Left Termination proof of /tmp/tmpHiuueL/pplus2.pl

Left Termination of the query pattern plus_in_3(g, a, a) w.r.t. the given Prolog program could not be shown:

↳ Prolog

  ↳ PrologToPiTRSProof

  ↳ PrologToPiTRSProof

Clauses:

p(s(0), 0).
p(s(s(X)), s(s(Y))) :- p(s(X), s(Y)).
plus(0, Y, Y).
plus(s(X), Y, s(Z)) :- ','(p(s(X), U), plus(U, Y, Z)).

Queries:

plus(g,a,a).

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

plus_in_gaa(0, Y, Y) → plus_out_gaa(0, Y, Y)
plus_in_gaa(s(X), Y, s(Z)) → U2_gaa(X, Y, Z, p_in_ga(s(X), U))
p_in_ga(s(0), 0) → p_out_ga(s(0), 0)
p_in_ga(s(s(X)), s(s(Y))) → U1_ga(X, Y, p_in_gg(s(X), s(Y)))
p_in_gg(s(0), 0) → p_out_gg(s(0), 0)
p_in_gg(s(s(X)), s(s(Y))) → U1_gg(X, Y, p_in_gg(s(X), s(Y)))
U1_gg(X, Y, p_out_gg(s(X), s(Y))) → p_out_gg(s(s(X)), s(s(Y)))
U1_ga(X, Y, p_out_gg(s(X), s(Y))) → p_out_ga(s(s(X)), s(s(Y)))
U2_gaa(X, Y, Z, p_out_ga(s(X), U)) → U3_gaa(X, Y, Z, plus_in_gaa(U, Y, Z))
U3_gaa(X, Y, Z, plus_out_gaa(U, Y, Z)) → plus_out_gaa(s(X), Y, s(Z))

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

  ↳ PrologToPiTRSProof

Pi-finite rewrite system:
The TRS R consists of the following rules:

plus_in_gaa(0, Y, Y) → plus_out_gaa(0, Y, Y)
plus_in_gaa(s(X), Y, s(Z)) → U2_gaa(X, Y, Z, p_in_ga(s(X), U))
p_in_ga(s(0), 0) → p_out_ga(s(0), 0)
p_in_ga(s(s(X)), s(s(Y))) → U1_ga(X, Y, p_in_gg(s(X), s(Y)))
p_in_gg(s(0), 0) → p_out_gg(s(0), 0)
p_in_gg(s(s(X)), s(s(Y))) → U1_gg(X, Y, p_in_gg(s(X), s(Y)))
U1_gg(X, Y, p_out_gg(s(X), s(Y))) → p_out_gg(s(s(X)), s(s(Y)))
U1_ga(X, Y, p_out_gg(s(X), s(Y))) → p_out_ga(s(s(X)), s(s(Y)))
U2_gaa(X, Y, Z, p_out_ga(s(X), U)) → U3_gaa(X, Y, Z, plus_in_gaa(U, Y, Z))
U3_gaa(X, Y, Z, plus_out_gaa(U, Y, Z)) → plus_out_gaa(s(X), Y, s(Z))

The argument filtering Pi contains the following mapping:
plus_in_gaa(x1, x2, x3) = plus_in_gaa(x1)
0 = 0
plus_out_gaa(x1, x2, x3) = plus_out_gaa
s(x1) = s
U2_gaa(x1, x2, x3, x4) = U2_gaa(x4)
p_in_ga(x1, x2) = p_in_ga(x1)
p_out_ga(x1, x2) = p_out_ga(x2)
U1_ga(x1, x2, x3) = U1_ga(x3)
p_in_gg(x1, x2) = p_in_gg(x1, x2)
p_out_gg(x1, x2) = p_out_gg
U1_gg(x1, x2, x3) = U1_gg(x3)
U3_gaa(x1, x2, x3, x4) = U3_gaa(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:

PLUS_IN_GAA(s(X), Y, s(Z)) → U2_GAA(X, Y, Z, p_in_ga(s(X), U))
PLUS_IN_GAA(s(X), Y, s(Z)) → P_IN_GA(s(X), U)
P_IN_GA(s(s(X)), s(s(Y))) → U1_GA(X, Y, p_in_gg(s(X), s(Y)))
P_IN_GA(s(s(X)), s(s(Y))) → P_IN_GG(s(X), s(Y))
P_IN_GG(s(s(X)), s(s(Y))) → U1_GG(X, Y, p_in_gg(s(X), s(Y)))
P_IN_GG(s(s(X)), s(s(Y))) → P_IN_GG(s(X), s(Y))
U2_GAA(X, Y, Z, p_out_ga(s(X), U)) → U3_GAA(X, Y, Z, plus_in_gaa(U, Y, Z))
U2_GAA(X, Y, Z, p_out_ga(s(X), U)) → PLUS_IN_GAA(U, Y, Z)

The TRS R consists of the following rules:

plus_in_gaa(0, Y, Y) → plus_out_gaa(0, Y, Y)
plus_in_gaa(s(X), Y, s(Z)) → U2_gaa(X, Y, Z, p_in_ga(s(X), U))
p_in_ga(s(0), 0) → p_out_ga(s(0), 0)
p_in_ga(s(s(X)), s(s(Y))) → U1_ga(X, Y, p_in_gg(s(X), s(Y)))
p_in_gg(s(0), 0) → p_out_gg(s(0), 0)
p_in_gg(s(s(X)), s(s(Y))) → U1_gg(X, Y, p_in_gg(s(X), s(Y)))
U1_gg(X, Y, p_out_gg(s(X), s(Y))) → p_out_gg(s(s(X)), s(s(Y)))
U1_ga(X, Y, p_out_gg(s(X), s(Y))) → p_out_ga(s(s(X)), s(s(Y)))
U2_gaa(X, Y, Z, p_out_ga(s(X), U)) → U3_gaa(X, Y, Z, plus_in_gaa(U, Y, Z))
U3_gaa(X, Y, Z, plus_out_gaa(U, Y, Z)) → plus_out_gaa(s(X), Y, s(Z))

