Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar4/RubioSLASHelimdupl.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:

eq₂(0, 0) -> true
eq₂(0, s₁(X)) -> false
eq₂(s₁(X), 0) -> false
eq₂(s₁(X), s₁(Y)) -> eq₂(X, Y)
rm₂(N, nil) -> nil
rm₂(N, add₂(M, X)) -> ifrm₃(eq₂(N, M), N, add₂(M, X))
ifrm₃(true, N, add₂(M, X)) -> rm₂(N, X)
ifrm₃(false, N, add₂(M, X)) -> add₂(M, rm₂(N, X))
purge₁(nil) -> nil
purge₁(add₂(N, X)) -> add₂(N, purge₁(rm₂(N, X)))

Q is empty.

↳ QTRS

  ↳ Non-Overlap Check

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

eq₂(0, 0) -> true
eq₂(0, s₁(X)) -> false
eq₂(s₁(X), 0) -> false
eq₂(s₁(X), s₁(Y)) -> eq₂(X, Y)
rm₂(N, nil) -> nil
rm₂(N, add₂(M, X)) -> ifrm₃(eq₂(N, M), N, add₂(M, X))
ifrm₃(true, N, add₂(M, X)) -> rm₂(N, X)
ifrm₃(false, N, add₂(M, X)) -> add₂(M, rm₂(N, X))
purge₁(nil) -> nil
purge₁(add₂(N, X)) -> add₂(N, purge₁(rm₂(N, X)))

Q is empty.

The TRS is non-overlapping. Hence, we can switch to innermost.

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

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

eq₂(0, 0) -> true
eq₂(0, s₁(X)) -> false
eq₂(s₁(X), 0) -> false
eq₂(s₁(X), s₁(Y)) -> eq₂(X, Y)
rm₂(N, nil) -> nil
rm₂(N, add₂(M, X)) -> ifrm₃(eq₂(N, M), N, add₂(M, X))
ifrm₃(true, N, add₂(M, X)) -> rm₂(N, X)
ifrm₃(false, N, add₂(M, X)) -> add₂(M, rm₂(N, X))
purge₁(nil) -> nil
purge₁(add₂(N, X)) -> add₂(N, purge₁(rm₂(N, X)))

The set Q consists of the following terms:

eq₂(0, 0)
eq₂(0, s₁(x0))
eq₂(s₁(x0), 0)
eq₂(s₁(x0), s₁(x1))
rm₂(x0, nil)
rm₂(x0, add₂(x1, x2))
ifrm₃(true, x0, add₂(x1, x2))
ifrm₃(false, x0, add₂(x1, x2))
purge₁(nil)
purge₁(add₂(x0, x1))

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

PURGE₁(add₂(N, X)) -> RM₂(N, X)
PURGE₁(add₂(N, X)) -> PURGE₁(rm₂(N, X))
EQ₂(s₁(X), s₁(Y)) -> EQ₂(X, Y)
IFRM₃(true, N, add₂(M, X)) -> RM₂(N, X)
IFRM₃(false, N, add₂(M, X)) -> RM₂(N, X)
RM₂(N, add₂(M, X)) -> EQ₂(N, M)
RM₂(N, add₂(M, X)) -> IFRM₃(eq₂(N, M), N, add₂(M, X))

The TRS R consists of the following rules:

eq₂(0, 0) -> true
eq₂(0, s₁(X)) -> false
eq₂(s₁(X), 0) -> false
eq₂(s₁(X), s₁(Y)) -> eq₂(X, Y)
rm₂(N, nil) -> nil
rm₂(N, add₂(M, X)) -> ifrm₃(eq₂(N, M), N, add₂(M, X))
ifrm₃(true, N, add₂(M, X)) -> rm₂(N, X)
ifrm₃(false, N, add₂(M, X)) -> add₂(M, rm₂(N, X))
purge₁(nil) -> nil
purge₁(add₂(N, X)) -> add₂(N, purge₁(rm₂(N, X)))

The set Q consists of the following terms:

eq₂(0, 0)
eq₂(0, s₁(x0))
eq₂(s₁(x0), 0)
eq₂(s₁(x0), s₁(x1))
rm₂(x0, nil)
rm₂(x0, add₂(x1, x2))
ifrm₃(true, x0, add₂(x1, x2))
ifrm₃(false, x0, add₂(x1, x2))
purge₁(nil)
purge₁(add₂(x0, x1))

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

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

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

PURGE₁(add₂(N, X)) -> RM₂(N, X)
PURGE₁(add₂(N, X)) -> PURGE₁(rm₂(N, X))
EQ₂(s₁(X), s₁(Y)) -> EQ₂(X, Y)
IFRM₃(true, N, add₂(M, X)) -> RM₂(N, X)
IFRM₃(false, N, add₂(M, X)) -> RM₂(N, X)
RM₂(N, add₂(M, X)) -> EQ₂(N, M)
RM₂(N, add₂(M, X)) -> IFRM₃(eq₂(N, M), N, add₂(M, X))

