Left Termination proof of /tmp/tmpdwqIfz/pl4.5.2.pl

Left Termination of the query pattern s2_in_2(g, a) w.r.t. the given Prolog program could not be shown:

↳ Prolog

  ↳ PrologToPiTRSProof

  ↳ PrologToPiTRSProof

Clauses:

s2(plus(A, plus(B, C)), D) :- s2(plus(plus(A, B), C), D).
s2(plus(A, B), C) :- s2(plus(B, A), C).
s2(plus(X, 0), X).
s2(plus(X, Y), Z) :- ','(s2(X, A), ','(s2(Y, B), s2(plus(A, B), Z))).
s2(plus(A, B), C) :- ','(isNat(A), ','(isNat(B), add(A, B, C))).
isNat(s(X)) :- isNat(X).
isNat(0).
add(s(X), Y, s(Z)) :- add(X, Y, Z).
add(0, X, X).

Queries:

s2(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:

s2_in(plus(A, B), C) → U6(A, B, C, isNat_in(A))
isNat_in(0) → isNat_out(0)
isNat_in(s(X)) → U9(X, isNat_in(X))
U9(X, isNat_out(X)) → isNat_out(s(X))
U6(A, B, C, isNat_out(A)) → U7(A, B, C, isNat_in(B))
U7(A, B, C, isNat_out(B)) → U8(A, B, C, add_in(A, B, C))
add_in(0, X, X) → add_out(0, X, X)
add_in(s(X), Y, s(Z)) → U10(X, Y, Z, add_in(X, Y, Z))
U10(X, Y, Z, add_out(X, Y, Z)) → add_out(s(X), Y, s(Z))
U8(A, B, C, add_out(A, B, C)) → s2_out(plus(A, B), C)
s2_in(plus(X, Y), Z) → U3(X, Y, Z, s2_in(X, A))
s2_in(plus(X, 0), X) → s2_out(plus(X, 0), X)
s2_in(plus(A, B), C) → U2(A, B, C, s2_in(plus(B, A), C))
s2_in(plus(A, plus(B, C)), D) → U1(A, B, C, D, s2_in(plus(plus(A, B), C), D))
U1(A, B, C, D, s2_out(plus(plus(A, B), C), D)) → s2_out(plus(A, plus(B, C)), D)
U2(A, B, C, s2_out(plus(B, A), C)) → s2_out(plus(A, B), C)
U3(X, Y, Z, s2_out(X, A)) → U4(X, Y, Z, A, s2_in(Y, B))
U4(X, Y, Z, A, s2_out(Y, B)) → U5(X, Y, Z, s2_in(plus(A, B), Z))
U5(X, Y, Z, s2_out(plus(A, B), Z)) → s2_out(plus(X, Y), 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:

s2_in(plus(A, B), C) → U6(A, B, C, isNat_in(A))
isNat_in(0) → isNat_out(0)
isNat_in(s(X)) → U9(X, isNat_in(X))
U9(X, isNat_out(X)) → isNat_out(s(X))
U6(A, B, C, isNat_out(A)) → U7(A, B, C, isNat_in(B))
U7(A, B, C, isNat_out(B)) → U8(A, B, C, add_in(A, B, C))
add_in(0, X, X) → add_out(0, X, X)
add_in(s(X), Y, s(Z)) → U10(X, Y, Z, add_in(X, Y, Z))
U10(X, Y, Z, add_out(X, Y, Z)) → add_out(s(X), Y, s(Z))
U8(A, B, C, add_out(A, B, C)) → s2_out(plus(A, B), C)
s2_in(plus(X, Y), Z) → U3(X, Y, Z, s2_in(X, A))
s2_in(plus(X, 0), X) → s2_out(plus(X, 0), X)
s2_in(plus(A, B), C) → U2(A, B, C, s2_in(plus(B, A), C))
s2_in(plus(A, plus(B, C)), D) → U1(A, B, C, D, s2_in(plus(plus(A, B), C), D))
U1(A, B, C, D, s2_out(plus(plus(A, B), C), D)) → s2_out(plus(A, plus(B, C)), D)
U2(A, B, C, s2_out(plus(B, A), C)) → s2_out(plus(A, B), C)
U3(X, Y, Z, s2_out(X, A)) → U4(X, Y, Z, A, s2_in(Y, B))
U4(X, Y, Z, A, s2_out(Y, B)) → U5(X, Y, Z, s2_in(plus(A, B), Z))
U5(X, Y, Z, s2_out(plus(A, B), Z)) → s2_out(plus(X, Y), Z)

S2_IN(plus(A, B), C) → U6¹(A, B, C, isNat_in(A))
S2_IN(plus(A, B), C) → ISNAT_IN(A)
ISNAT_IN(s(X)) → U9¹(X, isNat_in(X))
ISNAT_IN(s(X)) → ISNAT_IN(X)
U6¹(A, B, C, isNat_out(A)) → U7¹(A, B, C, isNat_in(B))
U6¹(A, B, C, isNat_out(A)) → ISNAT_IN(B)
U7¹(A, B, C, isNat_out(B)) → U8¹(A, B, C, add_in(A, B, C))
U7¹(A, B, C, isNat_out(B)) → ADD_IN(A, B, C)
ADD_IN(s(X), Y, s(Z)) → U10¹(X, Y, Z, add_in(X, Y, Z))
ADD_IN(s(X), Y, s(Z)) → ADD_IN(X, Y, Z)
S2_IN(plus(X, Y), Z) → U3¹(X, Y, Z, s2_in(X, A))
S2_IN(plus(X, Y), Z) → S2_IN(X, A)
S2_IN(plus(A, B), C) → U2¹(A, B, C, s2_in(plus(B, A), C))
S2_IN(plus(A, B), C) → S2_IN(plus(B, A), C)
S2_IN(plus(A, plus(B, C)), D) → U1¹(A, B, C, D, s2_in(plus(plus(A, B), C), D))
S2_IN(plus(A, plus(B, C)), D) → S2_IN(plus(plus(A, B), C), D)
U3¹(X, Y, Z, s2_out(X, A)) → U4¹(X, Y, Z, A, s2_in(Y, B))
U3¹(X, Y, Z, s2_out(X, A)) → S2_IN(Y, B)
U4¹(X, Y, Z, A, s2_out(Y, B)) → U5¹(X, Y, Z, s2_in(plus(A, B), Z))
U4¹(X, Y, Z, A, s2_out(Y, B)) → S2_IN(plus(A, B), Z)

The TRS R consists of the following rules:

s2_in(plus(A, B), C) → U6(A, B, C, isNat_in(A))
isNat_in(0) → isNat_out(0)
isNat_in(s(X)) → U9(X, isNat_in(X))
U9(X, isNat_out(X)) → isNat_out(s(X))
U6(A, B, C, isNat_out(A)) → U7(A, B, C, isNat_in(B))
U7(A, B, C, isNat_out(B)) → U8(A, B, C, add_in(A, B, C))
add_in(0, X, X) → add_out(0, X, X)
add_in(s(X), Y, s(Z)) → U10(X, Y, Z, add_in(X, Y, Z))
U10(X, Y, Z, add_out(X, Y, Z)) → add_out(s(X), Y, s(Z))
U8(A, B, C, add_out(A, B, C)) → s2_out(plus(A, B), C)
s2_in(plus(X, Y), Z) → U3(X, Y, Z, s2_in(X, A))
s2_in(plus(X, 0), X) → s2_out(plus(X, 0), X)
s2_in(plus(A, B), C) → U2(A, B, C, s2_in(plus(B, A), C))
s2_in(plus(A, plus(B, C)), D) → U1(A, B, C, D, s2_in(plus(plus(A, B), C), D))
U1(A, B, C, D, s2_out(plus(plus(A, B), C), D)) → s2_out(plus(A, plus(B, C)), D)
U2(A, B, C, s2_out(plus(B, A), C)) → s2_out(plus(A, B), C)
U3(X, Y, Z, s2_out(X, A)) → U4(X, Y, Z, A, s2_in(Y, B))
U4(X, Y, Z, A, s2_out(Y, B)) → U5(X, Y, Z, s2_in(plus(A, B), Z))
U5(X, Y, Z, s2_out(plus(A, B), Z)) → s2_out(plus(X, Y), Z)

The argument filtering Pi contains the following mapping:
s2_in(x1, x2) = s2_in(x1)
plus(x1, x2) = plus(x1, x2)
U6(x1, x2, x3, x4) = U6(x1, x2, x4)
isNat_in(x1) = isNat_in(x1)
0 = 0
isNat_out(x1) = isNat_out
s(x1) = s(x1)
U9(x1, x2) = U9(x2)
U7(x1, x2, x3, x4) = U7(x1, x2, x4)
U8(x1, x2, x3, x4) = U8(x4)
add_in(x1, x2, x3) = add_in(x1, x2)
add_out(x1, x2, x3) = add_out(x3)
U10(x1, x2, x3, x4) = U10(x4)
s2_out(x1, x2) = s2_out(x2)
U3(x1, x2, x3, x4) = U3(x2, x4)
U2(x1, x2, x3, x4) = U2(x4)
U1(x1, x2, x3, x4, x5) = U1(x5)
U4(x1, x2, x3, x4, x5) = U4(x4, x5)
U5(x1, x2, x3, x4) = U5(x4)
ADD_IN(x1, x2, x3) = ADD_IN(x1, x2)
U10¹(x1, x2, x3, x4) = U10¹(x4)
U2¹(x1, x2, x3, x4) = U2¹(x4)
ISNAT_IN(x1) = ISNAT_IN(x1)
U1¹(x1, x2, x3, x4, x5) = U1¹(x5)
U7¹(x1, x2, x3, x4) = U7¹(x1, x2, x4)
U5¹(x1, x2, x3, x4) = U5¹(x4)
U9¹(x1, x2) = U9¹(x2)
U3¹(x1, x2, x3, x4) = U3¹(x2, x4)
U4¹(x1, x2, x3, x4, x5) = U4¹(x4, x5)
U8¹(x1, x2, x3, x4) = U8¹(x4)
S2_IN(x1, x2) = S2_IN(x1)
U6¹(x1, x2, x3, x4) = U6¹(x1, x2, x4)