The argument filtering Pi contains the following mapping:
plus_in_gaa(x1, x2, x3) = plus_in_gaa(x1)
0 = 0
plus_out_gaa(x1, x2, x3) = plus_out_gaa
s(x1) = s
U2_gaa(x1, x2, x3, x4) = U2_gaa(x4)
p_in_ga(x1, x2) = p_in_ga(x1)
p_out_ga(x1, x2) = p_out_ga(x2)
U1_ga(x1, x2, x3) = U1_ga(x3)
p_in_gg(x1, x2) = p_in_gg(x1, x2)
p_out_gg(x1, x2) = p_out_gg
U1_gg(x1, x2, x3) = U1_gg(x3)
U3_gaa(x1, x2, x3, x4) = U3_gaa(x4)
P_IN_GA(x1, x2) = P_IN_GA(x1)
U1_GG(x1, x2, x3) = U1_GG(x3)
U3_GAA(x1, x2, x3, x4) = U3_GAA(x4)
P_IN_GG(x1, x2) = P_IN_GG(x1, x2)
U1_GA(x1, x2, x3) = U1_GA(x3)
U2_GAA(x1, x2, x3, x4) = U2_GAA(x4)
PLUS_IN_GAA(x1, x2, x3) = PLUS_IN_GAA(x1)

We have to consider all (P,R,Pi)-chains

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

  ↳ PrologToPiTRSProof

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

PLUS_IN_GAA(s(X), Y, s(Z)) → U2_GAA(X, Y, Z, p_in_ga(s(X), U))
PLUS_IN_GAA(s(X), Y, s(Z)) → P_IN_GA(s(X), U)
P_IN_GA(s(s(X)), s(s(Y))) → U1_GA(X, Y, p_in_gg(s(X), s(Y)))
P_IN_GA(s(s(X)), s(s(Y))) → P_IN_GG(s(X), s(Y))
P_IN_GG(s(s(X)), s(s(Y))) → U1_GG(X, Y, p_in_gg(s(X), s(Y)))
P_IN_GG(s(s(X)), s(s(Y))) → P_IN_GG(s(X), s(Y))
U2_GAA(X, Y, Z, p_out_ga(s(X), U)) → U3_GAA(X, Y, Z, plus_in_gaa(U, Y, Z))
U2_GAA(X, Y, Z, p_out_ga(s(X), U)) → PLUS_IN_GAA(U, Y, Z)

The TRS R consists of the following rules:

plus_in_gaa(0, Y, Y) → plus_out_gaa(0, Y, Y)
plus_in_gaa(s(X), Y, s(Z)) → U2_gaa(X, Y, Z, p_in_ga(s(X), U))
p_in_ga(s(0), 0) → p_out_ga(s(0), 0)
p_in_ga(s(s(X)), s(s(Y))) → U1_ga(X, Y, p_in_gg(s(X), s(Y)))
p_in_gg(s(0), 0) → p_out_gg(s(0), 0)
p_in_gg(s(s(X)), s(s(Y))) → U1_gg(X, Y, p_in_gg(s(X), s(Y)))
U1_gg(X, Y, p_out_gg(s(X), s(Y))) → p_out_gg(s(s(X)), s(s(Y)))
U1_ga(X, Y, p_out_gg(s(X), s(Y))) → p_out_ga(s(s(X)), s(s(Y)))
U2_gaa(X, Y, Z, p_out_ga(s(X), U)) → U3_gaa(X, Y, Z, plus_in_gaa(U, Y, Z))
U3_gaa(X, Y, Z, plus_out_gaa(U, Y, Z)) → plus_out_gaa(s(X), Y, s(Z))

The argument filtering Pi contains the following mapping:
plus_in_gaa(x1, x2, x3) = plus_in_gaa(x1)
0 = 0
plus_out_gaa(x1, x2, x3) = plus_out_gaa
s(x1) = s
U2_gaa(x1, x2, x3, x4) = U2_gaa(x4)
p_in_ga(x1, x2) = p_in_ga(x1)
p_out_ga(x1, x2) = p_out_ga(x2)
U1_ga(x1, x2, x3) = U1_ga(x3)
p_in_gg(x1, x2) = p_in_gg(x1, x2)
p_out_gg(x1, x2) = p_out_gg
U1_gg(x1, x2, x3) = U1_gg(x3)
U3_gaa(x1, x2, x3, x4) = U3_gaa(x4)
P_IN_GA(x1, x2) = P_IN_GA(x1)
U1_GG(x1, x2, x3) = U1_GG(x3)
U3_GAA(x1, x2, x3, x4) = U3_GAA(x4)
P_IN_GG(x1, x2) = P_IN_GG(x1, x2)
U1_GA(x1, x2, x3) = U1_GA(x3)
U2_GAA(x1, x2, x3, x4) = U2_GAA(x4)
PLUS_IN_GAA(x1, x2, x3) = PLUS_IN_GAA(x1)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph [30] contains 2 SCCs with 5 less nodes.

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

                ↳ UsableRulesProof

              ↳ PiDP

  ↳ PrologToPiTRSProof

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

P_IN_GG(s(s(X)), s(s(Y))) → P_IN_GG(s(X), s(Y))

The TRS R consists of the following rules:

plus_in_gaa(0, Y, Y) → plus_out_gaa(0, Y, Y)
plus_in_gaa(s(X), Y, s(Z)) → U2_gaa(X, Y, Z, p_in_ga(s(X), U))
p_in_ga(s(0), 0) → p_out_ga(s(0), 0)
p_in_ga(s(s(X)), s(s(Y))) → U1_ga(X, Y, p_in_gg(s(X), s(Y)))
p_in_gg(s(0), 0) → p_out_gg(s(0), 0)
p_in_gg(s(s(X)), s(s(Y))) → U1_gg(X, Y, p_in_gg(s(X), s(Y)))
U1_gg(X, Y, p_out_gg(s(X), s(Y))) → p_out_gg(s(s(X)), s(s(Y)))
U1_ga(X, Y, p_out_gg(s(X), s(Y))) → p_out_ga(s(s(X)), s(s(Y)))
U2_gaa(X, Y, Z, p_out_ga(s(X), U)) → U3_gaa(X, Y, Z, plus_in_gaa(U, Y, Z))
U3_gaa(X, Y, Z, plus_out_gaa(U, Y, Z)) → plus_out_gaa(s(X), Y, s(Z))

