Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar18/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))

Using Dependency Pairs [1,13] we result in the following initial DP problem:
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 [13,14,18] contains 3 SCCs with 2 less nodes.

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPOrderProof

              ↳ 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.
We use the reduction pair processor [13].

The following pairs can be strictly oriented and are deleted.

EQ₂(s₁(X), s₁(Y)) -> EQ₂(X, Y)

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
EQ₂(x1, x2) = x2
s₁(x1) = s₁(x1)

Lexicographic Path Order [19].
Precedence:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Non-Overlap Check

    ↳ 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:

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

                ↳ QDPOrderProof

              ↳ 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.
We use the reduction pair processor [13].

The following pairs can be strictly oriented and are deleted.

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

The remaining pairs can at least by weakly be oriented.

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

Used ordering: Combined order from the following AFS and order.
IFRM₃(x1, x2, x3) = x3
true = true
add₂(x1, x2) = add₂(x1, x2)
RM₂(x1, x2) = x2
false = false
eq₂(x1, x2) = x2
0 = 0
s₁(x1) = s₁(x1)

Lexicographic Path Order [19].
Precedence:

0 > true
0 > false
s₁ > false

The following usable rules [14] were oriented:

eq₂(0, 0) -> true
eq₂(0, s₁(X)) -> false
eq₂(s₁(X), 0) -> false
eq₂(s₁(X), s₁(Y)) -> eq₂(X, Y)

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ 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 [13,14,18] contains 0 SCCs with 1 less node.

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

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.
We use the reduction pair processor [13].

The following pairs can be strictly oriented and are deleted.

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

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
PURGE₁(x1) = PURGE₁(x1)
add₂(x1, x2) = add₂(x1, x2)
rm₂(x1, x2) = x2
nil = nil
ifrm₃(x1, x2, x3) = x3
true = true
eq₂(x1, x2) = eq₁(x1)
0 = 0
s₁(x1) = s₁(x1)
false = false

Lexicographic Path Order [19].
Precedence:

add₂ > PURGE₁ > nil
add₂ > eq₁ > true > nil
add₂ > eq₁ > false > nil
0 > nil
s₁ > eq₁ > true > nil
s₁ > eq₁ > false > nil

The following usable rules [14] were oriented:

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

↳ QTRS

  ↳ Non-Overlap Check

    ↳ 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:

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.