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.