The argument filtering Pi contains the following mapping:
plus_in_gaa(x1, x2, x3) = plus_in_gaa(x1)
0 = 0
plus_out_gaa(x1, x2, x3) = plus_out_gaa
s(x1) = s
U2_gaa(x1, x2, x3, x4) = U2_gaa(x4)
p_in_ga(x1, x2) = p_in_ga(x1)
p_out_ga(x1, x2) = p_out_ga(x2)
U1_ga(x1, x2, x3) = U1_ga(x3)
p_in_gg(x1, x2) = p_in_gg(x1, x2)
p_out_gg(x1, x2) = p_out_gg
U1_gg(x1, x2, x3) = U1_gg(x3)
U3_gaa(x1, x2, x3, x4) = U3_gaa(x4)
P_IN_GG(x1, x2) = P_IN_GG(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

            ↳ AND

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

              ↳ PiDP

  ↳ PrologToPiTRSProof

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

P_IN_GG(s(s(X)), s(s(Y))) → P_IN_GG(s(X), s(Y))

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

            ↳ AND

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ NonTerminationProof

              ↳ PiDP

  ↳ PrologToPiTRSProof

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

P_IN_GG(s, s) → P_IN_GG(s, s)

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
We used the non-termination processor [17] to show that the DP problem is infinite.
Found a loop by semiunifying a rule from P directly.

The TRS P consists of the following rules:

P_IN_GG(s, s) → P_IN_GG(s, s)

The TRS R consists of the following rules:none

s = P_IN_GG(s, s) evaluates to t =P_IN_GG(s, s)

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:

Semiunifier: [ ]
Matcher: [ ]

Rewriting sequence

The DP semiunifies directly so there is only one rewrite step from P_IN_GG(s, s) to P_IN_GG(s, s).

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

  ↳ PrologToPiTRSProof

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

U2_GAA(X, Y, Z, p_out_ga(s(X), U)) → PLUS_IN_GAA(U, Y, Z)
PLUS_IN_GAA(s(X), Y, s(Z)) → U2_GAA(X, Y, Z, p_in_ga(s(X), U))

The TRS R consists of the following rules:

plus_in_gaa(0, Y, Y) → plus_out_gaa(0, Y, Y)
plus_in_gaa(s(X), Y, s(Z)) → U2_gaa(X, Y, Z, p_in_ga(s(X), U))
p_in_ga(s(0), 0) → p_out_ga(s(0), 0)
p_in_ga(s(s(X)), s(s(Y))) → U1_ga(X, Y, p_in_gg(s(X), s(Y)))
p_in_gg(s(0), 0) → p_out_gg(s(0), 0)
p_in_gg(s(s(X)), s(s(Y))) → U1_gg(X, Y, p_in_gg(s(X), s(Y)))
U1_gg(X, Y, p_out_gg(s(X), s(Y))) → p_out_gg(s(s(X)), s(s(Y)))
U1_ga(X, Y, p_out_gg(s(X), s(Y))) → p_out_ga(s(s(X)), s(s(Y)))
U2_gaa(X, Y, Z, p_out_ga(s(X), U)) → U3_gaa(X, Y, Z, plus_in_gaa(U, Y, Z))
U3_gaa(X, Y, Z, plus_out_gaa(U, Y, Z)) → plus_out_gaa(s(X), Y, s(Z))

The argument filtering Pi contains the following mapping:
plus_in_gaa(x1, x2, x3) = plus_in_gaa(x1)
0 = 0
plus_out_gaa(x1, x2, x3) = plus_out_gaa
s(x1) = s
U2_gaa(x1, x2, x3, x4) = U2_gaa(x4)
p_in_ga(x1, x2) = p_in_ga(x1)
p_out_ga(x1, x2) = p_out_ga(x2)
U1_ga(x1, x2, x3) = U1_ga(x3)
p_in_gg(x1, x2) = p_in_gg(x1, x2)
p_out_gg(x1, x2) = p_out_gg
U1_gg(x1, x2, x3) = U1_gg(x3)
U3_gaa(x1, x2, x3, x4) = U3_gaa(x4)
U2_GAA(x1, x2, x3, x4) = U2_GAA(x4)
PLUS_IN_GAA(x1, x2, x3) = PLUS_IN_GAA(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

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

  ↳ PrologToPiTRSProof

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

U2_GAA(X, Y, Z, p_out_ga(s(X), U)) → PLUS_IN_GAA(U, Y, Z)
PLUS_IN_GAA(s(X), Y, s(Z)) → U2_GAA(X, Y, Z, p_in_ga(s(X), U))

The TRS R consists of the following rules:

p_in_ga(s(0), 0) → p_out_ga(s(0), 0)
p_in_ga(s(s(X)), s(s(Y))) → U1_ga(X, Y, p_in_gg(s(X), s(Y)))
U1_ga(X, Y, p_out_gg(s(X), s(Y))) → p_out_ga(s(s(X)), s(s(Y)))
p_in_gg(s(s(X)), s(s(Y))) → U1_gg(X, Y, p_in_gg(s(X), s(Y)))
U1_gg(X, Y, p_out_gg(s(X), s(Y))) → p_out_gg(s(s(X)), s(s(Y)))

The argument filtering Pi contains the following mapping:
0 = 0
s(x1) = s
p_in_ga(x1, x2) = p_in_ga(x1)
p_out_ga(x1, x2) = p_out_ga(x2)
U1_ga(x1, x2, x3) = U1_ga(x3)
p_in_gg(x1, x2) = p_in_gg(x1, x2)
p_out_gg(x1, x2) = p_out_gg
U1_gg(x1, x2, x3) = U1_gg(x3)
U2_GAA(x1, x2, x3, x4) = U2_GAA(x4)
PLUS_IN_GAA(x1, x2, x3) = PLUS_IN_GAA(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

