Left Termination proof of /tmp/tmpB-TMEE/der.pl

Left Termination of the query pattern der_in_2(g, a) w.r.t. the given Prolog program could successfully be proven:

↳ Prolog

  ↳ PrologToPiTRSProof

  ↳ PrologToPiTRSProof

Clauses:

der(d(e(t)), const(1)).
der(d(e(const(A))), const(0)).
der(d(e(+(X, Y))), +(DX, DY)) :- ','(der(d(e(X)), DX), der(d(e(Y)), DY)).
der(d(e(*(X, Y))), +(*(X, DY), *(Y, DX))) :- ','(der(d(e(X)), DX), der(d(e(Y)), DY)).
der(d(d(X)), DDX) :- ','(der(d(X), DX), der(d(e(DX)), DDX)).

Queries:

der(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:

der_in(d(d(X)), DDX) → U5(X, DDX, der_in(d(X), DX))
der_in(d(e(*(X, Y))), +(*(X, DY), *(Y, DX))) → U3(X, Y, DY, DX, der_in(d(e(X)), DX))
der_in(d(e(+(X, Y))), +(DX, DY)) → U1(X, Y, DX, DY, der_in(d(e(X)), DX))
der_in(d(e(const(A))), const(0)) → der_out(d(e(const(A))), const(0))
der_in(d(e(t)), const(1)) → der_out(d(e(t)), const(1))
U1(X, Y, DX, DY, der_out(d(e(X)), DX)) → U2(X, Y, DX, DY, der_in(d(e(Y)), DY))
U2(X, Y, DX, DY, der_out(d(e(Y)), DY)) → der_out(d(e(+(X, Y))), +(DX, DY))
U3(X, Y, DY, DX, der_out(d(e(X)), DX)) → U4(X, Y, DY, DX, der_in(d(e(Y)), DY))
U4(X, Y, DY, DX, der_out(d(e(Y)), DY)) → der_out(d(e(*(X, Y))), +(*(X, DY), *(Y, DX)))
U5(X, DDX, der_out(d(X), DX)) → U6(X, DDX, DX, der_in(d(e(DX)), DDX))
U6(X, DDX, DX, der_out(d(e(DX)), DDX)) → der_out(d(d(X)), DDX)

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:

der_in(d(d(X)), DDX) → U5(X, DDX, der_in(d(X), DX))
der_in(d(e(*(X, Y))), +(*(X, DY), *(Y, DX))) → U3(X, Y, DY, DX, der_in(d(e(X)), DX))
der_in(d(e(+(X, Y))), +(DX, DY)) → U1(X, Y, DX, DY, der_in(d(e(X)), DX))
der_in(d(e(const(A))), const(0)) → der_out(d(e(const(A))), const(0))
der_in(d(e(t)), const(1)) → der_out(d(e(t)), const(1))
U1(X, Y, DX, DY, der_out(d(e(X)), DX)) → U2(X, Y, DX, DY, der_in(d(e(Y)), DY))
U2(X, Y, DX, DY, der_out(d(e(Y)), DY)) → der_out(d(e(+(X, Y))), +(DX, DY))
U3(X, Y, DY, DX, der_out(d(e(X)), DX)) → U4(X, Y, DY, DX, der_in(d(e(Y)), DY))
U4(X, Y, DY, DX, der_out(d(e(Y)), DY)) → der_out(d(e(*(X, Y))), +(*(X, DY), *(Y, DX)))
U5(X, DDX, der_out(d(X), DX)) → U6(X, DDX, DX, der_in(d(e(DX)), DDX))
U6(X, DDX, DX, der_out(d(e(DX)), DDX)) → der_out(d(d(X)), DDX)