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:

S2_IN(plus(A, B), C) → U6¹(A, B, C, isNat_in(A))
S2_IN(plus(A, B), C) → ISNAT_IN(A)
ISNAT_IN(s(X)) → U9¹(X, isNat_in(X))
ISNAT_IN(s(X)) → ISNAT_IN(X)
U6¹(A, B, C, isNat_out(A)) → U7¹(A, B, C, isNat_in(B))
U6¹(A, B, C, isNat_out(A)) → ISNAT_IN(B)
U7¹(A, B, C, isNat_out(B)) → U8¹(A, B, C, add_in(A, B, C))
U7¹(A, B, C, isNat_out(B)) → ADD_IN(A, B, C)
ADD_IN(s(X), Y, s(Z)) → U10¹(X, Y, Z, add_in(X, Y, Z))
ADD_IN(s(X), Y, s(Z)) → ADD_IN(X, Y, Z)
S2_IN(plus(X, Y), Z) → U3¹(X, Y, Z, s2_in(X, A))
S2_IN(plus(X, Y), Z) → S2_IN(X, A)
S2_IN(plus(A, B), C) → U2¹(A, B, C, s2_in(plus(B, A), C))
S2_IN(plus(A, B), C) → S2_IN(plus(B, A), C)
S2_IN(plus(A, plus(B, C)), D) → U1¹(A, B, C, D, s2_in(plus(plus(A, B), C), D))
S2_IN(plus(A, plus(B, C)), D) → S2_IN(plus(plus(A, B), C), D)
U3¹(X, Y, Z, s2_out(X, A)) → U4¹(X, Y, Z, A, s2_in(Y, B))
U3¹(X, Y, Z, s2_out(X, A)) → S2_IN(Y, B)
U4¹(X, Y, Z, A, s2_out(Y, B)) → U5¹(X, Y, Z, s2_in(plus(A, B), Z))
U4¹(X, Y, Z, A, s2_out(Y, B)) → S2_IN(plus(A, B), Z)

The TRS R consists of the following rules:

s2_in(plus(A, B), C) → U6(A, B, C, isNat_in(A))
isNat_in(0) → isNat_out(0)
isNat_in(s(X)) → U9(X, isNat_in(X))
U9(X, isNat_out(X)) → isNat_out(s(X))
U6(A, B, C, isNat_out(A)) → U7(A, B, C, isNat_in(B))
U7(A, B, C, isNat_out(B)) → U8(A, B, C, add_in(A, B, C))
add_in(0, X, X) → add_out(0, X, X)
add_in(s(X), Y, s(Z)) → U10(X, Y, Z, add_in(X, Y, Z))
U10(X, Y, Z, add_out(X, Y, Z)) → add_out(s(X), Y, s(Z))
U8(A, B, C, add_out(A, B, C)) → s2_out(plus(A, B), C)
s2_in(plus(X, Y), Z) → U3(X, Y, Z, s2_in(X, A))
s2_in(plus(X, 0), X) → s2_out(plus(X, 0), X)
s2_in(plus(A, B), C) → U2(A, B, C, s2_in(plus(B, A), C))
s2_in(plus(A, plus(B, C)), D) → U1(A, B, C, D, s2_in(plus(plus(A, B), C), D))
U1(A, B, C, D, s2_out(plus(plus(A, B), C), D)) → s2_out(plus(A, plus(B, C)), D)
U2(A, B, C, s2_out(plus(B, A), C)) → s2_out(plus(A, B), C)
U3(X, Y, Z, s2_out(X, A)) → U4(X, Y, Z, A, s2_in(Y, B))
U4(X, Y, Z, A, s2_out(Y, B)) → U5(X, Y, Z, s2_in(plus(A, B), Z))
U5(X, Y, Z, s2_out(plus(A, B), Z)) → s2_out(plus(X, Y), Z)

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

                ↳ UsableRulesProof

              ↳ PiDP

              ↳ PiDP

  ↳ PrologToPiTRSProof

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

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

The TRS R consists of the following rules:

s2_in(plus(A, B), C) → U6(A, B, C, isNat_in(A))
isNat_in(0) → isNat_out(0)
isNat_in(s(X)) → U9(X, isNat_in(X))
U9(X, isNat_out(X)) → isNat_out(s(X))
U6(A, B, C, isNat_out(A)) → U7(A, B, C, isNat_in(B))
U7(A, B, C, isNat_out(B)) → U8(A, B, C, add_in(A, B, C))
add_in(0, X, X) → add_out(0, X, X)
add_in(s(X), Y, s(Z)) → U10(X, Y, Z, add_in(X, Y, Z))
U10(X, Y, Z, add_out(X, Y, Z)) → add_out(s(X), Y, s(Z))
U8(A, B, C, add_out(A, B, C)) → s2_out(plus(A, B), C)
s2_in(plus(X, Y), Z) → U3(X, Y, Z, s2_in(X, A))
s2_in(plus(X, 0), X) → s2_out(plus(X, 0), X)
s2_in(plus(A, B), C) → U2(A, B, C, s2_in(plus(B, A), C))
s2_in(plus(A, plus(B, C)), D) → U1(A, B, C, D, s2_in(plus(plus(A, B), C), D))
U1(A, B, C, D, s2_out(plus(plus(A, B), C), D)) → s2_out(plus(A, plus(B, C)), D)
U2(A, B, C, s2_out(plus(B, A), C)) → s2_out(plus(A, B), C)
U3(X, Y, Z, s2_out(X, A)) → U4(X, Y, Z, A, s2_in(Y, B))
U4(X, Y, Z, A, s2_out(Y, B)) → U5(X, Y, Z, s2_in(plus(A, B), Z))
U5(X, Y, Z, s2_out(plus(A, B), Z)) → s2_out(plus(X, Y), Z)

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

              ↳ PiDP

              ↳ PiDP

  ↳ PrologToPiTRSProof

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

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

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

            ↳ AND

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ QDPSizeChangeProof

              ↳ PiDP

              ↳ PiDP

  ↳ PrologToPiTRSProof

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

ADD_IN(s(X), Y) → ADD_IN(X, 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:

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

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

              ↳ PiDP

  ↳ PrologToPiTRSProof

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

ISNAT_IN(s(X)) → ISNAT_IN(X)

The TRS R consists of the following rules:

s2_in(plus(A, B), C) → U6(A, B, C, isNat_in(A))
isNat_in(0) → isNat_out(0)
isNat_in(s(X)) → U9(X, isNat_in(X))
U9(X, isNat_out(X)) → isNat_out(s(X))
U6(A, B, C, isNat_out(A)) → U7(A, B, C, isNat_in(B))
U7(A, B, C, isNat_out(B)) → U8(A, B, C, add_in(A, B, C))
add_in(0, X, X) → add_out(0, X, X)
add_in(s(X), Y, s(Z)) → U10(X, Y, Z, add_in(X, Y, Z))
U10(X, Y, Z, add_out(X, Y, Z)) → add_out(s(X), Y, s(Z))
U8(A, B, C, add_out(A, B, C)) → s2_out(plus(A, B), C)
s2_in(plus(X, Y), Z) → U3(X, Y, Z, s2_in(X, A))
s2_in(plus(X, 0), X) → s2_out(plus(X, 0), X)
s2_in(plus(A, B), C) → U2(A, B, C, s2_in(plus(B, A), C))
s2_in(plus(A, plus(B, C)), D) → U1(A, B, C, D, s2_in(plus(plus(A, B), C), D))
U1(A, B, C, D, s2_out(plus(plus(A, B), C), D)) → s2_out(plus(A, plus(B, C)), D)
U2(A, B, C, s2_out(plus(B, A), C)) → s2_out(plus(A, B), C)
U3(X, Y, Z, s2_out(X, A)) → U4(X, Y, Z, A, s2_in(Y, B))
U4(X, Y, Z, A, s2_out(Y, B)) → U5(X, Y, Z, s2_in(plus(A, B), Z))
U5(X, Y, Z, s2_out(plus(A, B), Z)) → s2_out(plus(X, Y), Z)

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

              ↳ PiDP

  ↳ PrologToPiTRSProof

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

ISNAT_IN(s(X)) → ISNAT_IN(X)

R is empty.
Pi is empty.
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

                        ↳ QDPSizeChangeProof

              ↳ PiDP

  ↳ PrologToPiTRSProof

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

ISNAT_IN(s(X)) → ISNAT_IN(X)

From the DPs we obtained the following set of size-change graphs:

ISNAT_IN(s(X)) → ISNAT_IN(X)
The graph contains the following edges 1 > 1

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

                ↳ PiDPToQDPProof

  ↳ PrologToPiTRSProof

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

U3¹(X, Y, Z, s2_out(X, A)) → S2_IN(Y, B)
S2_IN(plus(X, Y), Z) → U3¹(X, Y, Z, s2_in(X, A))
U4¹(X, Y, Z, A, s2_out(Y, B)) → S2_IN(plus(A, B), Z)
S2_IN(plus(A, B), C) → S2_IN(plus(B, A), C)
U3¹(X, Y, Z, s2_out(X, A)) → U4¹(X, Y, Z, A, s2_in(Y, B))
S2_IN(plus(A, plus(B, C)), D) → S2_IN(plus(plus(A, B), C), D)
S2_IN(plus(X, Y), Z) → S2_IN(X, A)

