Left Termination proof of /tmp/tmpR6hdaz/pplus2.pl

Left Termination of the query pattern plus_in_3(g, a, a) w.r.t. the given Prolog program could successfully be proven:

↳ Prolog

  ↳ 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)
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_ga(s(X), s(Y)))
U1_ga(X, Y, p_out_ga(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

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_ga(s(X), s(Y)))
U1_ga(X, Y, p_out_ga(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(x1)
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)
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_ga(s(X), s(Y)))
P_IN_GA(s(s(X)), s(s(Y))) → P_IN_GA(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_ga(s(X), s(Y)))
U1_ga(X, Y, p_out_ga(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(x1)
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)
U3_gaa(x1, x2, x3, x4) = U3_gaa(x4)
P_IN_GA(x1, x2) = P_IN_GA(x1)
U3_GAA(x1, x2, x3, x4) = U3_GAA(x4)
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

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_ga(s(X), s(Y)))
P_IN_GA(s(s(X)), s(s(Y))) → P_IN_GA(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_ga(s(X), s(Y)))
U1_ga(X, Y, p_out_ga(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(x1)
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)
U3_gaa(x1, x2, x3, x4) = U3_gaa(x4)
P_IN_GA(x1, x2) = P_IN_GA(x1)
U3_GAA(x1, x2, x3, x4) = U3_GAA(x4)
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 3 less nodes.

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

                ↳ UsableRulesProof

              ↳ PiDP

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

P_IN_GA(s(s(X)), s(s(Y))) → P_IN_GA(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_ga(s(X), s(Y)))
U1_ga(X, Y, p_out_ga(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(x1)
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)
U3_gaa(x1, x2, x3, x4) = U3_gaa(x4)
P_IN_GA(x1, x2) = P_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:

P_IN_GA(s(s(X)), s(s(Y))) → P_IN_GA(s(X), s(Y))

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

P_IN_GA(s(s(X))) → P_IN_GA(s(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:

P_IN_GA(s(s(X))) → P_IN_GA(s(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:

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_ga(s(X), s(Y)))
U1_ga(X, Y, p_out_ga(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(x1)
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)
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

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_ga(s(X), s(Y)))
U1_ga(X, Y, p_out_ga(s(X), s(Y))) → p_out_ga(s(s(X)), s(s(Y)))

The argument filtering Pi contains the following mapping:
0 = 0
s(x1) = s(x1)
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)
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

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

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

The TRS R consists of the following rules:

p_in_ga(s(0)) → p_out_ga(0)
p_in_ga(s(s(X))) → U1_ga(p_in_ga(s(X)))
U1_ga(p_out_ga(s(Y))) → p_out_ga(s(s(Y)))

The set Q consists of the following terms:

p_in_ga(x0)
U1_ga(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:

p_in_ga(s(0)) → p_out_ga(0)

Used ordering: POLO with Polynomial interpretation [25]:

POL(0) = 1
POL(PLUS_IN_GAA(x₁)) = 2·x₁
POL(U1_ga(x₁)) = 2·x₁
POL(U2_GAA(x₁)) = 2·x₁
POL(p_in_ga(x₁)) = x₁
POL(p_out_ga(x₁)) = x₁
POL(s(x₁)) = 2·x₁

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ RuleRemovalProof

                          ↳ QDP

                            ↳ RuleRemovalProof

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

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

The TRS R consists of the following rules:

p_in_ga(s(s(X))) → U1_ga(p_in_ga(s(X)))
U1_ga(p_out_ga(s(Y))) → p_out_ga(s(s(Y)))

The set Q consists of the following terms:

p_in_ga(x0)
U1_ga(x0)

U2_GAA(p_out_ga(U)) → PLUS_IN_GAA(U)

Strictly oriented rules of the TRS R:

U1_ga(p_out_ga(s(Y))) → p_out_ga(s(s(Y)))

Used ordering: POLO with Polynomial interpretation [25]:

POL(PLUS_IN_GAA(x₁)) = 2·x₁
POL(U1_ga(x₁)) = 2·x₁
POL(U2_GAA(x₁)) = x₁
POL(p_in_ga(x₁)) = x₁
POL(p_out_ga(x₁)) = 2 + 2·x₁
POL(s(x₁)) = 2·x₁

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ RuleRemovalProof

                          ↳ QDP

                            ↳ RuleRemovalProof

                              ↳ QDP

                                ↳ DependencyGraphProof

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

PLUS_IN_GAA(s(X)) → U2_GAA(p_in_ga(s(X)))

The TRS R consists of the following rules:

p_in_ga(s(s(X))) → U1_ga(p_in_ga(s(X)))

The set Q consists of the following terms:

p_in_ga(x0)
U1_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.