DER_IN(d(d(X)), DDX) → U5¹(X, DDX, der_in(d(X), DX))
DER_IN(d(d(X)), DDX) → DER_IN(d(X), DX)
DER_IN(d(e(*(X, Y))), +(*(X, DY), *(Y, DX))) → U3¹(X, Y, DY, DX, der_in(d(e(X)), DX))
DER_IN(d(e(*(X, Y))), +(*(X, DY), *(Y, DX))) → DER_IN(d(e(X)), DX)
DER_IN(d(e(+(X, Y))), +(DX, DY)) → U1¹(X, Y, DX, DY, der_in(d(e(X)), DX))
DER_IN(d(e(+(X, Y))), +(DX, DY)) → DER_IN(d(e(X)), DX)
U1¹(X, Y, DX, DY, der_out(d(e(X)), DX)) → U2¹(X, Y, DX, DY, der_in(d(e(Y)), DY))
U1¹(X, Y, DX, DY, der_out(d(e(X)), DX)) → DER_IN(d(e(Y)), DY)
U3¹(X, Y, DY, DX, der_out(d(e(X)), DX)) → U4¹(X, Y, DY, DX, der_in(d(e(Y)), DY))
U3¹(X, Y, DY, DX, der_out(d(e(X)), DX)) → DER_IN(d(e(Y)), DY)
U5¹(X, DDX, der_out(d(X), DX)) → U6¹(X, DDX, DX, der_in(d(e(DX)), DDX))
U5¹(X, DDX, der_out(d(X), DX)) → DER_IN(d(e(DX)), DDX)

The TRS R consists of the following rules:

der_in(d(d(X)), DDX) → U5(X, DDX, der_in(d(X), DX))
der_in(d(e(*(X, Y))), +(*(X, DY), *(Y, DX))) → U3(X, Y, DY, DX, der_in(d(e(X)), DX))
der_in(d(e(+(X, Y))), +(DX, DY)) → U1(X, Y, DX, DY, der_in(d(e(X)), DX))
der_in(d(e(const(A))), const(0)) → der_out(d(e(const(A))), const(0))
der_in(d(e(t)), const(1)) → der_out(d(e(t)), const(1))
U1(X, Y, DX, DY, der_out(d(e(X)), DX)) → U2(X, Y, DX, DY, der_in(d(e(Y)), DY))
U2(X, Y, DX, DY, der_out(d(e(Y)), DY)) → der_out(d(e(+(X, Y))), +(DX, DY))
U3(X, Y, DY, DX, der_out(d(e(X)), DX)) → U4(X, Y, DY, DX, der_in(d(e(Y)), DY))
U4(X, Y, DY, DX, der_out(d(e(Y)), DY)) → der_out(d(e(*(X, Y))), +(*(X, DY), *(Y, DX)))
U5(X, DDX, der_out(d(X), DX)) → U6(X, DDX, DX, der_in(d(e(DX)), DDX))
U6(X, DDX, DX, der_out(d(e(DX)), DDX)) → der_out(d(d(X)), DDX)

The argument filtering Pi contains the following mapping:
der_in(x1, x2) = der_in(x1)
d(x1) = d(x1)
U5(x1, x2, x3) = U5(x3)
e(x1) = e(x1)
*(x1, x2) = *(x1, x2)
+(x1, x2) = +(x1, x2)
U3(x1, x2, x3, x4, x5) = U3(x1, x2, x5)
U1(x1, x2, x3, x4, x5) = U1(x2, x5)
const(x1) = const(x1)
0 = 0
der_out(x1, x2) = der_out(x2)
t = t
1 = 1
U2(x1, x2, x3, x4, x5) = U2(x3, x5)
U4(x1, x2, x3, x4, x5) = U4(x1, x2, x4, x5)
U6(x1, x2, x3, x4) = U6(x4)
U5¹(x1, x2, x3) = U5¹(x3)
U4¹(x1, x2, x3, x4, x5) = U4¹(x1, x2, x4, x5)
U3¹(x1, x2, x3, x4, x5) = U3¹(x1, x2, x5)
U2¹(x1, x2, x3, x4, x5) = U2¹(x3, x5)
U1¹(x1, x2, x3, x4, x5) = U1¹(x2, x5)
U6¹(x1, x2, x3, x4) = U6¹(x4)
DER_IN(x1, x2) = DER_IN(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:

