Termination w.r.t. Q proof of /tmp/tmpIHzslF/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

  ↳ Overlay + Local Confluence

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.

The TRS is overlay and locally confluent. By [19] we can switch to innermost.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ 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)))

The set Q consists of the following terms:

eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
rm(x0, nil)
rm(x0, add(x1, x2))
ifrm(true, x0, add(x1, x2))
ifrm(false, x0, add(x1, x2))
purge(nil)
purge(add(x0, x1))

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)))

The set Q consists of the following terms:

eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
rm(x0, nil)
rm(x0, add(x1, x2))
ifrm(true, x0, add(x1, x2))
ifrm(false, x0, add(x1, x2))
purge(nil)
purge(add(x0, x1))

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ 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)))

The set Q consists of the following terms:

eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
rm(x0, nil)
rm(x0, add(x1, x2))
ifrm(true, x0, add(x1, x2))
ifrm(false, x0, add(x1, x2))
purge(nil)
purge(add(x0, x1))

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

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ UsableRulesProof

              ↳ 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)))

The set Q consists of the following terms:

eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
rm(x0, nil)
rm(x0, add(x1, x2))
ifrm(true, x0, add(x1, x2))
ifrm(false, x0, add(x1, x2))
purge(nil)
purge(add(x0, x1))

We have to consider all minimal (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.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

              ↳ QDP

              ↳ QDP

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

EQ(s(X), s(Y)) → EQ(X, Y)

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

eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
rm(x0, nil)
rm(x0, add(x1, x2))
ifrm(true, x0, add(x1, x2))
ifrm(false, x0, add(x1, x2))
purge(nil)
purge(add(x0, x1))

We have to consider all minimal (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.

eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
rm(x0, nil)
rm(x0, add(x1, x2))
ifrm(true, x0, add(x1, x2))
ifrm(false, x0, add(x1, x2))
purge(nil)
purge(add(x0, x1))

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ QDPSizeChangeProof

              ↳ QDP

              ↳ QDP

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

EQ(s(X), s(Y)) → EQ(X, Y)

R is empty.
Q is empty.
We have to consider all minimal (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:

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

              ↳ 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)))

The set Q consists of the following terms:

eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
rm(x0, nil)
rm(x0, add(x1, x2))
ifrm(true, x0, add(x1, x2))
ifrm(false, x0, add(x1, x2))
purge(nil)
purge(add(x0, x1))

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

              ↳ 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)

The set Q consists of the following terms:

eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
rm(x0, nil)
rm(x0, add(x1, x2))
ifrm(true, x0, add(x1, x2))
ifrm(false, x0, add(x1, x2))
purge(nil)
purge(add(x0, x1))

We have to consider all minimal (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.

rm(x0, nil)
rm(x0, add(x1, x2))
ifrm(true, x0, add(x1, x2))
ifrm(false, x0, add(x1, x2))
purge(nil)
purge(add(x0, x1))

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ QDPSizeChangeProof

              ↳ 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)

The set Q consists of the following terms:

eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))

We have to consider all minimal (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:

RM(N, add(M, X)) → IFRM(eq(N, M), N, add(M, X))
The graph contains the following edges 1 >= 2, 2 >= 3
IFRM(false, N, add(M, X)) → RM(N, X)
The graph contains the following edges 2 >= 1, 3 > 2
IFRM(true, N, add(M, X)) → RM(N, X)
The graph contains the following edges 2 >= 1, 3 > 2

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

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)))

The set Q consists of the following terms:

eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
rm(x0, nil)
rm(x0, add(x1, x2))
ifrm(true, x0, add(x1, x2))
ifrm(false, x0, add(x1, x2))
purge(nil)
purge(add(x0, x1))

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

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:

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

The set Q consists of the following terms:

eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
rm(x0, nil)
rm(x0, add(x1, x2))
ifrm(true, x0, add(x1, x2))
ifrm(false, x0, add(x1, x2))
purge(nil)
purge(add(x0, x1))

We have to consider all minimal (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.

purge(nil)
purge(add(x0, x1))

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ 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:

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

The set Q consists of the following terms:

eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
rm(x0, nil)
rm(x0, add(x1, x2))
ifrm(true, x0, add(x1, x2))
ifrm(false, x0, add(x1, x2))

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]:

POL(0) = 0
POL(PURGE(x₁)) = x₁
POL(add(x₁, x₂)) = 1 + x₂
POL(eq(x₁, x₂)) = 0
POL(false) = 0
POL(ifrm(x₁, x₂, x₃)) = x₃
POL(nil) = 0
POL(rm(x₁, x₂)) = x₂
POL(s(x₁)) = 0
POL(true) = 0

The following usable rules [17] were oriented:

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ QDPOrderProof

                          ↳ QDP

                            ↳ PisEmptyProof

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

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

The set Q consists of the following terms:

eq(0, 0)
eq(0, s(x0))
eq(s(x0), 0)
eq(s(x0), s(x1))
rm(x0, nil)
rm(x0, add(x1, x2))
ifrm(true, x0, add(x1, x2))
ifrm(false, x0, add(x1, x2))

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