The TRS R consists of the following rules:

s2_in(plus(A, B), C) → U6(A, B, C, isNat_in(A))
isNat_in(0) → isNat_out(0)
isNat_in(s(X)) → U9(X, isNat_in(X))
U9(X, isNat_out(X)) → isNat_out(s(X))
U6(A, B, C, isNat_out(A)) → U7(A, B, C, isNat_in(B))
U7(A, B, C, isNat_out(B)) → U8(A, B, C, add_in(A, B, C))
add_in(0, X, X) → add_out(0, X, X)
add_in(s(X), Y, s(Z)) → U10(X, Y, Z, add_in(X, Y, Z))
U10(X, Y, Z, add_out(X, Y, Z)) → add_out(s(X), Y, s(Z))
U8(A, B, C, add_out(A, B, C)) → s2_out(plus(A, B), C)
s2_in(plus(X, Y), Z) → U3(X, Y, Z, s2_in(X, A))
s2_in(plus(X, 0), X) → s2_out(plus(X, 0), X)
s2_in(plus(A, B), C) → U2(A, B, C, s2_in(plus(B, A), C))
s2_in(plus(A, plus(B, C)), D) → U1(A, B, C, D, s2_in(plus(plus(A, B), C), D))
U1(A, B, C, D, s2_out(plus(plus(A, B), C), D)) → s2_out(plus(A, plus(B, C)), D)
U2(A, B, C, s2_out(plus(B, A), C)) → s2_out(plus(A, B), C)
U3(X, Y, Z, s2_out(X, A)) → U4(X, Y, Z, A, s2_in(Y, B))
U4(X, Y, Z, A, s2_out(Y, B)) → U5(X, Y, Z, s2_in(plus(A, B), Z))
U5(X, Y, Z, s2_out(plus(A, B), Z)) → s2_out(plus(X, Y), Z)

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

                ↳ PiDPToQDPProof

                  ↳ QDP

                    ↳ QDPOrderProof

  ↳ PrologToPiTRSProof

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

U3¹(Y, s2_out(A)) → S2_IN(Y)
U4¹(A, s2_out(B)) → S2_IN(plus(A, B))
U3¹(Y, s2_out(A)) → U4¹(A, s2_in(Y))
S2_IN(plus(X, Y)) → U3¹(Y, s2_in(X))
S2_IN(plus(X, Y)) → S2_IN(X)
S2_IN(plus(A, plus(B, C))) → S2_IN(plus(plus(A, B), C))
S2_IN(plus(A, B)) → S2_IN(plus(B, A))

The TRS R consists of the following rules:

s2_in(plus(A, B)) → U6(A, B, isNat_in(A))
isNat_in(0) → isNat_out
isNat_in(s(X)) → U9(isNat_in(X))
U9(isNat_out) → isNat_out
U6(A, B, isNat_out) → U7(A, B, isNat_in(B))
U7(A, B, isNat_out) → U8(add_in(A, B))
add_in(0, X) → add_out(X)
add_in(s(X), Y) → U10(add_in(X, Y))
U10(add_out(Z)) → add_out(s(Z))
U8(add_out(C)) → s2_out(C)
s2_in(plus(X, Y)) → U3(Y, s2_in(X))
s2_in(plus(X, 0)) → s2_out(X)
s2_in(plus(A, B)) → U2(s2_in(plus(B, A)))
s2_in(plus(A, plus(B, C))) → U1(s2_in(plus(plus(A, B), C)))
U1(s2_out(D)) → s2_out(D)
U2(s2_out(C)) → s2_out(C)
U3(Y, s2_out(A)) → U4(A, s2_in(Y))
U4(A, s2_out(B)) → U5(s2_in(plus(A, B)))
U5(s2_out(Z)) → s2_out(Z)

The set Q consists of the following terms:

s2_in(x0)
isNat_in(x0)
U9(x0)
U6(x0, x1, x2)
U7(x0, x1, x2)
add_in(x0, x1)
U10(x0)
U8(x0)
U1(x0)
U2(x0)
U3(x0, x1)
U4(x0, x1)
U5(x0)

We have to consider all (P,Q,R)-chains.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

U3¹(Y, s2_out(A)) → S2_IN(Y)
S2_IN(plus(X, Y)) → S2_IN(X)

The remaining pairs can at least be oriented weakly.

U4¹(A, s2_out(B)) → S2_IN(plus(A, B))
U3¹(Y, s2_out(A)) → U4¹(A, s2_in(Y))
S2_IN(plus(X, Y)) → U3¹(Y, s2_in(X))
S2_IN(plus(A, plus(B, C))) → S2_IN(plus(plus(A, B), C))
S2_IN(plus(A, B)) → S2_IN(plus(B, A))

Used ordering: Polynomial interpretation [25]:

POL(0) = 0
POL(S2_IN(x₁)) = x₁
POL(U1(x₁)) = x₁
POL(U10(x₁)) = x₁
POL(U2(x₁)) = x₁
POL(U3(x₁, x₂)) = 1 + x₁ + x₂
POL(U3¹(x₁, x₂)) = 1 + x₁ + x₂
POL(U4(x₁, x₂)) = 1 + x₁ + x₂
POL(U4¹(x₁, x₂)) = 1 + x₁ + x₂
POL(U5(x₁)) = x₁
POL(U6(x₁, x₂, x₃)) = x₁ + x₂
POL(U7(x₁, x₂, x₃)) = x₂
POL(U8(x₁)) = x₁
POL(U9(x₁)) = 1
POL(add_in(x₁, x₂)) = x₂
POL(add_out(x₁)) = x₁
POL(isNat_in(x₁)) = 1 + x₁
POL(isNat_out) = 0
POL(plus(x₁, x₂)) = 1 + x₁ + x₂
POL(s(x₁)) = x₁
POL(s2_in(x₁)) = x₁
POL(s2_out(x₁)) = x₁

The following usable rules [17] were oriented:

s2_in(plus(X, Y)) → U3(Y, s2_in(X))
U7(A, B, isNat_out) → U8(add_in(A, B))
s2_in(plus(A, B)) → U6(A, B, isNat_in(A))
U4(A, s2_out(B)) → U5(s2_in(plus(A, B)))
s2_in(plus(A, plus(B, C))) → U1(s2_in(plus(plus(A, B), C)))
U8(add_out(C)) → s2_out(C)
U6(A, B, isNat_out) → U7(A, B, isNat_in(B))
U10(add_out(Z)) → add_out(s(Z))
U5(s2_out(Z)) → s2_out(Z)
U1(s2_out(D)) → s2_out(D)
U2(s2_out(C)) → s2_out(C)
add_in(s(X), Y) → U10(add_in(X, Y))
s2_in(plus(X, 0)) → s2_out(X)
U3(Y, s2_out(A)) → U4(A, s2_in(Y))
add_in(0, X) → add_out(X)
s2_in(plus(A, B)) → U2(s2_in(plus(B, A)))

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

                ↳ PiDPToQDPProof

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ QDPOrderProof

  ↳ PrologToPiTRSProof

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

U4¹(A, s2_out(B)) → S2_IN(plus(A, B))
S2_IN(plus(X, Y)) → U3¹(Y, s2_in(X))
U3¹(Y, s2_out(A)) → U4¹(A, s2_in(Y))
S2_IN(plus(A, plus(B, C))) → S2_IN(plus(plus(A, B), C))
S2_IN(plus(A, B)) → S2_IN(plus(B, A))

The TRS R consists of the following rules:

s2_in(plus(A, B)) → U6(A, B, isNat_in(A))
isNat_in(0) → isNat_out
isNat_in(s(X)) → U9(isNat_in(X))
U9(isNat_out) → isNat_out
U6(A, B, isNat_out) → U7(A, B, isNat_in(B))
U7(A, B, isNat_out) → U8(add_in(A, B))
add_in(0, X) → add_out(X)
add_in(s(X), Y) → U10(add_in(X, Y))
U10(add_out(Z)) → add_out(s(Z))
U8(add_out(C)) → s2_out(C)
s2_in(plus(X, Y)) → U3(Y, s2_in(X))
s2_in(plus(X, 0)) → s2_out(X)
s2_in(plus(A, B)) → U2(s2_in(plus(B, A)))
s2_in(plus(A, plus(B, C))) → U1(s2_in(plus(plus(A, B), C)))
U1(s2_out(D)) → s2_out(D)
U2(s2_out(C)) → s2_out(C)
U3(Y, s2_out(A)) → U4(A, s2_in(Y))
U4(A, s2_out(B)) → U5(s2_in(plus(A, B)))
U5(s2_out(Z)) → s2_out(Z)

The set Q consists of the following terms:

s2_in(x0)
isNat_in(x0)
U9(x0)
U6(x0, x1, x2)
U7(x0, x1, x2)
add_in(x0, x1)
U10(x0)
U8(x0)
U1(x0)
U2(x0)
U3(x0, x1)
U4(x0, x1)
U5(x0)

We have to consider all (P,Q,R)-chains.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

U4¹(A, s2_out(B)) → S2_IN(plus(A, B))
U3¹(Y, s2_out(A)) → U4¹(A, s2_in(Y))

The remaining pairs can at least be oriented weakly.

S2_IN(plus(X, Y)) → U3¹(Y, s2_in(X))
S2_IN(plus(A, plus(B, C))) → S2_IN(plus(plus(A, B), C))
S2_IN(plus(A, B)) → S2_IN(plus(B, A))

Used ordering: Polynomial interpretation [25]:

POL(0) = 1
POL(S2_IN(x₁)) = 1 + x₁
POL(U1(x₁)) = x₁
POL(U10(x₁)) = 1 + x₁
POL(U2(x₁)) = x₁
POL(U3(x₁, x₂)) = x₁ + x₂
POL(U3¹(x₁, x₂)) = 1 + x₁ + x₂
POL(U4(x₁, x₂)) = 1 + x₁ + x₂
POL(U4¹(x₁, x₂)) = 1 + x₁ + x₂
POL(U5(x₁)) = 1 + x₁
POL(U6(x₁, x₂, x₃)) = x₁ + x₂
POL(U7(x₁, x₂, x₃)) = x₁ + x₂
POL(U8(x₁)) = x₁
POL(U9(x₁)) = 1
POL(add_in(x₁, x₂)) = x₁ + x₂
POL(add_out(x₁)) = 1 + x₁
POL(isNat_in(x₁)) = x₁
POL(isNat_out) = 0
POL(plus(x₁, x₂)) = x₁ + x₂
POL(s(x₁)) = 1 + x₁
POL(s2_in(x₁)) = x₁
POL(s2_out(x₁)) = 1 + x₁