DER_IN(d(d(X)), DDX) → U5¹(X, DDX, der_in(d(X), DX))
DER_IN(d(d(X)), DDX) → DER_IN(d(X), DX)
DER_IN(d(e(*(X, Y))), +(*(X, DY), *(Y, DX))) → U3¹(X, Y, DY, DX, der_in(d(e(X)), DX))
DER_IN(d(e(*(X, Y))), +(*(X, DY), *(Y, DX))) → DER_IN(d(e(X)), DX)
DER_IN(d(e(+(X, Y))), +(DX, DY)) → U1¹(X, Y, DX, DY, der_in(d(e(X)), DX))
DER_IN(d(e(+(X, Y))), +(DX, DY)) → DER_IN(d(e(X)), DX)
U1¹(X, Y, DX, DY, der_out(d(e(X)), DX)) → U2¹(X, Y, DX, DY, der_in(d(e(Y)), DY))
U1¹(X, Y, DX, DY, der_out(d(e(X)), DX)) → DER_IN(d(e(Y)), DY)
U3¹(X, Y, DY, DX, der_out(d(e(X)), DX)) → U4¹(X, Y, DY, DX, der_in(d(e(Y)), DY))
U3¹(X, Y, DY, DX, der_out(d(e(X)), DX)) → DER_IN(d(e(Y)), DY)
U5¹(X, DDX, der_out(d(X), DX)) → U6¹(X, DDX, DX, der_in(d(e(DX)), DDX))
U5¹(X, DDX, der_out(d(X), DX)) → DER_IN(d(e(DX)), DDX)

The TRS R consists of the following rules:

der_in(d(d(X)), DDX) → U5(X, DDX, der_in(d(X), DX))
der_in(d(e(*(X, Y))), +(*(X, DY), *(Y, DX))) → U3(X, Y, DY, DX, der_in(d(e(X)), DX))
der_in(d(e(+(X, Y))), +(DX, DY)) → U1(X, Y, DX, DY, der_in(d(e(X)), DX))
der_in(d(e(const(A))), const(0)) → der_out(d(e(const(A))), const(0))
der_in(d(e(t)), const(1)) → der_out(d(e(t)), const(1))
U1(X, Y, DX, DY, der_out(d(e(X)), DX)) → U2(X, Y, DX, DY, der_in(d(e(Y)), DY))
U2(X, Y, DX, DY, der_out(d(e(Y)), DY)) → der_out(d(e(+(X, Y))), +(DX, DY))
U3(X, Y, DY, DX, der_out(d(e(X)), DX)) → U4(X, Y, DY, DX, der_in(d(e(Y)), DY))
U4(X, Y, DY, DX, der_out(d(e(Y)), DY)) → der_out(d(e(*(X, Y))), +(*(X, DY), *(Y, DX)))
U5(X, DDX, der_out(d(X), DX)) → U6(X, DDX, DX, der_in(d(e(DX)), DDX))
U6(X, DDX, DX, der_out(d(e(DX)), DDX)) → der_out(d(d(X)), DDX)

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

                ↳ UsableRulesProof

              ↳ PiDP

  ↳ PrologToPiTRSProof

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

DER_IN(d(e(+(X, Y))), +(DX, DY)) → U1¹(X, Y, DX, DY, der_in(d(e(X)), DX))
DER_IN(d(e(*(X, Y))), +(*(X, DY), *(Y, DX))) → DER_IN(d(e(X)), DX)
U1¹(X, Y, DX, DY, der_out(d(e(X)), DX)) → DER_IN(d(e(Y)), DY)
U3¹(X, Y, DY, DX, der_out(d(e(X)), DX)) → DER_IN(d(e(Y)), DY)
DER_IN(d(e(+(X, Y))), +(DX, DY)) → DER_IN(d(e(X)), DX)
DER_IN(d(e(*(X, Y))), +(*(X, DY), *(Y, DX))) → U3¹(X, Y, DY, DX, der_in(d(e(X)), DX))