The TRS R consists of the following rules:

eq₂(0, 0) -> true
eq₂(0, s₁(X)) -> false
eq₂(s₁(X), 0) -> false
eq₂(s₁(X), s₁(Y)) -> eq₂(X, Y)
rm₂(N, nil) -> nil
rm₂(N, add₂(M, X)) -> ifrm₃(eq₂(N, M), N, add₂(M, X))
ifrm₃(true, N, add₂(M, X)) -> rm₂(N, X)
ifrm₃(false, N, add₂(M, X)) -> add₂(M, rm₂(N, X))
purge₁(nil) -> nil
purge₁(add₂(N, X)) -> add₂(N, purge₁(rm₂(N, X)))

The set Q consists of the following terms:

eq₂(0, 0)
eq₂(0, s₁(x0))
eq₂(s₁(x0), 0)
eq₂(s₁(x0), s₁(x1))
rm₂(x0, nil)
rm₂(x0, add₂(x1, x2))
ifrm₃(true, x0, add₂(x1, x2))
ifrm₃(false, x0, add₂(x1, x2))
purge₁(nil)
purge₁(add₂(x0, x1))

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

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPAfsSolverProof

              ↳ QDP

              ↳ QDP

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

EQ₂(s₁(X), s₁(Y)) -> EQ₂(X, Y)

The TRS R consists of the following rules:

eq₂(0, 0) -> true
eq₂(0, s₁(X)) -> false
eq₂(s₁(X), 0) -> false
eq₂(s₁(X), s₁(Y)) -> eq₂(X, Y)
rm₂(N, nil) -> nil
rm₂(N, add₂(M, X)) -> ifrm₃(eq₂(N, M), N, add₂(M, X))
ifrm₃(true, N, add₂(M, X)) -> rm₂(N, X)
ifrm₃(false, N, add₂(M, X)) -> add₂(M, rm₂(N, X))
purge₁(nil) -> nil
purge₁(add₂(N, X)) -> add₂(N, purge₁(rm₂(N, X)))

The set Q consists of the following terms:

eq₂(0, 0)
eq₂(0, s₁(x0))
eq₂(s₁(x0), 0)
eq₂(s₁(x0), s₁(x1))
rm₂(x0, nil)
rm₂(x0, add₂(x1, x2))
ifrm₃(true, x0, add₂(x1, x2))
ifrm₃(false, x0, add₂(x1, x2))
purge₁(nil)
purge₁(add₂(x0, x1))

We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

EQ₂(s₁(X), s₁(Y)) -> EQ₂(X, Y)

Used argument filtering: EQ₂(x1, x2) = x2
s₁(x1) = s₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ PisEmptyProof

              ↳ QDP

              ↳ QDP

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

eq₂(0, 0) -> true
eq₂(0, s₁(X)) -> false
eq₂(s₁(X), 0) -> false
eq₂(s₁(X), s₁(Y)) -> eq₂(X, Y)
rm₂(N, nil) -> nil
rm₂(N, add₂(M, X)) -> ifrm₃(eq₂(N, M), N, add₂(M, X))
ifrm₃(true, N, add₂(M, X)) -> rm₂(N, X)
ifrm₃(false, N, add₂(M, X)) -> add₂(M, rm₂(N, X))
purge₁(nil) -> nil
purge₁(add₂(N, X)) -> add₂(N, purge₁(rm₂(N, X)))

The set Q consists of the following terms:

eq₂(0, 0)
eq₂(0, s₁(x0))
eq₂(s₁(x0), 0)
eq₂(s₁(x0), s₁(x1))
rm₂(x0, nil)
rm₂(x0, add₂(x1, x2))
ifrm₃(true, x0, add₂(x1, x2))
ifrm₃(false, x0, add₂(x1, x2))
purge₁(nil)
purge₁(add₂(x0, x1))

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

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPAfsSolverProof

              ↳ QDP

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

IFRM₃(true, N, add₂(M, X)) -> RM₂(N, X)
IFRM₃(false, N, add₂(M, X)) -> RM₂(N, X)
RM₂(N, add₂(M, X)) -> IFRM₃(eq₂(N, M), N, add₂(M, X))

The TRS R consists of the following rules:

eq₂(0, 0) -> true
eq₂(0, s₁(X)) -> false
eq₂(s₁(X), 0) -> false
eq₂(s₁(X), s₁(Y)) -> eq₂(X, Y)
rm₂(N, nil) -> nil
rm₂(N, add₂(M, X)) -> ifrm₃(eq₂(N, M), N, add₂(M, X))
ifrm₃(true, N, add₂(M, X)) -> rm₂(N, X)
ifrm₃(false, N, add₂(M, X)) -> add₂(M, rm₂(N, X))
purge₁(nil) -> nil
purge₁(add₂(N, X)) -> add₂(N, purge₁(rm₂(N, X)))