The following usable rules [17] were oriented:

s2_in(plus(X, Y)) → U3(Y, s2_in(X))
U7(A, B, isNat_out) → U8(add_in(A, B))
s2_in(plus(A, B)) → U6(A, B, isNat_in(A))
U4(A, s2_out(B)) → U5(s2_in(plus(A, B)))
s2_in(plus(A, plus(B, C))) → U1(s2_in(plus(plus(A, B), C)))
U8(add_out(C)) → s2_out(C)
U6(A, B, isNat_out) → U7(A, B, isNat_in(B))
U10(add_out(Z)) → add_out(s(Z))
U5(s2_out(Z)) → s2_out(Z)
U1(s2_out(D)) → s2_out(D)
U2(s2_out(C)) → s2_out(C)
add_in(s(X), Y) → U10(add_in(X, Y))
s2_in(plus(X, 0)) → s2_out(X)
U3(Y, s2_out(A)) → U4(A, s2_in(Y))
add_in(0, X) → add_out(X)
s2_in(plus(A, B)) → U2(s2_in(plus(B, A)))

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

                ↳ PiDPToQDPProof

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ QDPOrderProof

                          ↳ QDP

                            ↳ DependencyGraphProof

  ↳ PrologToPiTRSProof

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

S2_IN(plus(X, Y)) → U3¹(Y, s2_in(X))
S2_IN(plus(A, plus(B, C))) → S2_IN(plus(plus(A, B), C))
S2_IN(plus(A, B)) → S2_IN(plus(B, A))

The TRS R consists of the following rules:

s2_in(plus(A, B)) → U6(A, B, isNat_in(A))
isNat_in(0) → isNat_out
isNat_in(s(X)) → U9(isNat_in(X))
U9(isNat_out) → isNat_out
U6(A, B, isNat_out) → U7(A, B, isNat_in(B))
U7(A, B, isNat_out) → U8(add_in(A, B))
add_in(0, X) → add_out(X)
add_in(s(X), Y) → U10(add_in(X, Y))
U10(add_out(Z)) → add_out(s(Z))
U8(add_out(C)) → s2_out(C)
s2_in(plus(X, Y)) → U3(Y, s2_in(X))
s2_in(plus(X, 0)) → s2_out(X)
s2_in(plus(A, B)) → U2(s2_in(plus(B, A)))
s2_in(plus(A, plus(B, C))) → U1(s2_in(plus(plus(A, B), C)))
U1(s2_out(D)) → s2_out(D)
U2(s2_out(C)) → s2_out(C)
U3(Y, s2_out(A)) → U4(A, s2_in(Y))
U4(A, s2_out(B)) → U5(s2_in(plus(A, B)))
U5(s2_out(Z)) → s2_out(Z)

The set Q consists of the following terms:

s2_in(x0)
isNat_in(x0)
U9(x0)
U6(x0, x1, x2)
U7(x0, x1, x2)
add_in(x0, x1)
U10(x0)
U8(x0)
U1(x0)
U2(x0)
U3(x0, x1)
U4(x0, x1)
U5(x0)

We have to consider all (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 1 less node.

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

                ↳ PiDPToQDPProof

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ QDPOrderProof

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ UsableRulesProof

  ↳ PrologToPiTRSProof

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

S2_IN(plus(A, plus(B, C))) → S2_IN(plus(plus(A, B), C))
S2_IN(plus(A, B)) → S2_IN(plus(B, A))

The TRS R consists of the following rules:

s2_in(plus(A, B)) → U6(A, B, isNat_in(A))
isNat_in(0) → isNat_out
isNat_in(s(X)) → U9(isNat_in(X))
U9(isNat_out) → isNat_out
U6(A, B, isNat_out) → U7(A, B, isNat_in(B))
U7(A, B, isNat_out) → U8(add_in(A, B))
add_in(0, X) → add_out(X)
add_in(s(X), Y) → U10(add_in(X, Y))
U10(add_out(Z)) → add_out(s(Z))
U8(add_out(C)) → s2_out(C)
s2_in(plus(X, Y)) → U3(Y, s2_in(X))
s2_in(plus(X, 0)) → s2_out(X)
s2_in(plus(A, B)) → U2(s2_in(plus(B, A)))
s2_in(plus(A, plus(B, C))) → U1(s2_in(plus(plus(A, B), C)))
U1(s2_out(D)) → s2_out(D)
U2(s2_out(C)) → s2_out(C)
U3(Y, s2_out(A)) → U4(A, s2_in(Y))
U4(A, s2_out(B)) → U5(s2_in(plus(A, B)))
U5(s2_out(Z)) → s2_out(Z)

The set Q consists of the following terms:

s2_in(x0)
isNat_in(x0)
U9(x0)
U6(x0, x1, x2)
U7(x0, x1, x2)
add_in(x0, x1)
U10(x0)
U8(x0)
U1(x0)
U2(x0)
U3(x0, x1)
U4(x0, x1)
U5(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

              ↳ PiDP

                ↳ PiDPToQDPProof

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ QDPOrderProof

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ UsableRulesProof

                                  ↳ QDP

                                    ↳ QReductionProof

  ↳ PrologToPiTRSProof

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

S2_IN(plus(A, plus(B, C))) → S2_IN(plus(plus(A, B), C))
S2_IN(plus(A, B)) → S2_IN(plus(B, A))

R is empty.
The set Q consists of the following terms:

s2_in(x0)
isNat_in(x0)
U9(x0)
U6(x0, x1, x2)
U7(x0, x1, x2)
add_in(x0, x1)
U10(x0)
U8(x0)
U1(x0)
U2(x0)
U3(x0, x1)
U4(x0, x1)
U5(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.

s2_in(x0)
isNat_in(x0)
U9(x0)
U6(x0, x1, x2)
U7(x0, x1, x2)
add_in(x0, x1)
U10(x0)
U8(x0)
U1(x0)
U2(x0)
U3(x0, x1)
U4(x0, x1)
U5(x0)

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

                ↳ PiDPToQDPProof

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ QDPOrderProof

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ UsableRulesProof

                                  ↳ QDP

                                    ↳ QReductionProof

                                      ↳ QDP

                                        ↳ NonTerminationProof

  ↳ PrologToPiTRSProof

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

S2_IN(plus(A, plus(B, C))) → S2_IN(plus(plus(A, B), C))
S2_IN(plus(A, B)) → S2_IN(plus(B, A))

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:

S2_IN(plus(A, plus(B, C))) → S2_IN(plus(plus(A, B), C))
S2_IN(plus(A, B)) → S2_IN(plus(B, A))

The TRS R consists of the following rules:none

s = S2_IN(plus(A, B)) evaluates to t =S2_IN(plus(B, A))

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

Matcher: [B / A, A / B]
Semiunifier: [ ]

Rewriting sequence

The DP semiunifies directly so there is only one rewrite step from S2_IN(plus(A, B)) to S2_IN(plus(B, 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:

s2_in(plus(A, B), C) → U6(A, B, C, isNat_in(A))
isNat_in(0) → isNat_out(0)
isNat_in(s(X)) → U9(X, isNat_in(X))
U9(X, isNat_out(X)) → isNat_out(s(X))
U6(A, B, C, isNat_out(A)) → U7(A, B, C, isNat_in(B))
U7(A, B, C, isNat_out(B)) → U8(A, B, C, add_in(A, B, C))
add_in(0, X, X) → add_out(0, X, X)
add_in(s(X), Y, s(Z)) → U10(X, Y, Z, add_in(X, Y, Z))
U10(X, Y, Z, add_out(X, Y, Z)) → add_out(s(X), Y, s(Z))
U8(A, B, C, add_out(A, B, C)) → s2_out(plus(A, B), C)
s2_in(plus(X, Y), Z) → U3(X, Y, Z, s2_in(X, A))
s2_in(plus(X, 0), X) → s2_out(plus(X, 0), X)
s2_in(plus(A, B), C) → U2(A, B, C, s2_in(plus(B, A), C))
s2_in(plus(A, plus(B, C)), D) → U1(A, B, C, D, s2_in(plus(plus(A, B), C), D))
U1(A, B, C, D, s2_out(plus(plus(A, B), C), D)) → s2_out(plus(A, plus(B, C)), D)
U2(A, B, C, s2_out(plus(B, A), C)) → s2_out(plus(A, B), C)
U3(X, Y, Z, s2_out(X, A)) → U4(X, Y, Z, A, s2_in(Y, B))
U4(X, Y, Z, A, s2_out(Y, B)) → U5(X, Y, Z, s2_in(plus(A, B), Z))
U5(X, Y, Z, s2_out(plus(A, B), Z)) → s2_out(plus(X, Y), 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:

s2_in(plus(A, B), C) → U6(A, B, C, isNat_in(A))
isNat_in(0) → isNat_out(0)
isNat_in(s(X)) → U9(X, isNat_in(X))
U9(X, isNat_out(X)) → isNat_out(s(X))
U6(A, B, C, isNat_out(A)) → U7(A, B, C, isNat_in(B))
U7(A, B, C, isNat_out(B)) → U8(A, B, C, add_in(A, B, C))
add_in(0, X, X) → add_out(0, X, X)
add_in(s(X), Y, s(Z)) → U10(X, Y, Z, add_in(X, Y, Z))
U10(X, Y, Z, add_out(X, Y, Z)) → add_out(s(X), Y, s(Z))
U8(A, B, C, add_out(A, B, C)) → s2_out(plus(A, B), C)
s2_in(plus(X, Y), Z) → U3(X, Y, Z, s2_in(X, A))
s2_in(plus(X, 0), X) → s2_out(plus(X, 0), X)
s2_in(plus(A, B), C) → U2(A, B, C, s2_in(plus(B, A), C))
s2_in(plus(A, plus(B, C)), D) → U1(A, B, C, D, s2_in(plus(plus(A, B), C), D))
U1(A, B, C, D, s2_out(plus(plus(A, B), C), D)) → s2_out(plus(A, plus(B, C)), D)
U2(A, B, C, s2_out(plus(B, A), C)) → s2_out(plus(A, B), C)
U3(X, Y, Z, s2_out(X, A)) → U4(X, Y, Z, A, s2_in(Y, B))
U4(X, Y, Z, A, s2_out(Y, B)) → U5(X, Y, Z, s2_in(plus(A, B), Z))
U5(X, Y, Z, s2_out(plus(A, B), Z)) → s2_out(plus(X, Y), Z)