The TRS R consists of the following rules:

der_in(d(d(X)), DDX) → U5(X, DDX, der_in(d(X), DX))
der_in(d(e(*(X, Y))), +(*(X, DY), *(Y, DX))) → U3(X, Y, DY, DX, der_in(d(e(X)), DX))
der_in(d(e(+(X, Y))), +(DX, DY)) → U1(X, Y, DX, DY, der_in(d(e(X)), DX))
der_in(d(e(const(A))), const(0)) → der_out(d(e(const(A))), const(0))
der_in(d(e(t)), const(1)) → der_out(d(e(t)), const(1))
U1(X, Y, DX, DY, der_out(d(e(X)), DX)) → U2(X, Y, DX, DY, der_in(d(e(Y)), DY))
U2(X, Y, DX, DY, der_out(d(e(Y)), DY)) → der_out(d(e(+(X, Y))), +(DX, DY))
U3(X, Y, DY, DX, der_out(d(e(X)), DX)) → U4(X, Y, DY, DX, der_in(d(e(Y)), DY))
U4(X, Y, DY, DX, der_out(d(e(Y)), DY)) → der_out(d(e(*(X, Y))), +(*(X, DY), *(Y, DX)))
U5(X, DDX, der_out(d(X), DX)) → U6(X, DDX, DX, der_in(d(e(DX)), DDX))
U6(X, DDX, DX, der_out(d(e(DX)), DDX)) → der_out(d(d(X)), DDX)

The argument filtering Pi contains the following mapping:
der_in(x1, x2) = der_in(x1)
d(x1) = d(x1)
U5(x1, x2, x3) = U5(x3)
e(x1) = e(x1)
*(x1, x2) = *(x1, x2)
+(x1, x2) = +(x1, x2)
U3(x1, x2, x3, x4, x5) = U3(x1, x2, x5)
U1(x1, x2, x3, x4, x5) = U1(x2, x5)
const(x1) = const(x1)
0 = 0
der_out(x1, x2) = der_out(x2)
t = t
1 = 1
U2(x1, x2, x3, x4, x5) = U2(x3, x5)
U4(x1, x2, x3, x4, x5) = U4(x1, x2, x4, x5)
U6(x1, x2, x3, x4) = U6(x4)
U3¹(x1, x2, x3, x4, x5) = U3¹(x1, x2, x5)
U1¹(x1, x2, x3, x4, x5) = U1¹(x2, x5)
DER_IN(x1, x2) = DER_IN(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

  ↳ PrologToPiTRSProof

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

DER_IN(d(e(+(X, Y))), +(DX, DY)) → U1¹(X, Y, DX, DY, der_in(d(e(X)), DX))
DER_IN(d(e(*(X, Y))), +(*(X, DY), *(Y, DX))) → DER_IN(d(e(X)), DX)
U1¹(X, Y, DX, DY, der_out(d(e(X)), DX)) → DER_IN(d(e(Y)), DY)
U3¹(X, Y, DY, DX, der_out(d(e(X)), DX)) → DER_IN(d(e(Y)), DY)
DER_IN(d(e(+(X, Y))), +(DX, DY)) → DER_IN(d(e(X)), DX)
DER_IN(d(e(*(X, Y))), +(*(X, DY), *(Y, DX))) → U3¹(X, Y, DY, DX, der_in(d(e(X)), DX))

The TRS R consists of the following rules:

der_in(d(e(*(X, Y))), +(*(X, DY), *(Y, DX))) → U3(X, Y, DY, DX, der_in(d(e(X)), DX))
der_in(d(e(+(X, Y))), +(DX, DY)) → U1(X, Y, DX, DY, der_in(d(e(X)), DX))
der_in(d(e(const(A))), const(0)) → der_out(d(e(const(A))), const(0))
der_in(d(e(t)), const(1)) → der_out(d(e(t)), const(1))
U3(X, Y, DY, DX, der_out(d(e(X)), DX)) → U4(X, Y, DY, DX, der_in(d(e(Y)), DY))
U1(X, Y, DX, DY, der_out(d(e(X)), DX)) → U2(X, Y, DX, DY, der_in(d(e(Y)), DY))
U4(X, Y, DY, DX, der_out(d(e(Y)), DY)) → der_out(d(e(*(X, Y))), +(*(X, DY), *(Y, DX)))
U2(X, Y, DX, DY, der_out(d(e(Y)), DY)) → der_out(d(e(+(X, Y))), +(DX, DY))

