Termination w.r.t. Q proof of /tmp/tmpTQi1zQ/elimdupl.trs

Termination w.r.t. Q of the following Term Rewriting System could be proven:

Q restricted rewrite system:
The TRS R consists of the following rules:

eq(0, 0) → true
eq(0, s(X)) → false
eq(s(X), 0) → false
eq(s(X), s(Y)) → eq(X, Y)
rm(N, nil) → nil
rm(N, add(M, X)) → ifrm(eq(N, M), N, add(M, X))
ifrm(true, N, add(M, X)) → rm(N, X)
ifrm(false, N, add(M, X)) → add(M, rm(N, X))
purge(nil) → nil
purge(add(N, X)) → add(N, purge(rm(N, X)))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

Q restricted rewrite system:
The TRS R consists of the following rules:

eq(0, 0) → true
eq(0, s(X)) → false
eq(s(X), 0) → false
eq(s(X), s(Y)) → eq(X, Y)
rm(N, nil) → nil
rm(N, add(M, X)) → ifrm(eq(N, M), N, add(M, X))
ifrm(true, N, add(M, X)) → rm(N, X)
ifrm(false, N, add(M, X)) → add(M, rm(N, X))
purge(nil) → nil
purge(add(N, X)) → add(N, purge(rm(N, X)))

Q is empty.

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

IFRM(false, N, add(M, X)) → RM(N, X)
PURGE(add(N, X)) → PURGE(rm(N, X))
EQ(s(X), s(Y)) → EQ(X, Y)
RM(N, add(M, X)) → EQ(N, M)
RM(N, add(M, X)) → IFRM(eq(N, M), N, add(M, X))
PURGE(add(N, X)) → RM(N, X)
IFRM(true, N, add(M, X)) → RM(N, X)

The TRS R consists of the following rules:

eq(0, 0) → true
eq(0, s(X)) → false
eq(s(X), 0) → false
eq(s(X), s(Y)) → eq(X, Y)
rm(N, nil) → nil
rm(N, add(M, X)) → ifrm(eq(N, M), N, add(M, X))
ifrm(true, N, add(M, X)) → rm(N, X)
ifrm(false, N, add(M, X)) → add(M, rm(N, X))
purge(nil) → nil
purge(add(N, X)) → add(N, purge(rm(N, X)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

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

IFRM(false, N, add(M, X)) → RM(N, X)
PURGE(add(N, X)) → PURGE(rm(N, X))
EQ(s(X), s(Y)) → EQ(X, Y)
RM(N, add(M, X)) → EQ(N, M)
RM(N, add(M, X)) → IFRM(eq(N, M), N, add(M, X))
PURGE(add(N, X)) → RM(N, X)
IFRM(true, N, add(M, X)) → RM(N, X)

The TRS R consists of the following rules:

eq(0, 0) → true
eq(0, s(X)) → false
eq(s(X), 0) → false
eq(s(X), s(Y)) → eq(X, Y)
rm(N, nil) → nil
rm(N, add(M, X)) → ifrm(eq(N, M), N, add(M, X))
ifrm(true, N, add(M, X)) → rm(N, X)
ifrm(false, N, add(M, X)) → add(M, rm(N, X))
purge(nil) → nil
purge(add(N, X)) → add(N, purge(rm(N, X)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 3 SCCs with 2 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

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

EQ(s(X), s(Y)) → EQ(X, Y)

The TRS R consists of the following rules:

eq(0, 0) → true
eq(0, s(X)) → false
eq(s(X), 0) → false
eq(s(X), s(Y)) → eq(X, Y)
rm(N, nil) → nil
rm(N, add(M, X)) → ifrm(eq(N, M), N, add(M, X))
ifrm(true, N, add(M, X)) → rm(N, X)
ifrm(false, N, add(M, X)) → add(M, rm(N, X))
purge(nil) → nil
purge(add(N, X)) → add(N, purge(rm(N, X)))

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

The following pairs can be oriented strictly and are deleted.