S2_IN(plus(A, B), C) → U6¹(A, B, C, isNat_in(A))
S2_IN(plus(A, B), C) → ISNAT_IN(A)
ISNAT_IN(s(X)) → U9¹(X, isNat_in(X))
ISNAT_IN(s(X)) → ISNAT_IN(X)
U6¹(A, B, C, isNat_out(A)) → U7¹(A, B, C, isNat_in(B))
U6¹(A, B, C, isNat_out(A)) → ISNAT_IN(B)
U7¹(A, B, C, isNat_out(B)) → U8¹(A, B, C, add_in(A, B, C))
U7¹(A, B, C, isNat_out(B)) → ADD_IN(A, B, C)
ADD_IN(s(X), Y, s(Z)) → U10¹(X, Y, Z, add_in(X, Y, Z))
ADD_IN(s(X), Y, s(Z)) → ADD_IN(X, Y, Z)
S2_IN(plus(X, Y), Z) → U3¹(X, Y, Z, s2_in(X, A))
S2_IN(plus(X, Y), Z) → S2_IN(X, A)
S2_IN(plus(A, B), C) → U2¹(A, B, C, s2_in(plus(B, A), C))
S2_IN(plus(A, B), C) → S2_IN(plus(B, A), C)
S2_IN(plus(A, plus(B, C)), D) → U1¹(A, B, C, D, s2_in(plus(plus(A, B), C), D))
S2_IN(plus(A, plus(B, C)), D) → S2_IN(plus(plus(A, B), C), D)
U3¹(X, Y, Z, s2_out(X, A)) → U4¹(X, Y, Z, A, s2_in(Y, B))
U3¹(X, Y, Z, s2_out(X, A)) → S2_IN(Y, B)
U4¹(X, Y, Z, A, s2_out(Y, B)) → U5¹(X, Y, Z, s2_in(plus(A, B), Z))
U4¹(X, Y, Z, A, s2_out(Y, B)) → S2_IN(plus(A, B), Z)

The TRS R consists of the following rules:

s2_in(plus(A, B), C) → U6(A, B, C, isNat_in(A))
isNat_in(0) → isNat_out(0)
isNat_in(s(X)) → U9(X, isNat_in(X))
U9(X, isNat_out(X)) → isNat_out(s(X))
U6(A, B, C, isNat_out(A)) → U7(A, B, C, isNat_in(B))
U7(A, B, C, isNat_out(B)) → U8(A, B, C, add_in(A, B, C))
add_in(0, X, X) → add_out(0, X, X)
add_in(s(X), Y, s(Z)) → U10(X, Y, Z, add_in(X, Y, Z))
U10(X, Y, Z, add_out(X, Y, Z)) → add_out(s(X), Y, s(Z))
U8(A, B, C, add_out(A, B, C)) → s2_out(plus(A, B), C)
s2_in(plus(X, Y), Z) → U3(X, Y, Z, s2_in(X, A))
s2_in(plus(X, 0), X) → s2_out(plus(X, 0), X)
s2_in(plus(A, B), C) → U2(A, B, C, s2_in(plus(B, A), C))
s2_in(plus(A, plus(B, C)), D) → U1(A, B, C, D, s2_in(plus(plus(A, B), C), D))
U1(A, B, C, D, s2_out(plus(plus(A, B), C), D)) → s2_out(plus(A, plus(B, C)), D)
U2(A, B, C, s2_out(plus(B, A), C)) → s2_out(plus(A, B), C)
U3(X, Y, Z, s2_out(X, A)) → U4(X, Y, Z, A, s2_in(Y, B))
U4(X, Y, Z, A, s2_out(Y, B)) → U5(X, Y, Z, s2_in(plus(A, B), Z))
U5(X, Y, Z, s2_out(plus(A, B), Z)) → s2_out(plus(X, Y), Z)

The argument filtering Pi contains the following mapping:
s2_in(x1, x2) = s2_in(x1)
plus(x1, x2) = plus(x1, x2)
U6(x1, x2, x3, x4) = U6(x1, x2, x4)
isNat_in(x1) = isNat_in(x1)
0 = 0
isNat_out(x1) = isNat_out(x1)
s(x1) = s(x1)
U9(x1, x2) = U9(x1, x2)
U7(x1, x2, x3, x4) = U7(x1, x2, x4)
U8(x1, x2, x3, x4) = U8(x1, x2, x4)
add_in(x1, x2, x3) = add_in(x1, x2)
add_out(x1, x2, x3) = add_out(x1, x2, x3)
U10(x1, x2, x3, x4) = U10(x1, x2, x4)
s2_out(x1, x2) = s2_out(x1, x2)
U3(x1, x2, x3, x4) = U3(x1, x2, x4)
U2(x1, x2, x3, x4) = U2(x1, x2, x4)
U1(x1, x2, x3, x4, x5) = U1(x1, x2, x3, x5)
U4(x1, x2, x3, x4, x5) = U4(x1, x2, x4, x5)
U5(x1, x2, x3, x4) = U5(x1, x2, x4)
ADD_IN(x1, x2, x3) = ADD_IN(x1, x2)
U10¹(x1, x2, x3, x4) = U10¹(x1, x2, x4)
U2¹(x1, x2, x3, x4) = U2¹(x1, x2, x4)
ISNAT_IN(x1) = ISNAT_IN(x1)
U1¹(x1, x2, x3, x4, x5) = U1¹(x1, x2, x3, x5)
U7¹(x1, x2, x3, x4) = U7¹(x1, x2, x4)
U5¹(x1, x2, x3, x4) = U5¹(x1, x2, x4)
U9¹(x1, x2) = U9¹(x1, x2)
U3¹(x1, x2, x3, x4) = U3¹(x1, x2, x4)
U4¹(x1, x2, x3, x4, x5) = U4¹(x1, x2, x4, x5)
U8¹(x1, x2, x3, x4) = U8¹(x1, x2, x4)
S2_IN(x1, x2) = S2_IN(x1)
U6¹(x1, x2, x3, x4) = U6¹(x1, x2, x4)

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:

S2_IN(plus(A, B), C) → U6¹(A, B, C, isNat_in(A))
S2_IN(plus(A, B), C) → ISNAT_IN(A)
ISNAT_IN(s(X)) → U9¹(X, isNat_in(X))
ISNAT_IN(s(X)) → ISNAT_IN(X)
U6¹(A, B, C, isNat_out(A)) → U7¹(A, B, C, isNat_in(B))
U6¹(A, B, C, isNat_out(A)) → ISNAT_IN(B)
U7¹(A, B, C, isNat_out(B)) → U8¹(A, B, C, add_in(A, B, C))
U7¹(A, B, C, isNat_out(B)) → ADD_IN(A, B, C)
ADD_IN(s(X), Y, s(Z)) → U10¹(X, Y, Z, add_in(X, Y, Z))
ADD_IN(s(X), Y, s(Z)) → ADD_IN(X, Y, Z)
S2_IN(plus(X, Y), Z) → U3¹(X, Y, Z, s2_in(X, A))
S2_IN(plus(X, Y), Z) → S2_IN(X, A)
S2_IN(plus(A, B), C) → U2¹(A, B, C, s2_in(plus(B, A), C))
S2_IN(plus(A, B), C) → S2_IN(plus(B, A), C)
S2_IN(plus(A, plus(B, C)), D) → U1¹(A, B, C, D, s2_in(plus(plus(A, B), C), D))
S2_IN(plus(A, plus(B, C)), D) → S2_IN(plus(plus(A, B), C), D)
U3¹(X, Y, Z, s2_out(X, A)) → U4¹(X, Y, Z, A, s2_in(Y, B))
U3¹(X, Y, Z, s2_out(X, A)) → S2_IN(Y, B)
U4¹(X, Y, Z, A, s2_out(Y, B)) → U5¹(X, Y, Z, s2_in(plus(A, B), Z))
U4¹(X, Y, Z, A, s2_out(Y, B)) → S2_IN(plus(A, B), Z)

The TRS R consists of the following rules:

s2_in(plus(A, B), C) → U6(A, B, C, isNat_in(A))
isNat_in(0) → isNat_out(0)
isNat_in(s(X)) → U9(X, isNat_in(X))
U9(X, isNat_out(X)) → isNat_out(s(X))
U6(A, B, C, isNat_out(A)) → U7(A, B, C, isNat_in(B))
U7(A, B, C, isNat_out(B)) → U8(A, B, C, add_in(A, B, C))
add_in(0, X, X) → add_out(0, X, X)
add_in(s(X), Y, s(Z)) → U10(X, Y, Z, add_in(X, Y, Z))
U10(X, Y, Z, add_out(X, Y, Z)) → add_out(s(X), Y, s(Z))
U8(A, B, C, add_out(A, B, C)) → s2_out(plus(A, B), C)
s2_in(plus(X, Y), Z) → U3(X, Y, Z, s2_in(X, A))
s2_in(plus(X, 0), X) → s2_out(plus(X, 0), X)
s2_in(plus(A, B), C) → U2(A, B, C, s2_in(plus(B, A), C))
s2_in(plus(A, plus(B, C)), D) → U1(A, B, C, D, s2_in(plus(plus(A, B), C), D))
U1(A, B, C, D, s2_out(plus(plus(A, B), C), D)) → s2_out(plus(A, plus(B, C)), D)
U2(A, B, C, s2_out(plus(B, A), C)) → s2_out(plus(A, B), C)
U3(X, Y, Z, s2_out(X, A)) → U4(X, Y, Z, A, s2_in(Y, B))
U4(X, Y, Z, A, s2_out(Y, B)) → U5(X, Y, Z, s2_in(plus(A, B), Z))
U5(X, Y, Z, s2_out(plus(A, B), Z)) → s2_out(plus(X, Y), Z)

