)), s_1₁(0_0)) -> avg_3_out_gga₃(0_0, s_1₁(s_1₁(0_0)), s_1₁(0_0))
if_avg_3_in_2_gga₄(X, Y, Z, avg_3_out_gga₃(s_1₁(X), Y, Z)) -> avg_3_out_gga₃(X, s_1₁(s_1₁(s_1₁(Y))), s_1₁(Z))
if_avg_3_in_1_gga₄(X, Y, Z, avg_3_out_gga₃(X, s_1₁(Y), Z)) -> avg_3_out_gga₃(s_1₁(X), Y, Z)

The argument filtering Pi contains the following mapping:
avg_3_in_gga₃(x1, x2, x3) = avg_3_in_gga₂(x1, x2)
s_1₁(x1) = s_1₁(x1)
0_0 = 0_0
if_avg_3_in_1_gga₄(x1, x2, x3, x4) = if_avg_3_in_1_gga₁(x4)
if_avg_3_in_2_gga₄(x1, x2, x3, x4) = if_avg_3_in_2_gga₁(x4)
avg_3_out_gga₃(x1, x2, x3) = avg_3_out_gga₁(x3)
AVG_3_IN_GGA₃(x1, x2, x3) = AVG_3_IN_GGA₂(x1, x2)
IF_AVG_3_IN_1_GGA₄(x1, x2, x3, x4) = IF_AVG_3_IN_1_GGA₁(x4)
IF_AVG_3_IN_2_GGA₄(x1, x2, x3, x4) = IF_AVG_3_IN_2_GGA₁(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_3_IN_GGA₃(s_1₁(X), Y, Z) -> IF_AVG_3_IN_1_GGA₄(X, Y, Z, avg_3_in_gga₃(X, s_1₁(Y), Z))
AVG_3_IN_GGA₃(s_1₁(X), Y, Z) -> AVG_3_IN_GGA₃(X, s_1₁(Y), Z)
AVG_3_IN_GGA₃(X, s_1₁(s_1₁(s_1₁(Y))), s_1₁(Z)) -> IF_AVG_3_IN_2_GGA₄(X, Y, Z, avg_3_in_gga₃(s_1₁(X), Y, Z))
AVG_3_IN_GGA₃(X, s_1₁(s_1₁(s_1₁(Y))), s_1₁(Z)) -> AVG_3_IN_GGA₃(s_1₁(X), Y, Z)

The TRS R consists of the following rules:

avg_3_in_gga₃(s_1₁(X), Y, Z) -> if_avg_3_in_1_gga₄(X, Y, Z, avg_3_in_gga₃(X, s_1₁(Y), Z))
avg_3_in_gga₃(X, s_1₁(s_1₁(s_1₁(Y))), s_1₁(Z)) -> if_avg_3_in_2_gga₄(X, Y, Z, avg_3_in_gga₃(s_1₁(X), Y, Z))
avg_3_in_gga₃(0_0, 0_0, 0_0) -> avg_3_out_gga₃(0_0, 0_0, 0_0)
avg_3_in_gga₃(0_0, s_1₁(0_0), 0_0) -> avg_3_out_gga₃(0_0, s_1₁(0_0), 0_0)
avg_3_in_gga₃(0_0, s_1₁(s_1₁(0_0)), s_1₁(0_0)) -> avg_3_out_gga₃(0_0, s_1₁(s_1₁(0_0)), s_1₁(0_0))
if_avg_3_in_2_gga₄(X, Y, Z, avg_3_out_gga₃(s_1₁(X), Y, Z)) -> avg_3_out_gga₃(X, s_1₁(s_1₁(s_1₁(Y))), s_1₁(Z))
if_avg_3_in_1_gga₄(X, Y, Z, avg_3_out_gga₃(X, s_1₁(Y), Z)) -> avg_3_out_gga₃(s_1₁(X), Y, Z)

↳ PROLOG

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ PiDP

              ↳ UsableRulesProof

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

AVG_3_IN_GGA₃(X, s_1₁(s_1₁(s_1₁(Y))), s_1₁(Z)) -> AVG_3_IN_GGA₃(s_1₁(X), Y, Z)
AVG_3_IN_GGA₃(s_1₁(X), Y, Z) -> AVG_3_IN_GGA₃(X, s_1₁(Y), Z)