The set Q consists of the following terms:

eq₂(0, 0)
eq₂(0, s₁(x0))
eq₂(s₁(x0), 0)
eq₂(s₁(x0), s₁(x1))
rm₂(x0, nil)
rm₂(x0, add₂(x1, x2))
ifrm₃(true, x0, add₂(x1, x2))
ifrm₃(false, x0, add₂(x1, x2))
purge₁(nil)
purge₁(add₂(x0, x1))

We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

IFRM₃(true, N, add₂(M, X)) -> RM₂(N, X)
IFRM₃(false, N, add₂(M, X)) -> RM₂(N, X)

Used argument filtering: IFRM₃(x1, x2, x3) = x3
add₂(x1, x2) = add₁(x2)
RM₂(x1, x2) = x2
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ DependencyGraphProof

              ↳ QDP

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

RM₂(N, add₂(M, X)) -> IFRM₃(eq₂(N, M), N, add₂(M, X))

The TRS R consists of the following rules:

eq₂(0, 0) -> true
eq₂(0, s₁(X)) -> false
eq₂(s₁(X), 0) -> false
eq₂(s₁(X), s₁(Y)) -> eq₂(X, Y)
rm₂(N, nil) -> nil
rm₂(N, add₂(M, X)) -> ifrm₃(eq₂(N, M), N, add₂(M, X))
ifrm₃(true, N, add₂(M, X)) -> rm₂(N, X)
ifrm₃(false, N, add₂(M, X)) -> add₂(M, rm₂(N, X))
purge₁(nil) -> nil
purge₁(add₂(N, X)) -> add₂(N, purge₁(rm₂(N, X)))

The set Q consists of the following terms:

eq₂(0, 0)
eq₂(0, s₁(x0))
eq₂(s₁(x0), 0)
eq₂(s₁(x0), s₁(x1))
rm₂(x0, nil)
rm₂(x0, add₂(x1, x2))
ifrm₃(true, x0, add₂(x1, x2))
ifrm₃(false, x0, add₂(x1, x2))
purge₁(nil)
purge₁(add₂(x0, x1))

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

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPAfsSolverProof

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

PURGE₁(add₂(N, X)) -> PURGE₁(rm₂(N, X))

The TRS R consists of the following rules:

eq₂(0, 0) -> true
eq₂(0, s₁(X)) -> false
eq₂(s₁(X), 0) -> false
eq₂(s₁(X), s₁(Y)) -> eq₂(X, Y)
rm₂(N, nil) -> nil
rm₂(N, add₂(M, X)) -> ifrm₃(eq₂(N, M), N, add₂(M, X))
ifrm₃(true, N, add₂(M, X)) -> rm₂(N, X)
ifrm₃(false, N, add₂(M, X)) -> add₂(M, rm₂(N, X))
purge₁(nil) -> nil
purge₁(add₂(N, X)) -> add₂(N, purge₁(rm₂(N, X)))

The set Q consists of the following terms:

eq₂(0, 0)
eq₂(0, s₁(x0))
eq₂(s₁(x0), 0)
eq₂(s₁(x0), s₁(x1))
rm₂(x0, nil)
rm₂(x0, add₂(x1, x2))
ifrm₃(true, x0, add₂(x1, x2))
ifrm₃(false, x0, add₂(x1, x2))
purge₁(nil)
purge₁(add₂(x0, x1))

We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

PURGE₁(add₂(N, X)) -> PURGE₁(rm₂(N, X))

Used argument filtering: PURGE₁(x1) = x1
add₂(x1, x2) = add₁(x2)
rm₂(x1, x2) = x2
nil = nil
ifrm₃(x1, x2, x3) = x3
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ PisEmptyProof

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

eq₂(0, 0) -> true
eq₂(0, s₁(X)) -> false
eq₂(s₁(X), 0) -> false
eq₂(s₁(X), s₁(Y)) -> eq₂(X, Y)
rm₂(N, nil) -> nil
rm₂(N, add₂(M, X)) -> ifrm₃(eq₂(N, M), N, add₂(M, X))
ifrm₃(true, N, add₂(M, X)) -> rm₂(N, X)
ifrm₃(false, N, add₂(M, X)) -> add₂(M, rm₂(N, X))
purge₁(nil) -> nil
purge₁(add₂(N, X)) -> add₂(N, purge₁(rm₂(N, X)))

The set Q consists of the following terms:

eq₂(0, 0)
eq₂(0, s₁(x0))
eq₂(s₁(x0), 0)
eq₂(s₁(x0), s₁(x1))
rm₂(x0, nil)
rm₂(x0, add₂(x1, x2))
ifrm₃(true, x0, add₂(x1, x2))
ifrm₃(false, x0, add₂(x1, x2))
purge₁(nil)
purge₁(add₂(x0, x1))

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