The argument filtering Pi contains the following mapping:
s2_in(x1, x2) = s2_in(x1)
plus(x1, x2) = plus(x1, x2)
U6(x1, x2, x3, x4) = U6(x1, x2, x4)
isNat_in(x1) = isNat_in(x1)
0 = 0
isNat_out(x1) = isNat_out(x1)
s(x1) = s(x1)
U9(x1, x2) = U9(x1, x2)
U7(x1, x2, x3, x4) = U7(x1, x2, x4)
U8(x1, x2, x3, x4) = U8(x1, x2, x4)
add_in(x1, x2, x3) = add_in(x1, x2)
add_out(x1, x2, x3) = add_out(x1, x2, x3)
U10(x1, x2, x3, x4) = U10(x1, x2, x4)
s2_out(x1, x2) = s2_out(x1, x2)
U3(x1, x2, x3, x4) = U3(x1, x2, x4)
U2(x1, x2, x3, x4) = U2(x1, x2, x4)
U1(x1, x2, x3, x4, x5) = U1(x1, x2, x3, x5)
U4(x1, x2, x3, x4, x5) = U4(x1, x2, x4, x5)
U5(x1, x2, x3, x4) = U5(x1, x2, x4)
ADD_IN(x1, x2, x3) = ADD_IN(x1, x2)
U10¹(x1, x2, x3, x4) = U10¹(x1, x2, x4)
U2¹(x1, x2, x3, x4) = U2¹(x1, x2, x4)
ISNAT_IN(x1) = ISNAT_IN(x1)
U1¹(x1, x2, x3, x4, x5) = U1¹(x1, x2, x3, x5)
U7¹(x1, x2, x3, x4) = U7¹(x1, x2, x4)
U5¹(x1, x2, x3, x4) = U5¹(x1, x2, x4)
U9¹(x1, x2) = U9¹(x1, x2)
U3¹(x1, x2, x3, x4) = U3¹(x1, x2, x4)
U4¹(x1, x2, x3, x4, x5) = U4¹(x1, x2, x4, x5)
U8¹(x1, x2, x3, x4) = U8¹(x1, x2, x4)
S2_IN(x1, x2) = S2_IN(x1)
U6¹(x1, x2, x3, x4) = U6¹(x1, x2, x4)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph [30] contains 3 SCCs with 11 less nodes.

↳ Prolog

  ↳ PrologToPiTRSProof

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

                ↳ UsableRulesProof

              ↳ PiDP

              ↳ PiDP

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

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

The TRS R consists of the following rules:

s2_in(plus(A, B), C) → U6(A, B, C, isNat_in(A))
isNat_in(0) → isNat_out(0)
isNat_in(s(X)) → U9(X, isNat_in(X))
U9(X, isNat_out(X)) → isNat_out(s(X))
U6(A, B, C, isNat_out(A)) → U7(A, B, C, isNat_in(B))
U7(A, B, C, isNat_out(B)) → U8(A, B, C, add_in(A, B, C))
add_in(0, X, X) → add_out(0, X, X)
add_in(s(X), Y, s(Z)) → U10(X, Y, Z, add_in(X, Y, Z))
U10(X, Y, Z, add_out(X, Y, Z)) → add_out(s(X), Y, s(Z))
U8(A, B, C, add_out(A, B, C)) → s2_out(plus(A, B), C)
s2_in(plus(X, Y), Z) → U3(X, Y, Z, s2_in(X, A))
s2_in(plus(X, 0), X) → s2_out(plus(X, 0), X)
s2_in(plus(A, B), C) → U2(A, B, C, s2_in(plus(B, A), C))
s2_in(plus(A, plus(B, C)), D) → U1(A, B, C, D, s2_in(plus(plus(A, B), C), D))
U1(A, B, C, D, s2_out(plus(plus(A, B), C), D)) → s2_out(plus(A, plus(B, C)), D)
U2(A, B, C, s2_out(plus(B, A), C)) → s2_out(plus(A, B), C)
U3(X, Y, Z, s2_out(X, A)) → U4(X, Y, Z, A, s2_in(Y, B))
U4(X, Y, Z, A, s2_out(Y, B)) → U5(X, Y, Z, s2_in(plus(A, B), Z))
U5(X, Y, Z, s2_out(plus(A, B), Z)) → s2_out(plus(X, Y), Z)

↳ Prolog

  ↳ PrologToPiTRSProof

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

              ↳ PiDP

              ↳ PiDP

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

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

↳ Prolog

  ↳ PrologToPiTRSProof

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ QDPSizeChangeProof

              ↳ PiDP

              ↳ PiDP

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

ADD_IN(s(X), Y) → ADD_IN(X, Y)

From the DPs we obtained the following set of size-change graphs:

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

↳ Prolog

  ↳ PrologToPiTRSProof

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

              ↳ PiDP

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

ISNAT_IN(s(X)) → ISNAT_IN(X)

The TRS R consists of the following rules:

s2_in(plus(A, B), C) → U6(A, B, C, isNat_in(A))
isNat_in(0) → isNat_out(0)
isNat_in(s(X)) → U9(X, isNat_in(X))
U9(X, isNat_out(X)) → isNat_out(s(X))
U6(A, B, C, isNat_out(A)) → U7(A, B, C, isNat_in(B))
U7(A, B, C, isNat_out(B)) → U8(A, B, C, add_in(A, B, C))
add_in(0, X, X) → add_out(0, X, X)
add_in(s(X), Y, s(Z)) → U10(X, Y, Z, add_in(X, Y, Z))
U10(X, Y, Z, add_out(X, Y, Z)) → add_out(s(X), Y, s(Z))
U8(A, B, C, add_out(A, B, C)) → s2_out(plus(A, B), C)
s2_in(plus(X, Y), Z) → U3(X, Y, Z, s2_in(X, A))
s2_in(plus(X, 0), X) → s2_out(plus(X, 0), X)
s2_in(plus(A, B), C) → U2(A, B, C, s2_in(plus(B, A), C))
s2_in(plus(A, plus(B, C)), D) → U1(A, B, C, D, s2_in(plus(plus(A, B), C), D))
U1(A, B, C, D, s2_out(plus(plus(A, B), C), D)) → s2_out(plus(A, plus(B, C)), D)
U2(A, B, C, s2_out(plus(B, A), C)) → s2_out(plus(A, B), C)
U3(X, Y, Z, s2_out(X, A)) → U4(X, Y, Z, A, s2_in(Y, B))
U4(X, Y, Z, A, s2_out(Y, B)) → U5(X, Y, Z, s2_in(plus(A, B), Z))
U5(X, Y, Z, s2_out(plus(A, B), Z)) → s2_out(plus(X, Y), Z)

↳ Prolog

  ↳ PrologToPiTRSProof

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

              ↳ PiDP

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

ISNAT_IN(s(X)) → ISNAT_IN(X)

R is empty.
Pi is empty.
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

                        ↳ QDPSizeChangeProof

              ↳ PiDP

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

ISNAT_IN(s(X)) → ISNAT_IN(X)

From the DPs we obtained the following set of size-change graphs:

ISNAT_IN(s(X)) → ISNAT_IN(X)
The graph contains the following edges 1 > 1

↳ Prolog

  ↳ PrologToPiTRSProof

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

                ↳ PiDPToQDPProof

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

U3¹(X, Y, Z, s2_out(X, A)) → S2_IN(Y, B)
S2_IN(plus(X, Y), Z) → U3¹(X, Y, Z, s2_in(X, A))
U4¹(X, Y, Z, A, s2_out(Y, B)) → S2_IN(plus(A, B), Z)
S2_IN(plus(A, B), C) → S2_IN(plus(B, A), C)
U3¹(X, Y, Z, s2_out(X, A)) → U4¹(X, Y, Z, A, s2_in(Y, B))
S2_IN(plus(A, plus(B, C)), D) → S2_IN(plus(plus(A, B), C), D)
S2_IN(plus(X, Y), Z) → S2_IN(X, A)

The TRS R consists of the following rules:

s2_in(plus(A, B), C) → U6(A, B, C, isNat_in(A))
isNat_in(0) → isNat_out(0)
isNat_in(s(X)) → U9(X, isNat_in(X))
U9(X, isNat_out(X)) → isNat_out(s(X))
U6(A, B, C, isNat_out(A)) → U7(A, B, C, isNat_in(B))
U7(A, B, C, isNat_out(B)) → U8(A, B, C, add_in(A, B, C))
add_in(0, X, X) → add_out(0, X, X)
add_in(s(X), Y, s(Z)) → U10(X, Y, Z, add_in(X, Y, Z))
U10(X, Y, Z, add_out(X, Y, Z)) → add_out(s(X), Y, s(Z))
U8(A, B, C, add_out(A, B, C)) → s2_out(plus(A, B), C)
s2_in(plus(X, Y), Z) → U3(X, Y, Z, s2_in(X, A))
s2_in(plus(X, 0), X) → s2_out(plus(X, 0), X)
s2_in(plus(A, B), C) → U2(A, B, C, s2_in(plus(B, A), C))
s2_in(plus(A, plus(B, C)), D) → U1(A, B, C, D, s2_in(plus(plus(A, B), C), D))
U1(A, B, C, D, s2_out(plus(plus(A, B), C), D)) → s2_out(plus(A, plus(B, C)), D)
U2(A, B, C, s2_out(plus(B, A), C)) → s2_out(plus(A, B), C)
U3(X, Y, Z, s2_out(X, A)) → U4(X, Y, Z, A, s2_in(Y, B))
U4(X, Y, Z, A, s2_out(Y, B)) → U5(X, Y, Z, s2_in(plus(A, B), Z))
U5(X, Y, Z, s2_out(plus(A, B), Z)) → s2_out(plus(X, Y), Z)