EQ(s(X), s(Y)) → EQ(X, Y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(EQ(x₁, x₂)) = (1/2)x_1
POL(s(x₁)) = 1/2 + (4)x_1

The value of delta used in the strict ordering is 1/4.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

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

eq(0, 0) → true
eq(0, s(X)) → false
eq(s(X), 0) → false
eq(s(X), s(Y)) → eq(X, Y)
rm(N, nil) → nil
rm(N, add(M, X)) → ifrm(eq(N, M), N, add(M, X))
ifrm(true, N, add(M, X)) → rm(N, X)
ifrm(false, N, add(M, X)) → add(M, rm(N, X))
purge(nil) → nil
purge(add(N, X)) → add(N, purge(rm(N, X)))

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

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

IFRM(false, N, add(M, X)) → RM(N, X)
RM(N, add(M, X)) → IFRM(eq(N, M), N, add(M, X))
IFRM(true, N, add(M, X)) → RM(N, X)

The TRS R consists of the following rules:

eq(0, 0) → true
eq(0, s(X)) → false
eq(s(X), 0) → false
eq(s(X), s(Y)) → eq(X, Y)
rm(N, nil) → nil
rm(N, add(M, X)) → ifrm(eq(N, M), N, add(M, X))
ifrm(true, N, add(M, X)) → rm(N, X)
ifrm(false, N, add(M, X)) → add(M, rm(N, X))
purge(nil) → nil
purge(add(N, X)) → add(N, purge(rm(N, X)))

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

The following pairs can be oriented strictly and are deleted.

IFRM(false, N, add(M, X)) → RM(N, X)
IFRM(true, N, add(M, X)) → RM(N, X)

The remaining pairs can at least be oriented weakly.

RM(N, add(M, X)) → IFRM(eq(N, M), N, add(M, X))

Used ordering: Polynomial interpretation [25,35]:

POL(add(x₁, x₂)) = 1/4 + (4)x_2
POL(RM(x₁, x₂)) = x_2
POL(eq(x₁, x₂)) = (1/4)x_2
POL(true) = 0
POL(false) = 0
POL(s(x₁)) = (2)x_1
POL(IFRM(x₁, x₂, x₃)) = x_3
POL(0) = 1/4

The value of delta used in the strict ordering is 1/4.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ DependencyGraphProof

          ↳ QDP

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

RM(N, add(M, X)) → IFRM(eq(N, M), N, add(M, X))

The TRS R consists of the following rules:

eq(0, 0) → true
eq(0, s(X)) → false
eq(s(X), 0) → false
eq(s(X), s(Y)) → eq(X, Y)
rm(N, nil) → nil
rm(N, add(M, X)) → ifrm(eq(N, M), N, add(M, X))
ifrm(true, N, add(M, X)) → rm(N, X)
ifrm(false, N, add(M, X)) → add(M, rm(N, X))
purge(nil) → nil
purge(add(N, X)) → add(N, purge(rm(N, X)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 1 less node.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

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

PURGE(add(N, X)) → PURGE(rm(N, X))

The TRS R consists of the following rules:

eq(0, 0) → true
eq(0, s(X)) → false
eq(s(X), 0) → false
eq(s(X), s(Y)) → eq(X, Y)
rm(N, nil) → nil
rm(N, add(M, X)) → ifrm(eq(N, M), N, add(M, X))
ifrm(true, N, add(M, X)) → rm(N, X)
ifrm(false, N, add(M, X)) → add(M, rm(N, X))
purge(nil) → nil
purge(add(N, X)) → add(N, purge(rm(N, X)))

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

The following pairs can be oriented strictly and are deleted.

PURGE(add(N, X)) → PURGE(rm(N, X))

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(add(x₁, x₂)) = 1/4 + (4)x_2
POL(eq(x₁, x₂)) = 0
POL(true) = 0
POL(rm(x₁, x₂)) = x_2
POL(false) = 0
POL(s(x₁)) = 0
POL(ifrm(x₁, x₂, x₃)) = x_3
POL(0) = 0
POL(nil) = 1/2
POL(PURGE(x₁)) = (1/4)x_1

The value of delta used in the strict ordering is 1/16.
The following usable rules [17] were oriented:

rm(N, nil) → nil
ifrm(true, N, add(M, X)) → rm(N, X)
rm(N, add(M, X)) → ifrm(eq(N, M), N, add(M, X))
ifrm(false, N, add(M, X)) → add(M, rm(N, X))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

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

eq(0, 0) → true
eq(0, s(X)) → false
eq(s(X), 0) → false
eq(s(X), s(Y)) → eq(X, Y)
rm(N, nil) → nil
rm(N, add(M, X)) → ifrm(eq(N, M), N, add(M, X))
ifrm(true, N, add(M, X)) → rm(N, X)
ifrm(false, N, add(M, X)) → add(M, rm(N, X))
purge(nil) → nil
purge(add(N, X)) → add(N, purge(rm(N, X)))

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