The argument filtering Pi contains the following mapping:
der_in(x1, x2) = der_in(x1)
d(x1) = d(x1)
e(x1) = e(x1)
*(x1, x2) = *(x1, x2)
+(x1, x2) = +(x1, x2)
U3(x1, x2, x3, x4, x5) = U3(x1, x2, x5)
U1(x1, x2, x3, x4, x5) = U1(x2, x5)
const(x1) = const(x1)
0 = 0
der_out(x1, x2) = der_out(x2)
t = t
1 = 1
U2(x1, x2, x3, x4, x5) = U2(x3, x5)
U4(x1, x2, x3, x4, x5) = U4(x1, x2, x4, x5)
U3¹(x1, x2, x3, x4, x5) = U3¹(x1, x2, x5)
U1¹(x1, x2, x3, x4, x5) = U1¹(x2, x5)
DER_IN(x1, x2) = DER_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

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ QDPOrderProof

              ↳ PiDP

  ↳ PrologToPiTRSProof

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

DER_IN(d(e(+(X, Y)))) → DER_IN(d(e(X)))
DER_IN(d(e(*(X, Y)))) → DER_IN(d(e(X)))
U3¹(X, Y, der_out(DX)) → DER_IN(d(e(Y)))
U1¹(Y, der_out(DX)) → DER_IN(d(e(Y)))
DER_IN(d(e(*(X, Y)))) → U3¹(X, Y, der_in(d(e(X))))
DER_IN(d(e(+(X, Y)))) → U1¹(Y, der_in(d(e(X))))

The TRS R consists of the following rules:

der_in(d(e(*(X, Y)))) → U3(X, Y, der_in(d(e(X))))
der_in(d(e(+(X, Y)))) → U1(Y, der_in(d(e(X))))
der_in(d(e(const(A)))) → der_out(const(0))
der_in(d(e(t))) → der_out(const(1))
U3(X, Y, der_out(DX)) → U4(X, Y, DX, der_in(d(e(Y))))
U1(Y, der_out(DX)) → U2(DX, der_in(d(e(Y))))
U4(X, Y, DX, der_out(DY)) → der_out(+(*(X, DY), *(Y, DX)))
U2(DX, der_out(DY)) → der_out(+(DX, DY))

The set Q consists of the following terms:

der_in(x0)
U3(x0, x1, x2)
U1(x0, x1)
U4(x0, x1, x2, x3)
U2(x0, x1)

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.

DER_IN(d(e(+(X, Y)))) → DER_IN(d(e(X)))
DER_IN(d(e(+(X, Y)))) → U1¹(Y, der_in(d(e(X))))

The remaining pairs can at least be oriented weakly.

DER_IN(d(e(*(X, Y)))) → DER_IN(d(e(X)))
U3¹(X, Y, der_out(DX)) → DER_IN(d(e(Y)))
U1¹(Y, der_out(DX)) → DER_IN(d(e(Y)))
DER_IN(d(e(*(X, Y)))) → U3¹(X, Y, der_in(d(e(X))))

Used ordering: Polynomial interpretation [25]:

POL(*(x₁, x₂)) = x₁ + x₂
POL(+(x₁, x₂)) = 1 + x₁ + x₂
POL(0) = 0
POL(1) = 0
POL(DER_IN(x₁)) = x₁
POL(U1(x₁, x₂)) = 0
POL(U1¹(x₁, x₂)) = x₁
POL(U2(x₁, x₂)) = 0
POL(U3(x₁, x₂, x₃)) = 0
POL(U3¹(x₁, x₂, x₃)) = x₂
POL(U4(x₁, x₂, x₃, x₄)) = 0
POL(const(x₁)) = 0
POL(d(x₁)) = x₁
POL(der_in(x₁)) = 0
POL(der_out(x₁)) = 0
POL(e(x₁)) = x₁
POL(t) = 0