The argument filtering Pi contains the following mapping:
s2_in(x1, x2) = s2_in(x1)
plus(x1, x2) = plus(x1, x2)
U6(x1, x2, x3, x4) = U6(x1, x2, x4)
isNat_in(x1) = isNat_in(x1)
0 = 0
isNat_out(x1) = isNat_out(x1)
s(x1) = s(x1)
U9(x1, x2) = U9(x1, x2)
U7(x1, x2, x3, x4) = U7(x1, x2, x4)
U8(x1, x2, x3, x4) = U8(x1, x2, x4)
add_in(x1, x2, x3) = add_in(x1, x2)
add_out(x1, x2, x3) = add_out(x1, x2, x3)
U10(x1, x2, x3, x4) = U10(x1, x2, x4)
s2_out(x1, x2) = s2_out(x1, x2)
U3(x1, x2, x3, x4) = U3(x1, x2, x4)
U2(x1, x2, x3, x4) = U2(x1, x2, x4)
U1(x1, x2, x3, x4, x5) = U1(x1, x2, x3, x5)
U4(x1, x2, x3, x4, x5) = U4(x1, x2, x4, x5)
U5(x1, x2, x3, x4) = U5(x1, x2, x4)
U3¹(x1, x2, x3, x4) = U3¹(x1, x2, x4)
U4¹(x1, x2, x3, x4, x5) = U4¹(x1, x2, x4, x5)
S2_IN(x1, x2) = S2_IN(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

              ↳ PiDP

                ↳ PiDPToQDPProof

                  ↳ QDP

                    ↳ QDPOrderProof

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

U4¹(X, Y, A, s2_out(Y, B)) → S2_IN(plus(A, B))
S2_IN(plus(X, Y)) → U3¹(X, Y, s2_in(X))
U3¹(X, Y, s2_out(X, A)) → U4¹(X, Y, A, s2_in(Y))
U3¹(X, Y, s2_out(X, A)) → S2_IN(Y)
S2_IN(plus(X, Y)) → S2_IN(X)
S2_IN(plus(A, plus(B, C))) → S2_IN(plus(plus(A, B), C))
S2_IN(plus(A, B)) → S2_IN(plus(B, A))

The TRS R consists of the following rules:

s2_in(plus(A, B)) → U6(A, B, isNat_in(A))
isNat_in(0) → isNat_out(0)
isNat_in(s(X)) → U9(X, isNat_in(X))
U9(X, isNat_out(X)) → isNat_out(s(X))
U6(A, B, isNat_out(A)) → U7(A, B, isNat_in(B))
U7(A, B, isNat_out(B)) → U8(A, B, add_in(A, B))
add_in(0, X) → add_out(0, X, X)
add_in(s(X), Y) → U10(X, Y, add_in(X, Y))
U10(X, Y, add_out(X, Y, Z)) → add_out(s(X), Y, s(Z))
U8(A, B, add_out(A, B, C)) → s2_out(plus(A, B), C)
s2_in(plus(X, Y)) → U3(X, Y, s2_in(X))
s2_in(plus(X, 0)) → s2_out(plus(X, 0), X)
s2_in(plus(A, B)) → U2(A, B, s2_in(plus(B, A)))
s2_in(plus(A, plus(B, C))) → U1(A, B, C, s2_in(plus(plus(A, B), C)))
U1(A, B, C, s2_out(plus(plus(A, B), C), D)) → s2_out(plus(A, plus(B, C)), D)
U2(A, B, s2_out(plus(B, A), C)) → s2_out(plus(A, B), C)
U3(X, Y, s2_out(X, A)) → U4(X, Y, A, s2_in(Y))
U4(X, Y, A, s2_out(Y, B)) → U5(X, Y, s2_in(plus(A, B)))
U5(X, Y, s2_out(plus(A, B), Z)) → s2_out(plus(X, Y), Z)

The set Q consists of the following terms:

s2_in(x0)
isNat_in(x0)
U9(x0, x1)
U6(x0, x1, x2)
U7(x0, x1, x2)
add_in(x0, x1)
U10(x0, x1, x2)
U8(x0, x1, x2)
U1(x0, x1, x2, x3)
U2(x0, x1, x2)
U3(x0, x1, x2)
U4(x0, x1, x2, x3)
U5(x0, x1, x2)

We have to consider all (P,Q,R)-chains.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

U3¹(X, Y, s2_out(X, A)) → S2_IN(Y)
S2_IN(plus(X, Y)) → S2_IN(X)

The remaining pairs can at least be oriented weakly.

U4¹(X, Y, A, s2_out(Y, B)) → S2_IN(plus(A, B))
S2_IN(plus(X, Y)) → U3¹(X, Y, s2_in(X))
U3¹(X, Y, s2_out(X, A)) → U4¹(X, Y, A, s2_in(Y))
S2_IN(plus(A, plus(B, C))) → S2_IN(plus(plus(A, B), C))
S2_IN(plus(A, B)) → S2_IN(plus(B, A))

Used ordering: Polynomial interpretation [25]:

POL(0) = 0
POL(S2_IN(x₁)) = x₁
POL(U1(x₁, x₂, x₃, x₄)) = x₄
POL(U10(x₁, x₂, x₃)) = 0
POL(U2(x₁, x₂, x₃)) = x₃
POL(U3(x₁, x₂, x₃)) = 1 + x₂ + x₃
POL(U3¹(x₁, x₂, x₃)) = 1 + x₂ + x₃
POL(U4(x₁, x₂, x₃, x₄)) = 1 + x₃ + x₄
POL(U4¹(x₁, x₂, x₃, x₄)) = 1 + x₃ + x₄
POL(U5(x₁, x₂, x₃)) = x₃
POL(U6(x₁, x₂, x₃)) = x₂
POL(U7(x₁, x₂, x₃)) = x₂
POL(U8(x₁, x₂, x₃)) = x₃
POL(U9(x₁, x₂)) = 0
POL(add_in(x₁, x₂)) = x₂
POL(add_out(x₁, x₂, x₃)) = x₃
POL(isNat_in(x₁)) = 0
POL(isNat_out(x₁)) = 0
POL(plus(x₁, x₂)) = 1 + x₁ + x₂
POL(s(x₁)) = 0
POL(s2_in(x₁)) = x₁
POL(s2_out(x₁, x₂)) = x₂

The following usable rules [17] were oriented:

U8(A, B, add_out(A, B, C)) → s2_out(plus(A, B), C)
s2_in(plus(A, plus(B, C))) → U1(A, B, C, s2_in(plus(plus(A, B), C)))
U4(X, Y, A, s2_out(Y, B)) → U5(X, Y, s2_in(plus(A, B)))
U7(A, B, isNat_out(B)) → U8(A, B, add_in(A, B))
add_in(0, X) → add_out(0, X, X)
s2_in(plus(A, B)) → U2(A, B, s2_in(plus(B, A)))
U3(X, Y, s2_out(X, A)) → U4(X, Y, A, s2_in(Y))
s2_in(plus(X, 0)) → s2_out(plus(X, 0), X)
add_in(s(X), Y) → U10(X, Y, add_in(X, Y))
U5(X, Y, s2_out(plus(A, B), Z)) → s2_out(plus(X, Y), Z)
s2_in(plus(A, B)) → U6(A, B, isNat_in(A))
U2(A, B, s2_out(plus(B, A), C)) → s2_out(plus(A, B), C)
s2_in(plus(X, Y)) → U3(X, Y, s2_in(X))
U10(X, Y, add_out(X, Y, Z)) → add_out(s(X), Y, s(Z))
U1(A, B, C, s2_out(plus(plus(A, B), C), D)) → s2_out(plus(A, plus(B, C)), D)
U6(A, B, isNat_out(A)) → U7(A, B, isNat_in(B))

↳ Prolog

  ↳ PrologToPiTRSProof

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

                ↳ PiDPToQDPProof

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ QDPOrderProof

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

U4¹(X, Y, A, s2_out(Y, B)) → S2_IN(plus(A, B))
S2_IN(plus(X, Y)) → U3¹(X, Y, s2_in(X))
U3¹(X, Y, s2_out(X, A)) → U4¹(X, Y, A, s2_in(Y))
S2_IN(plus(A, plus(B, C))) → S2_IN(plus(plus(A, B), C))
S2_IN(plus(A, B)) → S2_IN(plus(B, A))

The TRS R consists of the following rules:

s2_in(plus(A, B)) → U6(A, B, isNat_in(A))
isNat_in(0) → isNat_out(0)
isNat_in(s(X)) → U9(X, isNat_in(X))
U9(X, isNat_out(X)) → isNat_out(s(X))
U6(A, B, isNat_out(A)) → U7(A, B, isNat_in(B))
U7(A, B, isNat_out(B)) → U8(A, B, add_in(A, B))
add_in(0, X) → add_out(0, X, X)
add_in(s(X), Y) → U10(X, Y, add_in(X, Y))
U10(X, Y, add_out(X, Y, Z)) → add_out(s(X), Y, s(Z))
U8(A, B, add_out(A, B, C)) → s2_out(plus(A, B), C)
s2_in(plus(X, Y)) → U3(X, Y, s2_in(X))
s2_in(plus(X, 0)) → s2_out(plus(X, 0), X)
s2_in(plus(A, B)) → U2(A, B, s2_in(plus(B, A)))
s2_in(plus(A, plus(B, C))) → U1(A, B, C, s2_in(plus(plus(A, B), C)))
U1(A, B, C, s2_out(plus(plus(A, B), C), D)) → s2_out(plus(A, plus(B, C)), D)
U2(A, B, s2_out(plus(B, A), C)) → s2_out(plus(A, B), C)
U3(X, Y, s2_out(X, A)) → U4(X, Y, A, s2_in(Y))
U4(X, Y, A, s2_out(Y, B)) → U5(X, Y, s2_in(plus(A, B)))
U5(X, Y, s2_out(plus(A, B), Z)) → s2_out(plus(X, Y), Z)

The set Q consists of the following terms:

s2_in(x0)
isNat_in(x0)
U9(x0, x1)
U6(x0, x1, x2)
U7(x0, x1, x2)
add_in(x0, x1)
U10(x0, x1, x2)
U8(x0, x1, x2)
U1(x0, x1, x2, x3)
U2(x0, x1, x2)
U3(x0, x1, x2)
U4(x0, x1, x2, x3)
U5(x0, x1, x2)

We have to consider all (P,Q,R)-chains.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

U4¹(X, Y, A, s2_out(Y, B)) → S2_IN(plus(A, B))
S2_IN(plus(X, Y)) → U3¹(X, Y, s2_in(X))

The remaining pairs can at least be oriented weakly.

