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
  ↳ 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)))

Q is empty.

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)))

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:

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)))

Q is empty.
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
  ↳ 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)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


EQ2(s1(X), s1(Y)) -> EQ2(X, Y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(EQ2(x1, x2)) = x1   
POL(s1(x1)) = 1 + x1   

The following usable rules [14] 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:

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.
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:

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)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly 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 be oriented weakly.

RM2(N, add2(M, X)) -> IFRM3(eq2(N, M), N, add2(M, X))
Used ordering: Polynomial interpretation [21]:

POL(0) = 0   
POL(IFRM3(x1, x2, x3)) = x3   
POL(RM2(x1, x2)) = x2   
POL(add2(x1, x2)) = 1 + x2   
POL(eq2(x1, x2)) = 0   
POL(false) = 0   
POL(s1(x1)) = 0   
POL(true) = 0   

The following usable rules [14] 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:

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)))

Q is empty.
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
  ↳ 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)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


PURGE1(add2(N, X)) -> PURGE1(rm2(N, X))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(0) = 0   
POL(PURGE1(x1)) = 1 + x1   
POL(add2(x1, x2)) = 1 + x2   
POL(eq2(x1, x2)) = 0   
POL(false) = 0   
POL(ifrm3(x1, x2, x3)) = x3   
POL(nil) = 0   
POL(rm2(x1, x2)) = x2   
POL(s1(x1)) = 0   
POL(true) = 0   

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))
ifrm3(false, N, add2(M, X)) -> add2(M, rm2(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:

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