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ RuleRemovalProof

  ↳ PrologToPiTRSProof

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

PLUS_IN_GAA(s) → U2_GAA(p_in_ga(s))
U2_GAA(p_out_ga(U)) → PLUS_IN_GAA(U)

The TRS R consists of the following rules:

p_in_ga(s) → p_out_ga(0)
p_in_ga(s) → U1_ga(p_in_gg(s, s))
U1_ga(p_out_gg) → p_out_ga(s)
p_in_gg(s, s) → U1_gg(p_in_gg(s, s))
U1_gg(p_out_gg) → p_out_gg

The set Q consists of the following terms:

p_in_ga(x0)
U1_ga(x0)
p_in_gg(x0, x1)
U1_gg(x0)

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 rules of the TRS R:

U1_ga(p_out_gg) → p_out_ga(s)

Used ordering: POLO with Polynomial interpretation [25]:

POL(0) = 0
POL(PLUS_IN_GAA(x₁)) = 2·x₁
POL(U1_ga(x₁)) = x₁
POL(U1_gg(x₁)) = x₁
POL(U2_GAA(x₁)) = x₁
POL(p_in_ga(x₁)) = x₁
POL(p_in_gg(x₁, x₂)) = x₁ + x₂
POL(p_out_ga(x₁)) = 2·x₁
POL(p_out_gg) = 1
POL(s) = 0

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ RuleRemovalProof

                          ↳ QDP

                            ↳ RuleRemovalProof

  ↳ PrologToPiTRSProof

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

PLUS_IN_GAA(s) → U2_GAA(p_in_ga(s))
U2_GAA(p_out_ga(U)) → PLUS_IN_GAA(U)

The TRS R consists of the following rules:

p_in_ga(s) → p_out_ga(0)
p_in_ga(s) → U1_ga(p_in_gg(s, s))
p_in_gg(s, s) → U1_gg(p_in_gg(s, s))
U1_gg(p_out_gg) → p_out_gg

The set Q consists of the following terms:

p_in_ga(x0)
U1_ga(x0)
p_in_gg(x0, x1)
U1_gg(x0)

U1_gg(p_out_gg) → p_out_gg

Used ordering: POLO with Polynomial interpretation [25]:

POL(0) = 0
POL(PLUS_IN_GAA(x₁)) = 2·x₁
POL(U1_ga(x₁)) = x₁
POL(U1_gg(x₁)) = 2·x₁
POL(U2_GAA(x₁)) = x₁
POL(p_in_ga(x₁)) = x₁
POL(p_in_gg(x₁, x₂)) = x₁ + x₂
POL(p_out_ga(x₁)) = 2·x₁
POL(p_out_gg) = 2
POL(s) = 0

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ RuleRemovalProof

                          ↳ QDP

                            ↳ RuleRemovalProof

                              ↳ QDP

                                ↳ RuleRemovalProof

  ↳ PrologToPiTRSProof

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

PLUS_IN_GAA(s) → U2_GAA(p_in_ga(s))
U2_GAA(p_out_ga(U)) → PLUS_IN_GAA(U)

The TRS R consists of the following rules:

p_in_ga(s) → p_out_ga(0)
p_in_ga(s) → U1_ga(p_in_gg(s, s))
p_in_gg(s, s) → U1_gg(p_in_gg(s, s))

The set Q consists of the following terms:

p_in_ga(x0)
U1_ga(x0)
p_in_gg(x0, x1)
U1_gg(x0)

p_in_ga(s) → U1_ga(p_in_gg(s, s))

Used ordering: POLO with Polynomial interpretation [25]:

POL(0) = 0
POL(PLUS_IN_GAA(x₁)) = 1 + 2·x₁
POL(U1_ga(x₁)) = 2·x₁
POL(U1_gg(x₁)) = x₁
POL(U2_GAA(x₁)) = x₁
POL(p_in_ga(x₁)) = 1 + x₁
POL(p_in_gg(x₁, x₂)) = 2·x₁ + x₂
POL(p_out_ga(x₁)) = 1 + 2·x₁
POL(s) = 0

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ RuleRemovalProof

                          ↳ QDP

                            ↳ RuleRemovalProof

                              ↳ QDP

                                ↳ RuleRemovalProof

                                  ↳ QDP

                                    ↳ UsableRulesProof

  ↳ PrologToPiTRSProof

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

PLUS_IN_GAA(s) → U2_GAA(p_in_ga(s))
U2_GAA(p_out_ga(U)) → PLUS_IN_GAA(U)

The TRS R consists of the following rules:

p_in_ga(s) → p_out_ga(0)
p_in_gg(s, s) → U1_gg(p_in_gg(s, s))

The set Q consists of the following terms:

p_in_ga(x0)
U1_ga(x0)
p_in_gg(x0, x1)
U1_gg(x0)

We have to consider all (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ RuleRemovalProof

                          ↳ QDP

                            ↳ RuleRemovalProof

                              ↳ QDP

                                ↳ RuleRemovalProof

                                  ↳ QDP

                                    ↳ UsableRulesProof

                                      ↳ QDP

                                        ↳ QReductionProof

  ↳ PrologToPiTRSProof

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

PLUS_IN_GAA(s) → U2_GAA(p_in_ga(s))
U2_GAA(p_out_ga(U)) → PLUS_IN_GAA(U)

The TRS R consists of the following rules:

p_in_ga(s) → p_out_ga(0)

The set Q consists of the following terms:

p_in_ga(x0)
U1_ga(x0)
p_in_gg(x0, x1)
U1_gg(x0)

We have to consider all (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

U1_ga(x0)
p_in_gg(x0, x1)
U1_gg(x0)