U3¹(X, Y, s2_out(X, A)) → U4¹(X, Y, A, s2_in(Y))
S2_IN(plus(A, plus(B, C))) → S2_IN(plus(plus(A, B), C))
S2_IN(plus(A, B)) → S2_IN(plus(B, A))

Used ordering: Polynomial interpretation [25]:

POL(0) = 1
POL(S2_IN(x₁)) = x₁
POL(U1(x₁, x₂, x₃, x₄)) = x₄
POL(U10(x₁, x₂, x₃)) = 0
POL(U2(x₁, x₂, x₃)) = x₃
POL(U3(x₁, x₂, x₃)) = 1 + x₂ + x₃
POL(U3¹(x₁, x₂, x₃)) = x₂ + x₃
POL(U4(x₁, x₂, x₃, x₄)) = 1 + x₃ + x₄
POL(U4¹(x₁, x₂, x₃, x₄)) = 1 + x₃ + x₄
POL(U5(x₁, x₂, x₃)) = x₃
POL(U6(x₁, x₂, x₃)) = 1 + x₁ + x₂
POL(U7(x₁, x₂, x₃)) = 1 + x₁ + x₂
POL(U8(x₁, x₂, x₃)) = 1 + x₁ + x₃
POL(U9(x₁, x₂)) = 0
POL(add_in(x₁, x₂)) = x₂
POL(add_out(x₁, x₂, x₃)) = x₃
POL(isNat_in(x₁)) = 0
POL(isNat_out(x₁)) = 0
POL(plus(x₁, x₂)) = 1 + x₁ + x₂
POL(s(x₁)) = 0
POL(s2_in(x₁)) = x₁
POL(s2_out(x₁, x₂)) = 1 + x₂

The following usable rules [17] were oriented:

U8(A, B, add_out(A, B, C)) → s2_out(plus(A, B), C)
s2_in(plus(A, plus(B, C))) → U1(A, B, C, s2_in(plus(plus(A, B), C)))
U4(X, Y, A, s2_out(Y, B)) → U5(X, Y, s2_in(plus(A, B)))
U7(A, B, isNat_out(B)) → U8(A, B, add_in(A, B))
add_in(0, X) → add_out(0, X, X)
s2_in(plus(A, B)) → U2(A, B, s2_in(plus(B, A)))
U3(X, Y, s2_out(X, A)) → U4(X, Y, A, s2_in(Y))
s2_in(plus(X, 0)) → s2_out(plus(X, 0), X)
add_in(s(X), Y) → U10(X, Y, add_in(X, Y))
U5(X, Y, s2_out(plus(A, B), Z)) → s2_out(plus(X, Y), Z)
s2_in(plus(A, B)) → U6(A, B, isNat_in(A))
s2_in(plus(X, Y)) → U3(X, Y, s2_in(X))
U2(A, B, s2_out(plus(B, A), C)) → s2_out(plus(A, B), C)
U10(X, Y, add_out(X, Y, Z)) → add_out(s(X), Y, s(Z))
U1(A, B, C, s2_out(plus(plus(A, B), C), D)) → s2_out(plus(A, plus(B, C)), D)
U6(A, B, isNat_out(A)) → U7(A, B, isNat_in(B))

↳ Prolog

  ↳ PrologToPiTRSProof

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

                ↳ PiDPToQDPProof

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ QDPOrderProof

                          ↳ QDP

                            ↳ DependencyGraphProof

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

U3¹(X, Y, s2_out(X, A)) → U4¹(X, Y, A, s2_in(Y))
S2_IN(plus(A, plus(B, C))) → S2_IN(plus(plus(A, B), C))
S2_IN(plus(A, B)) → S2_IN(plus(B, A))

The TRS R consists of the following rules:

s2_in(plus(A, B)) → U6(A, B, isNat_in(A))
isNat_in(0) → isNat_out(0)
isNat_in(s(X)) → U9(X, isNat_in(X))
U9(X, isNat_out(X)) → isNat_out(s(X))
U6(A, B, isNat_out(A)) → U7(A, B, isNat_in(B))
U7(A, B, isNat_out(B)) → U8(A, B, add_in(A, B))
add_in(0, X) → add_out(0, X, X)
add_in(s(X), Y) → U10(X, Y, add_in(X, Y))
U10(X, Y, add_out(X, Y, Z)) → add_out(s(X), Y, s(Z))
U8(A, B, add_out(A, B, C)) → s2_out(plus(A, B), C)
s2_in(plus(X, Y)) → U3(X, Y, s2_in(X))
s2_in(plus(X, 0)) → s2_out(plus(X, 0), X)
s2_in(plus(A, B)) → U2(A, B, s2_in(plus(B, A)))
s2_in(plus(A, plus(B, C))) → U1(A, B, C, s2_in(plus(plus(A, B), C)))
U1(A, B, C, s2_out(plus(plus(A, B), C), D)) → s2_out(plus(A, plus(B, C)), D)
U2(A, B, s2_out(plus(B, A), C)) → s2_out(plus(A, B), C)
U3(X, Y, s2_out(X, A)) → U4(X, Y, A, s2_in(Y))
U4(X, Y, A, s2_out(Y, B)) → U5(X, Y, s2_in(plus(A, B)))
U5(X, Y, s2_out(plus(A, B), Z)) → s2_out(plus(X, Y), Z)

The set Q consists of the following terms:

s2_in(x0)
isNat_in(x0)
U9(x0, x1)
U6(x0, x1, x2)
U7(x0, x1, x2)
add_in(x0, x1)
U10(x0, x1, x2)
U8(x0, x1, x2)
U1(x0, x1, x2, x3)
U2(x0, x1, x2)
U3(x0, x1, x2)
U4(x0, x1, x2, x3)
U5(x0, x1, x2)

We have to consider all (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 1 less node.

↳ Prolog

  ↳ PrologToPiTRSProof

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

                ↳ PiDPToQDPProof

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ QDPOrderProof

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ UsableRulesProof

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

S2_IN(plus(A, plus(B, C))) → S2_IN(plus(plus(A, B), C))
S2_IN(plus(A, B)) → S2_IN(plus(B, A))

The TRS R consists of the following rules:

s2_in(plus(A, B)) → U6(A, B, isNat_in(A))
isNat_in(0) → isNat_out(0)
isNat_in(s(X)) → U9(X, isNat_in(X))
U9(X, isNat_out(X)) → isNat_out(s(X))
U6(A, B, isNat_out(A)) → U7(A, B, isNat_in(B))
U7(A, B, isNat_out(B)) → U8(A, B, add_in(A, B))
add_in(0, X) → add_out(0, X, X)
add_in(s(X), Y) → U10(X, Y, add_in(X, Y))
U10(X, Y, add_out(X, Y, Z)) → add_out(s(X), Y, s(Z))
U8(A, B, add_out(A, B, C)) → s2_out(plus(A, B), C)
s2_in(plus(X, Y)) → U3(X, Y, s2_in(X))
s2_in(plus(X, 0)) → s2_out(plus(X, 0), X)
s2_in(plus(A, B)) → U2(A, B, s2_in(plus(B, A)))
s2_in(plus(A, plus(B, C))) → U1(A, B, C, s2_in(plus(plus(A, B), C)))
U1(A, B, C, s2_out(plus(plus(A, B), C), D)) → s2_out(plus(A, plus(B, C)), D)
U2(A, B, s2_out(plus(B, A), C)) → s2_out(plus(A, B), C)
U3(X, Y, s2_out(X, A)) → U4(X, Y, A, s2_in(Y))
U4(X, Y, A, s2_out(Y, B)) → U5(X, Y, s2_in(plus(A, B)))
U5(X, Y, s2_out(plus(A, B), Z)) → s2_out(plus(X, Y), Z)

The set Q consists of the following terms:

s2_in(x0)
isNat_in(x0)
U9(x0, x1)
U6(x0, x1, x2)
U7(x0, x1, x2)
add_in(x0, x1)
U10(x0, x1, x2)
U8(x0, x1, x2)
U1(x0, x1, x2, x3)
U2(x0, x1, x2)
U3(x0, x1, x2)
U4(x0, x1, x2, x3)
U5(x0, x1, x2)

↳ Prolog

  ↳ PrologToPiTRSProof

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

                ↳ PiDPToQDPProof

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ QDPOrderProof

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ UsableRulesProof

                                  ↳ QDP

                                    ↳ QReductionProof

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

S2_IN(plus(A, plus(B, C))) → S2_IN(plus(plus(A, B), C))
S2_IN(plus(A, B)) → S2_IN(plus(B, A))

R is empty.
The set Q consists of the following terms:

s2_in(x0)
isNat_in(x0)
U9(x0, x1)
U6(x0, x1, x2)
U7(x0, x1, x2)
add_in(x0, x1)
U10(x0, x1, x2)
U8(x0, x1, x2)
U1(x0, x1, x2, x3)
U2(x0, x1, x2)
U3(x0, x1, x2)
U4(x0, x1, x2, x3)
U5(x0, x1, x2)

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.

s2_in(x0)
isNat_in(x0)
U9(x0, x1)
U6(x0, x1, x2)
U7(x0, x1, x2)
add_in(x0, x1)
U10(x0, x1, x2)
U8(x0, x1, x2)
U1(x0, x1, x2, x3)
U2(x0, x1, x2)
U3(x0, x1, x2)
U4(x0, x1, x2, x3)
U5(x0, x1, x2)

↳ Prolog

  ↳ PrologToPiTRSProof

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

                ↳ PiDPToQDPProof

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ QDPOrderProof

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ UsableRulesProof

                                  ↳ QDP

                                    ↳ QReductionProof

                                      ↳ QDP

                                        ↳ NonTerminationProof

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

S2_IN(plus(A, plus(B, C))) → S2_IN(plus(plus(A, B), C))
S2_IN(plus(A, B)) → S2_IN(plus(B, A))

The TRS R consists of the following rules:none

s = S2_IN(plus(A, B)) evaluates to t =S2_IN(plus(B, A))

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

Matcher: [B / A, A / B]
Semiunifier: [ ]

Rewriting sequence

The DP semiunifies directly so there is only one rewrite step from S2_IN(plus(A, B)) to S2_IN(plus(B, A)).