The following usable rules [17] were oriented: none

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ QDPOrderProof

                          ↳ QDP

                            ↳ DependencyGraphProof

              ↳ PiDP

  ↳ PrologToPiTRSProof

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

DER_IN(d(e(*(X, Y)))) → DER_IN(d(e(X)))
U1¹(Y, der_out(DX)) → DER_IN(d(e(Y)))
U3¹(X, Y, der_out(DX)) → DER_IN(d(e(Y)))
DER_IN(d(e(*(X, Y)))) → U3¹(X, Y, der_in(d(e(X))))

The TRS R consists of the following rules:

der_in(d(e(*(X, Y)))) → U3(X, Y, der_in(d(e(X))))
der_in(d(e(+(X, Y)))) → U1(Y, der_in(d(e(X))))
der_in(d(e(const(A)))) → der_out(const(0))
der_in(d(e(t))) → der_out(const(1))
U3(X, Y, der_out(DX)) → U4(X, Y, DX, der_in(d(e(Y))))
U1(Y, der_out(DX)) → U2(DX, der_in(d(e(Y))))
U4(X, Y, DX, der_out(DY)) → der_out(+(*(X, DY), *(Y, DX)))
U2(DX, der_out(DY)) → der_out(+(DX, DY))

The set Q consists of the following terms:

der_in(x0)
U3(x0, x1, x2)
U1(x0, x1)
U4(x0, x1, x2, x3)
U2(x0, x1)

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

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ QDPOrderProof

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ QDPOrderProof

              ↳ PiDP

  ↳ PrologToPiTRSProof

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

DER_IN(d(e(*(X, Y)))) → DER_IN(d(e(X)))
U3¹(X, Y, der_out(DX)) → DER_IN(d(e(Y)))
DER_IN(d(e(*(X, Y)))) → U3¹(X, Y, der_in(d(e(X))))

The TRS R consists of the following rules:

der_in(d(e(*(X, Y)))) → U3(X, Y, der_in(d(e(X))))
der_in(d(e(+(X, Y)))) → U1(Y, der_in(d(e(X))))
der_in(d(e(const(A)))) → der_out(const(0))
der_in(d(e(t))) → der_out(const(1))
U3(X, Y, der_out(DX)) → U4(X, Y, DX, der_in(d(e(Y))))
U1(Y, der_out(DX)) → U2(DX, der_in(d(e(Y))))
U4(X, Y, DX, der_out(DY)) → der_out(+(*(X, DY), *(Y, DX)))
U2(DX, der_out(DY)) → der_out(+(DX, DY))

The set Q consists of the following terms:

der_in(x0)
U3(x0, x1, x2)
U1(x0, x1)
U4(x0, x1, x2, x3)
U2(x0, x1)

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.

DER_IN(d(e(*(X, Y)))) → DER_IN(d(e(X)))
DER_IN(d(e(*(X, Y)))) → U3¹(X, Y, der_in(d(e(X))))

The remaining pairs can at least be oriented weakly.

U3¹(X, Y, der_out(DX)) → DER_IN(d(e(Y)))

Used ordering: Polynomial interpretation [25]:

POL(*(x₁, x₂)) = 1 + x₁ + x₂
POL(+(x₁, x₂)) = 0
POL(0) = 0
POL(1) = 0
POL(DER_IN(x₁)) = x₁
POL(U1(x₁, x₂)) = 0
POL(U2(x₁, x₂)) = 0
POL(U3(x₁, x₂, x₃)) = 0
POL(U3¹(x₁, x₂, x₃)) = x₁ + x₂
POL(U4(x₁, x₂, x₃, x₄)) = 0
POL(const(x₁)) = 0
POL(d(x₁)) = x₁
POL(der_in(x₁)) = 0
POL(der_out(x₁)) = 0
POL(e(x₁)) = x₁
POL(t) = 0