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ RuleRemovalProof

                          ↳ QDP

                            ↳ RuleRemovalProof

                              ↳ QDP

                                ↳ RuleRemovalProof

                                  ↳ QDP

                                    ↳ UsableRulesProof

                                      ↳ QDP

                                        ↳ QReductionProof

                                          ↳ QDP

                                            ↳ RuleRemovalProof

  ↳ PrologToPiTRSProof

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

PLUS_IN_GAA(s) → U2_GAA(p_in_ga(s))
U2_GAA(p_out_ga(U)) → PLUS_IN_GAA(U)

The TRS R consists of the following rules:

p_in_ga(s) → p_out_ga(0)

The set Q consists of the following terms:

p_in_ga(x0)

PLUS_IN_GAA(s) → U2_GAA(p_in_ga(s))

Used ordering: POLO with Polynomial interpretation [25]:

POL(0) = 1
POL(PLUS_IN_GAA(x₁)) = 2·x₁
POL(U2_GAA(x₁)) = x₁
POL(p_in_ga(x₁)) = x₁
POL(p_out_ga(x₁)) = 2·x₁
POL(s) = 2

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ RuleRemovalProof

                          ↳ QDP

                            ↳ RuleRemovalProof

                              ↳ QDP

                                ↳ RuleRemovalProof

                                  ↳ QDP

                                    ↳ UsableRulesProof

                                      ↳ QDP

                                        ↳ QReductionProof

                                          ↳ QDP

                                            ↳ RuleRemovalProof

                                              ↳ QDP

                                                ↳ DependencyGraphProof

  ↳ PrologToPiTRSProof

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

U2_GAA(p_out_ga(U)) → PLUS_IN_GAA(U)

The TRS R consists of the following rules:

p_in_ga(s) → p_out_ga(0)

The set Q consists of the following terms:

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

plus_in_gaa(0, Y, Y) → plus_out_gaa(0, Y, Y)
plus_in_gaa(s(X), Y, s(Z)) → U2_gaa(X, Y, Z, p_in_ga(s(X), U))
p_in_ga(s(0), 0) → p_out_ga(s(0), 0)
p_in_ga(s(s(X)), s(s(Y))) → U1_ga(X, Y, p_in_gg(s(X), s(Y)))
p_in_gg(s(0), 0) → p_out_gg(s(0), 0)
p_in_gg(s(s(X)), s(s(Y))) → U1_gg(X, Y, p_in_gg(s(X), s(Y)))
U1_gg(X, Y, p_out_gg(s(X), s(Y))) → p_out_gg(s(s(X)), s(s(Y)))
U1_ga(X, Y, p_out_gg(s(X), s(Y))) → p_out_ga(s(s(X)), s(s(Y)))
U2_gaa(X, Y, Z, p_out_ga(s(X), U)) → U3_gaa(X, Y, Z, plus_in_gaa(U, Y, Z))
U3_gaa(X, Y, Z, plus_out_gaa(U, Y, Z)) → plus_out_gaa(s(X), Y, s(Z))

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog

↳ Prolog

  ↳ PrologToPiTRSProof

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

Pi-finite rewrite system:
The TRS R consists of the following rules:

plus_in_gaa(0, Y, Y) → plus_out_gaa(0, Y, Y)
plus_in_gaa(s(X), Y, s(Z)) → U2_gaa(X, Y, Z, p_in_ga(s(X), U))
p_in_ga(s(0), 0) → p_out_ga(s(0), 0)
p_in_ga(s(s(X)), s(s(Y))) → U1_ga(X, Y, p_in_gg(s(X), s(Y)))
p_in_gg(s(0), 0) → p_out_gg(s(0), 0)
p_in_gg(s(s(X)), s(s(Y))) → U1_gg(X, Y, p_in_gg(s(X), s(Y)))
U1_gg(X, Y, p_out_gg(s(X), s(Y))) → p_out_gg(s(s(X)), s(s(Y)))
U1_ga(X, Y, p_out_gg(s(X), s(Y))) → p_out_ga(s(s(X)), s(s(Y)))
U2_gaa(X, Y, Z, p_out_ga(s(X), U)) → U3_gaa(X, Y, Z, plus_in_gaa(U, Y, Z))
U3_gaa(X, Y, Z, plus_out_gaa(U, Y, Z)) → plus_out_gaa(s(X), Y, s(Z))

The argument filtering Pi contains the following mapping:
plus_in_gaa(x1, x2, x3) = plus_in_gaa(x1)
0 = 0
plus_out_gaa(x1, x2, x3) = plus_out_gaa(x1)
s(x1) = s
U2_gaa(x1, x2, x3, x4) = U2_gaa(x4)
p_in_ga(x1, x2) = p_in_ga(x1)
p_out_ga(x1, x2) = p_out_ga(x1, x2)
U1_ga(x1, x2, x3) = U1_ga(x3)
p_in_gg(x1, x2) = p_in_gg(x1, x2)
p_out_gg(x1, x2) = p_out_gg(x1, x2)
U1_gg(x1, x2, x3) = U1_gg(x3)
U3_gaa(x1, x2, x3, x4) = U3_gaa(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:

PLUS_IN_GAA(s(X), Y, s(Z)) → U2_GAA(X, Y, Z, p_in_ga(s(X), U))
PLUS_IN_GAA(s(X), Y, s(Z)) → P_IN_GA(s(X), U)
P_IN_GA(s(s(X)), s(s(Y))) → U1_GA(X, Y, p_in_gg(s(X), s(Y)))
P_IN_GA(s(s(X)), s(s(Y))) → P_IN_GG(s(X), s(Y))
P_IN_GG(s(s(X)), s(s(Y))) → U1_GG(X, Y, p_in_gg(s(X), s(Y)))
P_IN_GG(s(s(X)), s(s(Y))) → P_IN_GG(s(X), s(Y))
U2_GAA(X, Y, Z, p_out_ga(s(X), U)) → U3_GAA(X, Y, Z, plus_in_gaa(U, Y, Z))
U2_GAA(X, Y, Z, p_out_ga(s(X), U)) → PLUS_IN_GAA(U, Y, Z)