The TRS R consists of the following rules:

avg_3_in_gga₃(s_1₁(X), Y, Z) -> if_avg_3_in_1_gga₄(X, Y, Z, avg_3_in_gga₃(X, s_1₁(Y), Z))
avg_3_in_gga₃(X, s_1₁(s_1₁(s_1₁(Y))), s_1₁(Z)) -> if_avg_3_in_2_gga₄(X, Y, Z, avg_3_in_gga₃(s_1₁(X), Y, Z))
avg_3_in_gga₃(0_0, 0_0, 0_0) -> avg_3_out_gga₃(0_0, 0_0, 0_0)
avg_3_in_gga₃(0_0, s_1₁(0_0), 0_0) -> avg_3_out_gga₃(0_0, s_1₁(0_0), 0_0)
avg_3_in_gga₃(0_0, s_1₁(s_1₁(0_0)), s_1₁(0_0)) -> avg_3_out_gga₃(0_0, s_1₁(s_1₁(0_0)), s_1₁(0_0))
if_avg_3_in_2_gga₄(X, Y, Z, avg_3_out_gga₃(s_1₁(X), Y, Z)) -> avg_3_out_gga₃(X, s_1₁(s_1₁(s_1₁(Y))), s_1₁(Z))
if_avg_3_in_1_gga₄(X, Y, Z, avg_3_out_gga₃(X, s_1₁(Y), Z)) -> avg_3_out_gga₃(s_1₁(X), Y, Z)

The argument filtering Pi contains the following mapping:
avg_3_in_gga₃(x1, x2, x3) = avg_3_in_gga₂(x1, x2)
s_1₁(x1) = s_1₁(x1)
0_0 = 0_0
if_avg_3_in_1_gga₄(x1, x2, x3, x4) = if_avg_3_in_1_gga₁(x4)
if_avg_3_in_2_gga₄(x1, x2, x3, x4) = if_avg_3_in_2_gga₁(x4)
avg_3_out_gga₃(x1, x2, x3) = avg_3_out_gga₁(x3)
AVG_3_IN_GGA₃(x1, x2, x3) = AVG_3_IN_GGA₂(x1, x2)

We have to consider all (P,R,Pi)-chains
For (infinitary) constructor rewriting 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_3_IN_GGA₃(X, s_1₁(s_1₁(s_1₁(Y))), s_1₁(Z)) -> AVG_3_IN_GGA₃(s_1₁(X), Y, Z)
AVG_3_IN_GGA₃(s_1₁(X), Y, Z) -> AVG_3_IN_GGA₃(X, s_1₁(Y), Z)

R is empty.
The argument filtering Pi contains the following mapping:
s_1₁(x1) = s_1₁(x1)
AVG_3_IN_GGA₃(x1, x2, x3) = AVG_3_IN_GGA₂(x1, x2)

We have to consider all (P,R,Pi)-chains
Transforming (infinitary) constructor rewriting Pi-DP problem into ordinary QDP problem 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:

AVG_3_IN_GGA₂(X, s_1₁(s_1₁(s_1₁(Y)))) -> AVG_3_IN_GGA₂(s_1₁(X), Y)
AVG_3_IN_GGA₂(s_1₁(X), Y) -> AVG_3_IN_GGA₂(X, s_1₁(Y))

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {AVG_3_IN_GGA₂}.
By using a polynomial ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

AVG_3_IN_GGA₂(X, s_1₁(s_1₁(s_1₁(Y)))) -> AVG_3_IN_GGA₂(s_1₁(X), Y)

Used ordering: POLO with Polynomial interpretation:

POL(AVG_3_IN_GGA₂(x₁, x₂)) = 1 + x₁ + x₂
POL(s_1₁(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:

AVG_3_IN_GGA₂(s_1₁(X), Y) -> AVG_3_IN_GGA₂(X, s_1₁(Y))

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {AVG_3_IN_GGA₂}.
By using the subterm criterion together with the size-change analysis 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_3_IN_GGA₂(s_1₁(X), Y) -> AVG_3_IN_GGA₂(X, s_1₁(Y))
The graph contains the following edges 1 > 1