The following usable rules [17] were oriented: none

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ QDPOrderProof

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ QDPOrderProof

                                  ↳ QDP

                                    ↳ DependencyGraphProof

              ↳ PiDP

  ↳ PrologToPiTRSProof

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

U3¹(X, Y, der_out(DX)) → DER_IN(d(e(Y)))

The TRS R consists of the following rules:

der_in(d(e(*(X, Y)))) → U3(X, Y, der_in(d(e(X))))
der_in(d(e(+(X, Y)))) → U1(Y, der_in(d(e(X))))
der_in(d(e(const(A)))) → der_out(const(0))
der_in(d(e(t))) → der_out(const(1))
U3(X, Y, der_out(DX)) → U4(X, Y, DX, der_in(d(e(Y))))
U1(Y, der_out(DX)) → U2(DX, der_in(d(e(Y))))
U4(X, Y, DX, der_out(DY)) → der_out(+(*(X, DY), *(Y, DX)))
U2(DX, der_out(DY)) → der_out(+(DX, DY))

The set Q consists of the following terms:

der_in(x0)
U3(x0, x1, x2)
U1(x0, x1)
U4(x0, x1, x2, x3)
U2(x0, x1)

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.

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

  ↳ PrologToPiTRSProof

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

DER_IN(d(d(X)), DDX) → DER_IN(d(X), DX)

The TRS R consists of the following rules:

der_in(d(d(X)), DDX) → U5(X, DDX, der_in(d(X), DX))
der_in(d(e(*(X, Y))), +(*(X, DY), *(Y, DX))) → U3(X, Y, DY, DX, der_in(d(e(X)), DX))
der_in(d(e(+(X, Y))), +(DX, DY)) → U1(X, Y, DX, DY, der_in(d(e(X)), DX))
der_in(d(e(const(A))), const(0)) → der_out(d(e(const(A))), const(0))
der_in(d(e(t)), const(1)) → der_out(d(e(t)), const(1))
U1(X, Y, DX, DY, der_out(d(e(X)), DX)) → U2(X, Y, DX, DY, der_in(d(e(Y)), DY))
U2(X, Y, DX, DY, der_out(d(e(Y)), DY)) → der_out(d(e(+(X, Y))), +(DX, DY))
U3(X, Y, DY, DX, der_out(d(e(X)), DX)) → U4(X, Y, DY, DX, der_in(d(e(Y)), DY))
U4(X, Y, DY, DX, der_out(d(e(Y)), DY)) → der_out(d(e(*(X, Y))), +(*(X, DY), *(Y, DX)))
U5(X, DDX, der_out(d(X), DX)) → U6(X, DDX, DX, der_in(d(e(DX)), DDX))
U6(X, DDX, DX, der_out(d(e(DX)), DDX)) → der_out(d(d(X)), DDX)

The argument filtering Pi contains the following mapping:
der_in(x1, x2) = der_in(x1)
d(x1) = d(x1)
U5(x1, x2, x3) = U5(x3)
e(x1) = e(x1)
*(x1, x2) = *(x1, x2)
+(x1, x2) = +(x1, x2)
U3(x1, x2, x3, x4, x5) = U3(x1, x2, x5)
U1(x1, x2, x3, x4, x5) = U1(x2, x5)
const(x1) = const(x1)
0 = 0
der_out(x1, x2) = der_out(x2)
t = t
1 = 1
U2(x1, x2, x3, x4, x5) = U2(x3, x5)
U4(x1, x2, x3, x4, x5) = U4(x1, x2, x4, x5)
U6(x1, x2, x3, x4) = U6(x4)
DER_IN(x1, x2) = DER_IN(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:

DER_IN(d(d(X)), DDX) → DER_IN(d(X), DX)

R is empty.
The argument filtering Pi contains the following mapping:
d(x1) = d(x1)
DER_IN(x1, x2) = DER_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