The TRS R consists of the following rules:

plus_in_gaa(0, Y, Y) → plus_out_gaa(0, Y, Y)
plus_in_gaa(s(X), Y, s(Z)) → U2_gaa(X, Y, Z, p_in_ga(s(X), U))
p_in_ga(s(0), 0) → p_out_ga(s(0), 0)
p_in_ga(s(s(X)), s(s(Y))) → U1_ga(X, Y, p_in_gg(s(X), s(Y)))
p_in_gg(s(0), 0) → p_out_gg(s(0), 0)
p_in_gg(s(s(X)), s(s(Y))) → U1_gg(X, Y, p_in_gg(s(X), s(Y)))
U1_gg(X, Y, p_out_gg(s(X), s(Y))) → p_out_gg(s(s(X)), s(s(Y)))
U1_ga(X, Y, p_out_gg(s(X), s(Y))) → p_out_ga(s(s(X)), s(s(Y)))
U2_gaa(X, Y, Z, p_out_ga(s(X), U)) → U3_gaa(X, Y, Z, plus_in_gaa(U, Y, Z))
U3_gaa(X, Y, Z, plus_out_gaa(U, Y, Z)) → plus_out_gaa(s(X), Y, s(Z))

The argument filtering Pi contains the following mapping:
plus_in_gaa(x1, x2, x3) = plus_in_gaa(x1)
0 = 0
plus_out_gaa(x1, x2, x3) = plus_out_gaa(x1)
s(x1) = s
U2_gaa(x1, x2, x3, x4) = U2_gaa(x4)
p_in_ga(x1, x2) = p_in_ga(x1)
p_out_ga(x1, x2) = p_out_ga(x1, x2)
U1_ga(x1, x2, x3) = U1_ga(x3)
p_in_gg(x1, x2) = p_in_gg(x1, x2)
p_out_gg(x1, x2) = p_out_gg(x1, x2)
U1_gg(x1, x2, x3) = U1_gg(x3)
U3_gaa(x1, x2, x3, x4) = U3_gaa(x4)
P_IN_GA(x1, x2) = P_IN_GA(x1)
U1_GG(x1, x2, x3) = U1_GG(x3)
U3_GAA(x1, x2, x3, x4) = U3_GAA(x4)
P_IN_GG(x1, x2) = P_IN_GG(x1, x2)
U1_GA(x1, x2, x3) = U1_GA(x3)
U2_GAA(x1, x2, x3, x4) = U2_GAA(x4)
PLUS_IN_GAA(x1, x2, x3) = PLUS_IN_GAA(x1)

We have to consider all (P,R,Pi)-chains

↳ Prolog

  ↳ PrologToPiTRSProof

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

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

PLUS_IN_GAA(s(X), Y, s(Z)) → U2_GAA(X, Y, Z, p_in_ga(s(X), U))
PLUS_IN_GAA(s(X), Y, s(Z)) → P_IN_GA(s(X), U)
P_IN_GA(s(s(X)), s(s(Y))) → U1_GA(X, Y, p_in_gg(s(X), s(Y)))
P_IN_GA(s(s(X)), s(s(Y))) → P_IN_GG(s(X), s(Y))
P_IN_GG(s(s(X)), s(s(Y))) → U1_GG(X, Y, p_in_gg(s(X), s(Y)))
P_IN_GG(s(s(X)), s(s(Y))) → P_IN_GG(s(X), s(Y))
U2_GAA(X, Y, Z, p_out_ga(s(X), U)) → U3_GAA(X, Y, Z, plus_in_gaa(U, Y, Z))
U2_GAA(X, Y, Z, p_out_ga(s(X), U)) → PLUS_IN_GAA(U, Y, Z)

The TRS R consists of the following rules:

plus_in_gaa(0, Y, Y) → plus_out_gaa(0, Y, Y)
plus_in_gaa(s(X), Y, s(Z)) → U2_gaa(X, Y, Z, p_in_ga(s(X), U))
p_in_ga(s(0), 0) → p_out_ga(s(0), 0)
p_in_ga(s(s(X)), s(s(Y))) → U1_ga(X, Y, p_in_gg(s(X), s(Y)))
p_in_gg(s(0), 0) → p_out_gg(s(0), 0)
p_in_gg(s(s(X)), s(s(Y))) → U1_gg(X, Y, p_in_gg(s(X), s(Y)))
U1_gg(X, Y, p_out_gg(s(X), s(Y))) → p_out_gg(s(s(X)), s(s(Y)))
U1_ga(X, Y, p_out_gg(s(X), s(Y))) → p_out_ga(s(s(X)), s(s(Y)))
U2_gaa(X, Y, Z, p_out_ga(s(X), U)) → U3_gaa(X, Y, Z, plus_in_gaa(U, Y, Z))
U3_gaa(X, Y, Z, plus_out_gaa(U, Y, Z)) → plus_out_gaa(s(X), Y, s(Z))

The argument filtering Pi contains the following mapping:
plus_in_gaa(x1, x2, x3) = plus_in_gaa(x1)
0 = 0
plus_out_gaa(x1, x2, x3) = plus_out_gaa(x1)
s(x1) = s
U2_gaa(x1, x2, x3, x4) = U2_gaa(x4)
p_in_ga(x1, x2) = p_in_ga(x1)
p_out_ga(x1, x2) = p_out_ga(x1, x2)
U1_ga(x1, x2, x3) = U1_ga(x3)
p_in_gg(x1, x2) = p_in_gg(x1, x2)
p_out_gg(x1, x2) = p_out_gg(x1, x2)
U1_gg(x1, x2, x3) = U1_gg(x3)
U3_gaa(x1, x2, x3, x4) = U3_gaa(x4)
P_IN_GA(x1, x2) = P_IN_GA(x1)
U1_GG(x1, x2, x3) = U1_GG(x3)
U3_GAA(x1, x2, x3, x4) = U3_GAA(x4)
P_IN_GG(x1, x2) = P_IN_GG(x1, x2)
U1_GA(x1, x2, x3) = U1_GA(x3)
U2_GAA(x1, x2, x3, x4) = U2_GAA(x4)
PLUS_IN_GAA(x1, x2, x3) = PLUS_IN_GAA(x1)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph [30] contains 2 SCCs with 5 less nodes.

↳ Prolog

  ↳ PrologToPiTRSProof

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

                ↳ UsableRulesProof

              ↳ PiDP

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

P_IN_GG(s(s(X)), s(s(Y))) → P_IN_GG(s(X), s(Y))

The TRS R consists of the following rules:

plus_in_gaa(0, Y, Y) → plus_out_gaa(0, Y, Y)
plus_in_gaa(s(X), Y, s(Z)) → U2_gaa(X, Y, Z, p_in_ga(s(X), U))
p_in_ga(s(0), 0) → p_out_ga(s(0), 0)
p_in_ga(s(s(X)), s(s(Y))) → U1_ga(X, Y, p_in_gg(s(X), s(Y)))
p_in_gg(s(0), 0) → p_out_gg(s(0), 0)
p_in_gg(s(s(X)), s(s(Y))) → U1_gg(X, Y, p_in_gg(s(X), s(Y)))
U1_gg(X, Y, p_out_gg(s(X), s(Y))) → p_out_gg(s(s(X)), s(s(Y)))
U1_ga(X, Y, p_out_gg(s(X), s(Y))) → p_out_ga(s(s(X)), s(s(Y)))
U2_gaa(X, Y, Z, p_out_ga(s(X), U)) → U3_gaa(X, Y, Z, plus_in_gaa(U, Y, Z))
U3_gaa(X, Y, Z, plus_out_gaa(U, Y, Z)) → plus_out_gaa(s(X), Y, s(Z))

The argument filtering Pi contains the following mapping:
plus_in_gaa(x1, x2, x3) = plus_in_gaa(x1)
0 = 0
plus_out_gaa(x1, x2, x3) = plus_out_gaa(x1)
s(x1) = s
U2_gaa(x1, x2, x3, x4) = U2_gaa(x4)
p_in_ga(x1, x2) = p_in_ga(x1)
p_out_ga(x1, x2) = p_out_ga(x1, x2)
U1_ga(x1, x2, x3) = U1_ga(x3)
p_in_gg(x1, x2) = p_in_gg(x1, x2)
p_out_gg(x1, x2) = p_out_gg(x1, x2)
U1_gg(x1, x2, x3) = U1_gg(x3)
U3_gaa(x1, x2, x3, x4) = U3_gaa(x4)
P_IN_GG(x1, x2) = P_IN_GG(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

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

              ↳ PiDP

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

P_IN_GG(s(s(X)), s(s(Y))) → P_IN_GG(s(X), s(Y))

↳ Prolog

  ↳ PrologToPiTRSProof

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ NonTerminationProof

              ↳ PiDP

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

P_IN_GG(s, s) → P_IN_GG(s, s)

The TRS R consists of the following rules:none

s = P_IN_GG(s, s) evaluates to t =P_IN_GG(s, s)

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:

Semiunifier: [ ]
Matcher: [ ]

Rewriting sequence

The DP semiunifies directly so there is only one rewrite step from P_IN_GG(s, s) to P_IN_GG(s, s).

↳ Prolog

  ↳ PrologToPiTRSProof

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

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

U2_GAA(X, Y, Z, p_out_ga(s(X), U)) → PLUS_IN_GAA(U, Y, Z)
PLUS_IN_GAA(s(X), Y, s(Z)) → U2_GAA(X, Y, Z, p_in_ga(s(X), U))

The TRS R consists of the following rules:

plus_in_gaa(0, Y, Y) → plus_out_gaa(0, Y, Y)
plus_in_gaa(s(X), Y, s(Z)) → U2_gaa(X, Y, Z, p_in_ga(s(X), U))
p_in_ga(s(0), 0) → p_out_ga(s(0), 0)
p_in_ga(s(s(X)), s(s(Y))) → U1_ga(X, Y, p_in_gg(s(X), s(Y)))
p_in_gg(s(0), 0) → p_out_gg(s(0), 0)
p_in_gg(s(s(X)), s(s(Y))) → U1_gg(X, Y, p_in_gg(s(X), s(Y)))
U1_gg(X, Y, p_out_gg(s(X), s(Y))) → p_out_gg(s(s(X)), s(s(Y)))
U1_ga(X, Y, p_out_gg(s(X), s(Y))) → p_out_ga(s(s(X)), s(s(Y)))
U2_gaa(X, Y, Z, p_out_ga(s(X), U)) → U3_gaa(X, Y, Z, plus_in_gaa(U, Y, Z))
U3_gaa(X, Y, Z, plus_out_gaa(U, Y, Z)) → plus_out_gaa(s(X), Y, s(Z))

The argument filtering Pi contains the following mapping:
plus_in_gaa(x1, x2, x3) = plus_in_gaa(x1)
0 = 0
plus_out_gaa(x1, x2, x3) = plus_out_gaa(x1)
s(x1) = s
U2_gaa(x1, x2, x3, x4) = U2_gaa(x4)
p_in_ga(x1, x2) = p_in_ga(x1)
p_out_ga(x1, x2) = p_out_ga(x1, x2)
U1_ga(x1, x2, x3) = U1_ga(x3)
p_in_gg(x1, x2) = p_in_gg(x1, x2)
p_out_gg(x1, x2) = p_out_gg(x1, x2)
U1_gg(x1, x2, x3) = U1_gg(x3)
U3_gaa(x1, x2, x3, x4) = U3_gaa(x4)
U2_GAA(x1, x2, x3, x4) = U2_GAA(x4)
PLUS_IN_GAA(x1, x2, x3) = PLUS_IN_GAA(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

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

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

U2_GAA(X, Y, Z, p_out_ga(s(X), U)) → PLUS_IN_GAA(U, Y, Z)
PLUS_IN_GAA(s(X), Y, s(Z)) → U2_GAA(X, Y, Z, p_in_ga(s(X), U))