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ QDPSizeChangeProof

  ↳ PrologToPiTRSProof

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

DER_IN(d(d(X))) → DER_IN(d(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:

DER_IN(d(d(X))) → DER_IN(d(X))
The graph contains the following edges 1 > 1

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:

der_in(d(d(X)), DDX) → U5(X, DDX, der_in(d(X), DX))
der_in(d(e(*(X, Y))), +(*(X, DY), *(Y, DX))) → U3(X, Y, DY, DX, der_in(d(e(X)), DX))
der_in(d(e(+(X, Y))), +(DX, DY)) → U1(X, Y, DX, DY, der_in(d(e(X)), DX))
der_in(d(e(const(A))), const(0)) → der_out(d(e(const(A))), const(0))
der_in(d(e(t)), const(1)) → der_out(d(e(t)), const(1))
U1(X, Y, DX, DY, der_out(d(e(X)), DX)) → U2(X, Y, DX, DY, der_in(d(e(Y)), DY))
U2(X, Y, DX, DY, der_out(d(e(Y)), DY)) → der_out(d(e(+(X, Y))), +(DX, DY))
U3(X, Y, DY, DX, der_out(d(e(X)), DX)) → U4(X, Y, DY, DX, der_in(d(e(Y)), DY))
U4(X, Y, DY, DX, der_out(d(e(Y)), DY)) → der_out(d(e(*(X, Y))), +(*(X, DY), *(Y, DX)))
U5(X, DDX, der_out(d(X), DX)) → U6(X, DDX, DX, der_in(d(e(DX)), DDX))
U6(X, DDX, DX, der_out(d(e(DX)), DDX)) → der_out(d(d(X)), DDX)

The argument filtering Pi contains the following mapping:
der_in(x1, x2) = der_in(x1)
d(x1) = d(x1)
U5(x1, x2, x3) = U5(x1, x3)
e(x1) = e(x1)
*(x1, x2) = *(x1, x2)
+(x1, x2) = +(x1, x2)
U3(x1, x2, x3, x4, x5) = U3(x1, x2, x5)
U1(x1, x2, x3, x4, x5) = U1(x1, x2, x5)
const(x1) = const(x1)
0 = 0
der_out(x1, x2) = der_out(x1, x2)
t = t
1 = 1
U2(x1, x2, x3, x4, x5) = U2(x1, x2, x3, x5)
U4(x1, x2, x3, x4, x5) = U4(x1, x2, x4, x5)
U6(x1, x2, x3, x4) = U6(x1, x4)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog

↳ Prolog

  ↳ PrologToPiTRSProof

  ↳ PrologToPiTRSProof

    ↳ PiTRS

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

der_in(d(d(X)), DDX) → U5(X, DDX, der_in(d(X), DX))
der_in(d(e(*(X, Y))), +(*(X, DY), *(Y, DX))) → U3(X, Y, DY, DX, der_in(d(e(X)), DX))
der_in(d(e(+(X, Y))), +(DX, DY)) → U1(X, Y, DX, DY, der_in(d(e(X)), DX))
der_in(d(e(const(A))), const(0)) → der_out(d(e(const(A))), const(0))
der_in(d(e(t)), const(1)) → der_out(d(e(t)), const(1))
U1(X, Y, DX, DY, der_out(d(e(X)), DX)) → U2(X, Y, DX, DY, der_in(d(e(Y)), DY))
U2(X, Y, DX, DY, der_out(d(e(Y)), DY)) → der_out(d(e(+(X, Y))), +(DX, DY))
U3(X, Y, DY, DX, der_out(d(e(X)), DX)) → U4(X, Y, DY, DX, der_in(d(e(Y)), DY))
U4(X, Y, DY, DX, der_out(d(e(Y)), DY)) → der_out(d(e(*(X, Y))), +(*(X, DY), *(Y, DX)))
U5(X, DDX, der_out(d(X), DX)) → U6(X, DDX, DX, der_in(d(e(DX)), DDX))
U6(X, DDX, DX, der_out(d(e(DX)), DDX)) → der_out(d(d(X)), DDX)