The TRS R consists of the following rules:

p_in_ga(s(0), 0) → p_out_ga(s(0), 0)
p_in_ga(s(s(X)), s(s(Y))) → U1_ga(X, Y, p_in_gg(s(X), s(Y)))
U1_ga(X, Y, p_out_gg(s(X), s(Y))) → p_out_ga(s(s(X)), s(s(Y)))
p_in_gg(s(s(X)), s(s(Y))) → U1_gg(X, Y, p_in_gg(s(X), s(Y)))
U1_gg(X, Y, p_out_gg(s(X), s(Y))) → p_out_gg(s(s(X)), s(s(Y)))

The argument filtering Pi contains the following mapping:
0 = 0
s(x1) = s
p_in_ga(x1, x2) = p_in_ga(x1)
p_out_ga(x1, x2) = p_out_ga(x1, x2)
U1_ga(x1, x2, x3) = U1_ga(x3)
p_in_gg(x1, x2) = p_in_gg(x1, x2)
p_out_gg(x1, x2) = p_out_gg(x1, x2)
U1_gg(x1, x2, x3) = U1_gg(x3)
U2_GAA(x1, x2, x3, x4) = U2_GAA(x4)
PLUS_IN_GAA(x1, x2, x3) = PLUS_IN_GAA(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

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ RuleRemovalProof

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

U2_GAA(p_out_ga(s, U)) → PLUS_IN_GAA(U)
PLUS_IN_GAA(s) → U2_GAA(p_in_ga(s))

The TRS R consists of the following rules:

p_in_ga(s) → p_out_ga(s, 0)
p_in_ga(s) → U1_ga(p_in_gg(s, s))
U1_ga(p_out_gg(s, s)) → p_out_ga(s, s)
p_in_gg(s, s) → U1_gg(p_in_gg(s, s))
U1_gg(p_out_gg(s, s)) → p_out_gg(s, s)

The set Q consists of the following terms:

p_in_ga(x0)
U1_ga(x0)
p_in_gg(x0, x1)
U1_gg(x0)

p_in_ga(s) → U1_ga(p_in_gg(s, s))
U1_gg(p_out_gg(s, s)) → p_out_gg(s, s)

Used ordering: POLO with Polynomial interpretation [25]:

POL(0) = 0
POL(PLUS_IN_GAA(x₁)) = 2 + 2·x₁
POL(U1_ga(x₁)) = 2·x₁
POL(U1_gg(x₁)) = 2·x₁
POL(U2_GAA(x₁)) = x₁
POL(p_in_ga(x₁)) = 2 + x₁
POL(p_in_gg(x₁, x₂)) = 2·x₁ + 2·x₂
POL(p_out_ga(x₁, x₂)) = 2 + 2·x₁ + 2·x₂
POL(p_out_gg(x₁, x₂)) = 1 + 2·x₁ + 2·x₂
POL(s) = 0

↳ Prolog

  ↳ PrologToPiTRSProof

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ RuleRemovalProof

                          ↳ QDP

                            ↳ UsableRulesProof

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

U2_GAA(p_out_ga(s, U)) → PLUS_IN_GAA(U)
PLUS_IN_GAA(s) → U2_GAA(p_in_ga(s))

The TRS R consists of the following rules:

p_in_ga(s) → p_out_ga(s, 0)
U1_ga(p_out_gg(s, s)) → p_out_ga(s, s)
p_in_gg(s, s) → U1_gg(p_in_gg(s, s))

The set Q consists of the following terms:

p_in_ga(x0)
U1_ga(x0)
p_in_gg(x0, x1)
U1_gg(x0)

↳ Prolog

  ↳ PrologToPiTRSProof

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ RuleRemovalProof

                          ↳ QDP

                            ↳ UsableRulesProof

                              ↳ QDP

                                ↳ QReductionProof

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

U2_GAA(p_out_ga(s, U)) → PLUS_IN_GAA(U)
PLUS_IN_GAA(s) → U2_GAA(p_in_ga(s))

The TRS R consists of the following rules:

p_in_ga(s) → p_out_ga(s, 0)

The set Q consists of the following terms:

p_in_ga(x0)
U1_ga(x0)
p_in_gg(x0, x1)
U1_gg(x0)

We have to consider all (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

U1_ga(x0)
p_in_gg(x0, x1)
U1_gg(x0)

↳ Prolog

  ↳ PrologToPiTRSProof

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ RuleRemovalProof

                          ↳ QDP

                            ↳ UsableRulesProof

                              ↳ QDP

                                ↳ QReductionProof

                                  ↳ QDP

                                    ↳ RuleRemovalProof

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

U2_GAA(p_out_ga(s, U)) → PLUS_IN_GAA(U)
PLUS_IN_GAA(s) → U2_GAA(p_in_ga(s))

The TRS R consists of the following rules:

p_in_ga(s) → p_out_ga(s, 0)

The set Q consists of the following terms:

p_in_ga(x0)

U2_GAA(p_out_ga(s, U)) → PLUS_IN_GAA(U)
PLUS_IN_GAA(s) → U2_GAA(p_in_ga(s))

Used ordering: POLO with Polynomial interpretation [25]:

POL(0) = 0
POL(PLUS_IN_GAA(x₁)) = 2 + 2·x₁
POL(U2_GAA(x₁)) = x₁
POL(p_in_ga(x₁)) = 1 + 2·x₁
POL(p_out_ga(x₁, x₂)) = 1 + 2·x₁ + 2·x₂
POL(s) = 1

↳ Prolog

  ↳ PrologToPiTRSProof

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ RuleRemovalProof

                          ↳ QDP

                            ↳ UsableRulesProof

                              ↳ QDP

                                ↳ QReductionProof

                                  ↳ QDP

                                    ↳ RuleRemovalProof

                                      ↳ QDP

                                        ↳ PisEmptyProof

Q DP problem:
P is empty.
The TRS R consists of the following rules:

p_in_ga(s) → p_out_ga(s, 0)

The set Q consists of the following terms:

p_in_ga(x0)

We have to consider all (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.